Download CHAPTER 1 - THE MOLE SECTION 1

CHAPTER 1 - THE MOLE
SECTION 1 - RELATIVE WEIGHTS
How do you weigh an atom? In modern jargon, I would suppose that the
answer would be "very carefully". An alternate, equally nonsensical answer
would be that you could get a very tiny person with very tiny scales
(balance, not fish) and very tiny forceps to do it for you.
How do you
weigh yourself? The answer will probably be with a scales. What good would
it do to know that you weighed 135 pounds, for example, if there were no one
around for comparison? How would you know if you were at a good weight or
not? There would be no insurance charts available. The weights of atoms
and molecules also need to be compared to some standard.
In fact, the
weights found on your periodic chart are comparison or relative weights.
If you have studied any physics or physical science, you will notice
that the correct term should be atomic mass rather than atomic weight. Much
of the older literature has also used atomic weight rather than mass.
Although several incorrect usages do not make a correct usage, the terms
will be used interchangeably here.
Objectives of this section:
1.
Given the weights (masses) of the elemental components
decomposition reaction, find the relative (weights) of
elements.
2.
Explain what is meant by relative weight (mass).
3.
Given a periodic chart, find the atomic weight (mass) of any given
element.
4.
Given a periodic chart, find the molecular (or formula) weight of
any given substance.
5.
Describe when the term formula weight is more appropriate
molecular weight.
CHAPTER 1
SECTION 1
in a
the
than
Pb, as was mentioned earlier in class, is a symbol for the element
lead. P4 is a formula for a molecule of the element phosphorus which, in
this case, consists of four atoms of phosphorus bonded together. H2O is a
formula for a molecule of the compound water.
What does the formula H2O
tell us? When you read H2O what are you really saying to yourself?
H2O indicates one molecule of water consisting of two atoms of hydrogen
bonded to one atom of oxygen.
You may want to read this as two parts
hydrogen combined chemically with one part oxygen but to do this you must
define "parts". It is not true that two grams of hydrogen combine with one
gram of oxygen to give three grams of water. The subscripts do not indicate
masses.
At this point we know nothing of the masses of these atoms.
A
molecule of water is much too small to see even with the most powerful
microscope today.
Even if we could see it, it would be so small that we
could not measure it directly with any of our instruments.
It is possible to find atomic and molecular masses by indirect means.
Look at the electrolysis of water apparatus.
Water is added to the
reservoir. The stopcocks to the collecting tubes must remain open so that
they will be filled with water completely.
Pure water does not conduct
electricity so a dilute solution of copper (II) sulfate or sulfuric acid
must be added.
The stopcocks should then be closed and the apparatus is
then connected to a battery. The current causes the water to split apart
into hydrogen and
oxygen. The positive hydrogen ions go to the negative electrode where they
combine to form hydrogen gas. The negative oxygen ions move to the positive
electrode and form oxygen gas. If the apparatus has been for a time, you
should observe that the water no longer fills the collecting tubes
completely. You will look at this apparatus again in another section and be
able to experiment then. For now, however, notice that one side has less
water than the other side. This side has hydrogen gas and the other side
has oxygen gas.
A lab worker carefully weighed two empty plastic bags and then
collected the gases in the tubes in the bags.
To avoid loss, he quickly
weighed the bags of gas. Here is the data he collected.
Gas
Hydrogen
Oxygen
Volume
200 mL
100 mL
Mass
0.0179 g
0.1429 g
Divide 0.1429 g of oxygen gas by 0.0179 g of hydrogen gas
get the relative mass ratio of oxygen to hydrogen.
0.1429 g oxygen gas
=
_____________
to
0.0179 g Hydrogen gas
What did you get? (Don't be lazy! Go back and do the problem.) Did you
put any units on your answer? For units you should have had only oxygen gas
over hydrogen gas since grams divides out in this case.
In 1808 John Dalton published one of the first tables of atomic masses
and actually listed the mass ratio of hydrogen to oxygen as 1 to 8 based
upon his observations. He thought that the formula for water was HO with
one atom of hydrogen to one atom of oxygen. In the above case the weight
ratio of the produced gases was 1 to 8 but our sample of hydrogen gas was
twice as large as our sample of oxygen gas. The formula for water is H 2O so
the mass ratio of one hydrogen atom to one oxygen atom is 0.5 to 8 (divide
the 1 by 2 since there are two hydrogens) which is also 1 to 16.
The original goal was to find the atomic masses of hydrogen and oxygen
but all that we really found was mass ratios of hydrogen and oxygen from a
sample of water.
Look at the periodic chart.
Find the atomic mass of
hydrogen. Use the key on the chart if you are not sure which number is the
atomic mass. Look up the atomic mass of oxygen also.
Atomic mass of hydrogen (to the nearest whole number) =______
Atomic mass of oxygen
=______
These are relative masses of the elements compared to some common
standard.
Each of the values on the chart has been determined
experimentally in much the same was as our example above. You can now say
one atom of hydrogen weighs 1, one atom of carbon weighs 12, one atom of
nitrogen weighs 14, and one atom of oxygen weighs 16. These are the values
from the chart. The next question to ask is 1, 12, 14, or 16 what? The
answer is that these are just numbers comparing the respective masses to a
set standard. Just as a person weighing 150 lbs is 1.5 times heavier than a
person weighing 100 lbs, oxygen (16) is 1.33 times heavier than carbon (12).
These atomic masses have no real units such as ounces or grams. Arbitrary
units of amu (atomic mass units) or au (atomic units) are assigned to these
numbers when the need for units arises.
Since it is possible to find relative atomic weights, it should also be
possible to compare masses of molecules relative to the same common
standard.
For water, there are two hydrogens with each weighing 1 for a
total of 2 and one oxygen weighing 16 for a total of 16.
Hydrogen
Oxygen
Water
2 atoms x 1.01 = 2.02
1 atom x 16.00 = 16.00
1 molecule H2O = 18.02
For most of our problems we will need to keep the number of decimal
places that the proper use of significant figures requires.
For now,
express your answers to the nearest hundredth.
Many textbooks will discuss the term molecular mass and the term
formula mass.
Although you learned to write the formula of table salt,
sodium chloride, as NaCl, it does not really occur as individual NaCl
molecules but in a crystalline form as shown.
Sodium atoms (ions actually) are connected to several chloride ions in a
network called a crystal lattice.
There is usually one chlorine for one
sodium so the formula is written as NaCl.
It would be correct to say
formula mass of NaCl rather than molecular mass.
The masses of those
compounds that occur in such crystalline structures should be called formula
masses.
In fact, the term "formula mass" can be used for any compound
including those that occur in individual molecular units.
Here are two more examples of finding formula masses.
The two are
calcium hydroxide, Ca(OH)2, and copper(II) sulfate pentahydrate, CuSO4 5H2O.
Each introduces something new.
Formula mass of Ca(OH)2:
Method 1
Ca
O
H
Method 2
1 atom x 40.08 = 40.08
2 atoms x 16.00 = 32.00
2 atoms x 1.01 = 2.02
74.10
Ca
OH
(OH)2
Formula mass of CuSO4
Method 1
Cu 1
S 1
O 4
H 10
O 5
1 atom x 40.08 = 40.08
1 atoms O x 16.00
1 atom H x 1.01
17.01
2 OH's x 17.01 =
34.02
74.10
5 H2O:
atom x 63.54 =
atom x 32.06 =
atoms x 16.00 =
atoms x 1.01 =
atoms x 16.00 =
63.54
32.06
64.00
10.10
80.00
249.70
Method 2
Cu 1 atom x 63.54 = 63.54
S 1 atom x 32.06 = 32.06
O 4 atoms x 16.00 = 64.00
159.60
Now try these exercises:
Find the molecular of formula masses:
H 2 atoms x 1.01 = 2.02
O 1 atom x 16.00 = 16.00
5 H2O molecules x 18.02 =
90.10
Total =
249.70
1.
2.
3.
4.
5.
6.
7.
8.
9.
HNO3, nitric acid
NaOH, sodium hydroxide
C12H22O11, sucrose or table sugar
C2H5OH, ethanol or ethyl alcohol
CCl4, carbon tetrachloride
Ca3(PO4)2, calcium phosphate
Na2CO3 10 H2O, sodium carbonate 10 hydrate
Cl2, chlorine gas
When is the term formula mass more correct than the term
molecular
mass?
10. What are the correct units for the mass of one atom of gold
(taken
from the periodic table)?
11. What does it mean to say that the atomic mass of lithium is
6.94 and
the atomic mass of copper is 63.54?
*12. In a chemical reaction 50 pounds of sulfur combined with 50
pounds of
oxygen to form a toxic pungent gas. The formula for the gas must be
either SO2 or SO3. Refer back to the
electrolysis experiment
and periodic chart to discover which
must be the correct compound
formed.
CHAPTER 1 - THE MOLE
SECTION 2 - WHAT IS A MOLE?
In the previous section we discovered that the atomic masses
on the periodic chart are really relative atomic masses or masses
compared to some standard. Nitrogen, for example is approximately
14 times heavier than hydrogen. In this section, we will see how
this ratio is very useful for working with very large numbers of
molecules.
Objectives:
1.
Describe why 23 pounds of sodium contain the same number of atoms
as 12 pounds of carbon, 31 pounds of phosphorus, 27 pounds of
aluminum, etc.
2.
Repeat the numerical value of Avogadro's Number.
3.
4.
5.
6.
7.
8.
9.
10.
Define the term "mole".
Given the number of particles of a substance, calculate the number
of moles of particles.
Given the number of moles of particles of a substance, calculate
the number of particles.
Differentiate between the units used for a particle of a substance
and a mole of particles of that substance.
Given the number of moles of particles of a substance, calculate
the mass of the particles.
Given the mass of a substance present, calculate the moles of
particles of that substance.
Given the mass and number of moles of a substance,
calculate the
molecular or formula mass.
Given the mass present of a compound, calculate the number of
atoms of an element within that compound.
In the electrolysis of water example and sulfur dioxide exercise we
were able to deduce the mass ratios of the atoms. It did not matter that
the first masses obtained were in grams and the second were in pounds. The
relative atomic and molecular masses are independent of the units used as
long as the units of the two substances being compared are the same.
The relative masses of some elements are hydrogen, 1; carbon, 12;
nitrogen, 14; and oxygen, 16. The masses of 10 atoms would be 10, 120, 149,
and 160 respectively. The mass of each atom was multiplied by 10. Divide
each mass by the greatest common divisor for the set to find that the
relative masses are still 1:12:14:16. This time we shall take 1000 atoms of
each.
The
masses
would
be:
hydrogen,_______;
carbon,______;
nitrogen,______; and oxygen,______. What is the greatest common divisor for
this set of numbers?___________
by the g.c.d.?
What are the relative weights when divided
We will look at one more example. This time we will take ______ atom.
The masses will be: carbon, 9000; fluorine, 14,250; and sodium 17,250. The
relative masses are 12 for C, 19 for F, and 23 for Na. If you have trouble
with this one, be sure to ask for help.
In the above example with carbon, fluorine, and sodium, the number of
atoms came out the same since each of the masses given was a multiple of
750. One gram of hydrogen should therefore contain the same number of atoms
as nineteen grams of fluorine atoms or twenty-three grams of sodium atoms.
Twelve ton of carbon should contain the same number of atoms as 197 tons of
gold.
By now you may be saying, "Fine, but how many atoms are in twenty-three
grams of sodium?". It was difficult enough to get 750 with simple math. It
has been experimentally determined that there are 6.02 x 1023 atoms in 12
grams of carbon-12
(a particular isotope of carbon).
There are three
questions that may come to mind concerning this. The answer to one is that
the methods used to find this number, except for an indirect approximate
method, are rather too complicated to be discussed her. A second question
might be that 6.02 x 1023 contains quite a bit of inaccuracy. You are right.
Chemists have actually been able to calculate the number more precisely, but
our calculations will not demand this.
A third question could be, "How large is 6.02 x 1023 ?".
602,000,000,000,000,000,000,000 certainly takes up more paper space than
does one million, 1,000,000 (1 x 106).
If we were to add up all of the
people who are now alive and have ever lived, we could not even come close
to that number. Only about ten times that number of sand grains exist on
earth!
Indiana is one of the world's largest popcorn producers. If you
were to pop 6.02 x
1023 kernels of popcorn, you could make a popcorn ball about one-tenth the
size of the moon.
The assumption is made, of course, that there are no
"old-maids" among the bunch.
How many carbons atoms in a two-ton load of coal?
Remember that 12
grams of carbon (about the mass of four pennies) contain 6.02 x 1023 atoms of
carbon. A two-ton load of soft Indiana coal is about 75% pure carbon so 2
ton x 0.75 = 1.50 ton of carbon from a two-ton load. Change the 1.50 ton of
carbon to grams of carbon:
1.50 tn x 2000 lb x
1 tn
1 kg
x 1000 g = 1.36 x 106g of carbon
2.205 lb
1 kg
Now we know that 12 grams of carbon contains 6.02 x 1023
Either 12 grams of carbon
6.02 x 1023 carbon atoms
carbon atoms.
or
6.02 x 1023 carbon atoms
12 grams of carbon
can be used in dimensional analysis problems just as the weight conversions
in the problem above. So far we have calculated that two ton of soft coal
contains about 1.36 x 106g of carbon. Since every 12 grams contains 6.02 x
1023 carbon atoms, dimensional analysis allows us to set up the problem as
follows:
1.36 x 106g of carbon x 6.02 x 1023 carbon atoms = 6.83 x1023
12 g of carbon
carbon atoms
That is a very large number of carbon atoms.
As you do more
calculations with extremely large numbers, keep in mind the comparisons on
this page. The size of a single atom must indeed be very small.
How many atoms are in 23 grams of sodium?_________________
Does this seem like a mystery question? After all that discussion on large
numbers, you may have forgotten that 23 grams of sodium should contain the
same number of atoms as 12 grams of carbon.
(You may wish to review the
first five paragraphs of this section.)
Did you say 6.02 x 1023 sodium
atoms? Good. If you are not sure, be sure to ask for help before you go
on.
Here is a new question for you. How many molecules are in 342 grams of
sucrose, C12H22O11?________________________
Up to this point the discussion has been about atoms rather than molecules.
In the previous section we were able to compare the weight of one atom to
the weight of one molecule. For example, the weight of one atom of oxygen
is 16 amu and the weight of one molecule of sucrose is 342 amu. 6.02 x 10 23
is the number of oxygen atoms in 16 grams of oxygen so 6.02 x 1023 should be
the number of molecules in 342 grams of sucrose. 6.02 x 1023 molecules of
water should weigh __________ grams.
The number 6.02 x 1023
is called Avogadro's Number after Amedeo
Avogadro, an Italian scientist who formulated some very important ideas
about matter in the 19th century. By now you should have read 6.02 x 10 23
enough times to know it and probably think that there should be some shorter
notation for it.
The term mole is used for this purpose.
One mole is
defined as 6.02 x 1023 of anything.
_______________________________
ONE MOLE = 6.02 X 1023 ANYTHING
_______________________________
One mole of carbon atoms (or oxygen atoms or hydrogen atoms) is 6.02 x
1023 carbon atoms. One mole of moles is 6.02 x 1023
moles. Can you imagine
what your garden would be like with a mole
of moles?
Just a note of interest, the word mole comes from the Latin word
"moles" which means a mass. The word molecule is a diminutive of
"moles" and means "a small mass". It is helpful to remember the
"cule" portion means small so that you do not confuse these two.
The correct abbreviation for mole is mol.
There is no
abbreviation for molecule.
Since one mole is equivalent to 6.03 x 1023, it can be used in
dimensional analysis. Always be sure to put the proper units on mole and
6.02 x 1023. Here are a few problems relating moles and molecules.
Example 1: 3.54 x 1023 molecules of C2H5OH, ethanol, is equal to how many
moles of ethanol?
3.54 x 1023 molecules C2H5OH x 1 mole C2H5OH
=
23
6.02 x 10 molecules C2H5OH
0.588 moles C2H5OH molecules
Example 2: 7.35 moles of sodium atoms is equal to how many sodium atoms?
7.35 moles Na x 6.02 x 1023 Na atoms = 4.42 x 1024 Na atoms
1 mole Na
The number of atoms or molecules is a huge number compared to the
number of moles. Many students try to put a number such as
1023 on the term mole when they are still attempting to understand the
concept. This really makes no sense if you think about the relative sizes
of these particles.
Exercises
Work the following exercises.
Be sure to use the correct
units.
1.
Calculate the number of molecules in 3.50 mole of hydrogen
(H2).
molecules
2.
Calculate the number of atoms in 3.50 moles of hydrogen atoms
3.
Calculate the number of moles of sodium atoms in 2.50 x 1023
atoms.
sodium
4.
Calculate the number of moles of methane, CH4, molecules in
1021 molecules of methane.
2.10 x
5.
Calculate the number of moles of carbon tetrachloride molecules (CCl4)
in 2.10 x 1025 molecules of carbon tetrachloride.
(H).
How much does one mole of gold atoms weigh?__________________
To answer this problem you must recall that one mole of gold is equivalent
to 6.02 x 1023 gold atoms. From the periodic chart it can be seen that the
atomic mass of gold is 196.97 amu. The mass of one mole of gold atoms is
therefore 196.97 grams. The relative mass of a carbon atoms is 12 amu. The
relative mass of an oxygen atom is 16 amu. Twelve grams of carbon contain
6.02 x 1023 atoms which is one mole. Sixteen grams of oxygen atoms contain
6.02 x 1023 atoms which is one mole.
12 g of C == 1 mole of C== 6.02 x 1023 atoms of C
16 g of O == 1 mole of O== 6.02 x 1023 atoms of O
196.97 g of Au == 1 mole of Au == 6.02 x 1023 atoms of Au
but
12 amu of C == 1 atom of C
16 amu of O == 1 atom of O
What is the mass of one mole of sodium hydroxide,NaOH?______
Here you would just find the molecular mass (actually formula mass) which
has either no units or units of amu's. The numerical value of the molecular
mass is the mass of one mole with grams as the unit.
Exercises
Work the following to check your understanding.
answers to the nearest hundredth.
1.
2.
3.
4.
5.
Express your
The mass of one mole of NaCl is ________________.
The molecular mass of water is _________________.
The atomic mass of iodine is ___________________.
The mass of one mole of fluorine atoms is _____________.
A chemist said that the molecular weight of sulfur dioxide was 64.06
grams. Is that correct? Explain your answer.
The idea of the mole is one of the most used ideas in chemistry
calculations. Dimensional analysis will again be very handy for this work.
You will need to practice a few problems using grams and moles.
Example 1: A new lab assistant who had never studied high school chemistry
was told to weigh out two and a half moles of salt for an experiment. After
an initial few minutes of panic, he grabbed a shovel, a large knife, and
asked another worker for a map of the nearby salt mines. After recovering
from a long bout of hysterical laughter the worker informed him that a mole
of salt is a number and not an animal that burrows in salt mines.
The
embarrassed new lab assistant then looked for a scales to weigh out 2.5
moles, but all he found were gram scales. He debated for a while whether he
should ask the other workers for help again or apply for unemployment
benefits.
Desperately he ran out into the hallway and grabbed the first
person he saw - you.
Since his name was mugger, everyone ignored your
screams. Your only choice was to help him solve the problem. Thankfully,
the lab did have a periodic chart.
Solution:
2.500 mole NaCl = ______________g NaCl
The molecular weight of NaCl = 22.99 + 35.45 = 58.44 amu.
The weight of one mole of NaCl = 58.44g
2.500 mol NaCl x 58.44 g NaCl = 146.10 g NaCl
1 mole NaCl
Example 2: The next day the lab assistant weighed out 45.53 grams of zinc
and wondered how many moles he had. From the periodic table he discovered
that zinc had an atomic mass of 65.37 grams.
Right away (thanks to your
good instruction from the previous day) he knew that he had less than one
mole but did not know how to proceed from there. Help him out.
Solution:
45.53 g Zn x 1 mole Zn
65.37 g Zn
= 0.6965 moles Zn
_______________
Exercises
Work the following exercises before proceeding to the next
section.
1.
Calculate the number of moles of oxygen atoms in 48.0 g of
atoms (O).
oxygen
2.
Calculate the number of moles of sulfuric acid in 0.200 grams
of sulfuric acid, H2SO4.
3.
Calculate the grams of nitrogen in 2.50 moles of nitrogen molecules,
N2.
4.
Calculate the pounds of potassium chloride in 10.0 mole of
potassium chloride, KCl.
*5.
Given that a sample of a compound contains 2.45 moles and
74.5 grams, calculate the molecular mass of the compound.
weighs
Moles, molecules, and grams can all be interrelated with moles serving as
the key.
In example 1, our desperate lab assistant could have been told to weigh
out 1.50 x 1024 molecules of salt. Since NaCl comes in crystals rather than
individual molecules, he would have had a very difficult time in deciding
exactly what a molecule was. Of course, he could have tried counting little
salt crystals. After the first thousand, he probably would have considered
jumping out the window. It is possible to weigh out 1.50 x 1024 NaCl units,
however, if this value is converted to grams.
1.50 x 1024 NaCl molecules x1 mole NaCl molecules
x 58.44 g
23
6.02 x 10 NaCl molecules 1 mole NaCl
molecules
= 146 g NaCl
__________
Exercises
1. Calculate the number of grams of methane, CH4, in 1.20 x 1021
molecules of methane.
2.
Calculate the number of molecules in 12.0 g of hydrogen gas,
3.
Calculate the number of milligrams of potassium chloride, KCl,
molecules in 1.50 x 1023 potassium chloride molecules.
4.
Calculate the number of molecules of sulfuric acid, H2SO4, in
60.0 pounds of sulfuric acid.
H2.
*5. What is the mass in grams of one molecule of H20?
How many oxygen atoms are in 75.0 grams of sucrose,
C12H22O11?
Everything needed to solve this problem has already been covered in this
unit. Give it a try on your own before you look at the answer.
One way to attack an unfamiliar problem is to list first what is given
and then what is to be found.
In this problem begin with 75.0 grams of
sucrose.
The is no direct relationship between grams and the number of
atoms in the sucrose molecule. List the relationships that you do know.
Known (or calculatable) relations:
75.0 grams of C12H22O11 =___________oxygen atoms
342 g C12H22O11 = 1 mole C12H22O11
1 mole C12H22O11 = 6.02 x 1023 molecules C12H22O11
1 molecule C12H22O11 contains 11 oxygen atoms
This last relationship is the step that you may have overlooked but is very
important. A very common mistake on this problem is to attempt to proceed
directly from mole to atom. The moles you calculate are moles of molecules
and should be labelled as such to prevent confusion. The steps in solving
the problem are shown below.
75.0 g C12H22O11x 1 mole C12H22O11 x 6.02 x 1023 molecules C12H22O11
342 g C12H22O11
1 mole C12H22O11
11 oxygen atoms
1 molecule C12H22O11
= 1.45 x 1023 oxygen atoms
________________________
Exercises
1. Give the number of hydrogen atoms in one molecule of cortisone,
C21H28O5.
2.
Calculate the number of oxygen atoms in 50.0 grams of carbon
dioxide, CO2.
*3. Given that there are 2.03 x 1024 oxygen atoms present in a
sample of nitric acid, HNO3, calculate the number of grams of
nitric acid present.
4.
Calculate the number of hydrogen atoms present in 2.50 moles
of sulfuric acid, H2SO4.
*5. Calculate the number of millimoles present in a sample of
sulfuric acid which contains 7.35 x 1022 atoms of hydrogen.
This chapter began with the question, "what does the formula H2O tell
us?"
At that time the answer was that H2O represented one
molecule of
water that contained two hydrogen atoms and one oxygen atom chemically
combined.
2 H2O would then represent two molecules of water with each
containing two hydrogen atoms and one oxygen atom for a total of four
hydrogen atoms and two oxygen atoms. 1/2 H2O would represent what?
Is it a molecule of water that has been cut in half? How do we slice
it? That is really a pretty silly question since a molecule is already the
smallest unit of compound water. Is 1/2 H2 O illogical or is it possible?
The formula H2O can represent one molecule of water but it can also
represent one mole of water.
Much of the work in chemistry uses moles
because they can be weighed. 2 H2O would then represent two moles of water
or two molecules of water depending upon the context of the problem. 1/2
H2O could not represent half of a molecule but it could easily represent
half a mole or 3.01 x 1023
molecules of water. How many grams of water
would it represent?_________ How many atoms of hydrogen and how many atoms
of oxygen?_____________________
If H2O represents one mole of water, what do the subscripts represent?
For a molecule they may represent atoms but you cannot say that one mole
(6.02 x 1023 molecules) of water contains two atoms of hydrogen and one of
oxygen. The size of those three atoms would have to be tremendous and would
cause all kinds of plumbing problems.
One mole of H2O molecules does
contain two moles of hydrogen atoms and one mole of oxygen atoms.
Exercises
1.
2.
3.
4.
5.
6.
One
One
One
One
One
One
molecule of H2SO4 contains____________ atoms.
mole of H2SO4 contains _________ molecules of H2SO4.
mole of H2SO4 contains _________ atoms.
mole of H2SO4 contains _________ H atoms.
mole of H2SO4 contains _________ S atoms.
mole of H2SO4 contains _________ moles of H atoms.
Section Exercises
Weigh out 342.30 g of sucrose, C12H22O11, and 58.44 g of sodium chloride,
NaCl. Use the specifically marked containers in the lab. Do not mix the
two and, unlike most lab activities, put the substances back when you are
finished with the following questions.
1.
Which of these substances contains the most formula units?
__________________________________________________________
2. How many formula units are in the sucrose?________________
3. How many moles are in the sucrose sample?_________________
4. How many oxygen atoms are in the sucrose sample?__________
5. How many moles of oxygen atoms are in the sucrose?________
*6. What is the percentage by weight of oxygen in sucrose?_______
7. How many formula units are in the weighed salt?___________
8. How many moles are in the salt sample?____________________
9. How many moles of chloride ions are in the weighed salt?_______
*10. What is the percentage by weight of the metallic ion in the
sample?____________
salt
CHAPTER 1 - THE MOLE
SECTION 3 - PERCENT COMPOSITION AND EMPIRICAL FORMULAS
Now that it is possible to interconvert grams, molecules, and atoms
through the use of moles, of what use is this skill?
This next section
shows how chemical formulas and moles can be useful in real lab problems.
All of the basic ideas have already been covered. This section will help
you put them together.
Objectives:
1. State the law of definite composition.
2. Given a formula of a compound, calculate the percent
composition by
weight of each elemental species.
3. Differentiate empirical (simplest) formula from molecular
formula.
4. Given the percent composition by weight of the elemental
species, determine the empirical formula.
5. Given the gram composition of the elemental species, determine
the molecular formula.
6. Given either the percent composition or gram composition and
the molecular weight of a compound, determine the molecular
formula.
This laboratory experiment should proceed Chapter 1, Section
An entire class period may be needed to perform this
Partial Weight Analysis of a Compound
3.
experiment.
The blue copper (II) sulfate found in the lab is an example of hydrate.
Hydrates have definite amounts of water incorporated chemically within the
crystalline structure. The five moles of water molecules per mole of copper
sulfate can be removed by heating. This will enable you to determine the
percentage of water of hydration present in a sample.
1.
2.
3.
4.
Weigh a clean, dry crucible to the nearest O.O1g.
Grind up crystalline hydrated copper (II) sulfate
Weigh out between 2.50 and 3.50 grams of the salt
crucible. Be sure to clean up any spilled copper
the balance pan before you record the weight.
Weight of sample and crucible
Weight of sample, calculated
Place the crucible and its contents on a pipestem
set on a ring clamped to a ring stand.
________g
in a mortar.
in the
sulfate from
_________g
_________g
triangle
Heat the system gently for 3 to 5 minutes to avoid splattering.
Continue to heat but more strongly for 10 more minutes. Record
any observations during this time.
Observations:
Obtain a desiccator from the lab or
make one from a large jar. The
bottom of the jar should have 1.2
inch solid anhydrous calcium
chloride.
A
commercial laboratory
desiccator may have the
lid coated
with petroleum jelly so do not set
the lid down flat on the table. The
top should be placed upon the jar as
soon as it is filled and not removed
until necessary. The crucible will
rest on a wire screen or plate placed
above the calcium chloride.
At the end of the 10-minute heating, allow the crucible to
the stand for 2 minutes and then use tongs to transfer it
desiccator.
5.
cool
to
After the crucible and contents have cooled to room temperature
on
the
in
the closed desiccator, weigh the crucible and solid. Weight
of crucible and residue.
______________g
6. Calculate the weight of the residue
______________g
7. Calculate the weight of the water loss by the original sample
______________g
8. Calculate the percentage of water of hydration in the sample
______________g
9. Place a few drops of water on the cooled, weighed residue.
Feel the
bottom of the crucible before and after doing this.
Record any
observations.
Observations:
Questions:
1.
If the hydrated copper (II) sulfate were heated strongly for a
longer
period of time, it would lose its water of hydration
and also begin to
decompose into black copper (II) oxide, CuO, and
colorless,
noxious,
pungent sulfur trioxide. Would this
cause your results to appear higher
than expected or lower
than expected? Explain your answer.
2.
The lid of a commercial desiccator is not fastened to the top.
it has been on a prepared desiccator for a few minutes, it
is
difficult to lift the lid. Why might this happen?
3.
Use a dictionary to find the meanings of hydrate, desiccate, and
anhydrous.
4.
What color is the hydrated copper (II) sulfate?
What color is the anhydrous copper (II) sulfate?
After
often
5.
Hydrated sodium carbonate (washing soda) is often found in
powdered
laundry detergents. An opened, unused box of the detergent may weigh less
than the shipping weight on the box.
Account for this discrepancy.
"A rose is a rose is a rose. . ." is a famous quote from Gertrude
Stein. It would also be correct to say H2O is H2O is
H2O no matter where it is found.
Water in the southwestern part of the
United States may have a bitter alkaline taste and sea water may have a
salty taste, but if all of the impurities were removed, there would be no
difference.
Salt found in a mine in Michigan has exactly the same
composition as salt found in a mine in Poland.
Commercially produced
Vitamin C (ascorbic acid) has the same composition as Vitamin C found in
orange juice.
The observation that a pure compound always has the same
elemental composition has long been known. It has been called the law of
constant composition or law of definite proportions.
When you learned to
write formulas, you had to assume that all table salt combined to from NaCl
rather than NaxCly with x and y representing numerical subscripts.
In the experiment with copper (II) sulfate five hydrate, you found the
percent by weight of water in the compound. The weight loss was divided by
the total weight of the compound.
It is possible to find the percent
composition by weight from the formula above.
To find the percentage of
hydrogen in water, take the weight of hydrogen (2 x 1.01) and divide by the
total molecular weight of water ((2 x 1.01 ) + (16.00)).
2.02 x 100 = 11.2% of hydrogen in water by weight
18.02
What does 11.2% hydrogen in water indicate? If 100 grams of water were
present, 11.2 grams of compound would be from hydrogen and 88.8 grams of the
total weight would be from oxygen.
Exercises
1.
Calculate the percent composition of carbon in ethyl chloride,
C2H5Cl.
2.
Calculate the percent composition of sulfur in aluminum sulfite,
Al2(SO3)3.
3.
Calculate the percent composition of each elemental species in acetic
acid, CH3COOH.
*4. Calculate the percent of metal ion in a compound where 0.350 g of metal
combines with 0.255 g of oxygen.
*5. From the formula for copper(II) sulfate 5 hydrate, CuSO4 5
H2O, calculate the percent water found in the hydrated compound.
An entire class period may be needed to perform this experiment.
Experiment:
The formula of a Compound
Each compound has its own distinct
formula, as was just seen in the
previous section, has its own
constant
composition. In this
experiment you are going
to make a
compound and then calculate its
simplest formula.
1. Weigh a clean, dry crucible and
record its weight to 0.01g ________
2. Obtain a clean piece of magnesium
ribbon about 35 cm in length. Use
a fine piece of steel wool to clean
the ribbon. Cut the ribbon into
pieces approximately 1 cm and add to the crucible. Weigh the crucible and
contents.
_______
3.
Weight of magnesium ribbon by difference.
4.
Calculate the moles of magnesium (M.W of Mg = 24.31)_______
5.
_______
Set up the ring, triangle, and ring
stand as in the previous experiment.
Heat the covered crucible (picture 1)
gently at first and then gradually
increase the heat intensity for 2
minutes.
With your tongs, carefully tilt the
lid (picture 2) and continue strong
heating of the partially uncovered
crucible for an additional 10 minutes.
At the end of this time, allow the
covered crucible and contents to cool.
Check to see that all of the magnesium
has reacted before covering. If there
is still unreacted magnesium, heat
strongly again.
Since the magnesium reacted with both the oxygen and nitrogen
in
the air, it is necessary to convert the mixture of compounds to only the
oxide compound.
Use a medicine dropper to add only enough distilled water to cover the
contents of the crucible. Carefully heat the
system. Observe the odor
of any vapor that has been produced.
When the water has evaporated, add
a little more and carefully
heat again. Continue to do this until there
is no longer a distinct odor produced upon heating.
When this has happened, continue careful heating until all of
the
water has evaporated. Strongly heat the uncovered
crucible
for
five
more minutes. Allow the crucible and its
product to cool.
Weigh the
crucible and its contents.
Weight of crucible and contents________
6.
Determine the weight of the magnesium oxide by difference_____
7.
Determine the weight of oxygen that combined with the original
weight of magnesium
(You know how much Mg is present in the compound.) ________
8.
Calculate the number of moles of combined oxygen
9.
Determine the simplest formula for magnesium oxide.
sure what to do here, read ahead in section 3.)
________
(If you are
not
When samples of unknowns are sent in for analyses, the results often
come back expressed as percentages.
An example might be that an unknown
nitrogen - oxygen compound was found to contain 63.6% N and 36.4% O weight.
From this information it is possible to calculate a simplest or empirical
formula. Remember that 63.6% N means that out of every hundred grams, there
are 63.6 g of nitrogen present. How does this help with the problem? Keep
in mind that we are trying to find the subscripts x and y in the formula
NxOy. It was mentioned in the last section that the subscripts can represent
not only the number or atoms of each elemental species per mole of
molecules.
If the grams of each species could be changed to moles, the
moles could serve as subscripts for the formula.
Example 1:
A compound consists of 63.6% N and 36.4% O.
the empirical formula of the compound?
Solution:
What is
63.6% N and 36.4% O by weight means
63.6 g N and 36.4 g O out of 100 g of
compound
sample.
Assume that we have 100 g of compound.
would have 63.6 g N and 36.4 g O in our
Change these gram values to moles.
63.6 g N x 1 mole N
14.0 g N
36.4 g O x 1 mole O
16.0 g O
= 4.54 mole N
= 2.28 mole O
We
This would give an empirical formula of
N4.5402.28. Having fractional subscripts would be
no problem if the subscripts represented
only moles. Since they can also represent
atoms, it is necessary that they be whole
numbers. To find the whole number ratio
represented by these fractional subscripts,
divide both of the numbers by the smaller
number to make the smaller value equal to zero.
In this case N4.54/2.28)2.28/2.28 nicely works out to
N2O1 which is N2O or the simplest formula of the
compound.
If this problem had state that 3.18 g of N and 1.82 g of O were present
in a nitrogen - oxygen compound, the moles of nitrogen and oxygen could be
calculate directly without using percents and assuming that 100 g sample was
present.
The subscripts in Example 1 worked out very nicely to a 2 to 1 ratio of
N to O. What do you do if the subscript ratios work out to something like 1
to 1.25?
The decimal portion of 0.25 is 1/4 in fractions.
If 1.25 is
multiplied by 4, a whole number will result. For a ratio such as 1 to 1.25,
when one number is multiplied by 4 the other number must also be multiplied
by 4. The ratio is now 4 to 5.
Example 2: A compound of arsenic and oxygen was found to contain
75.8% arsenic. What is the empirical formula of the
compound?
Solution:
Work out the first part of this problem yourself?
The ratio of moles of atoms comes out approximately
1 As to 1.5 O. Multiply both numbers by 2 to get
the final
answer of As2O3 for the empirical
formula.
When is the empirical or simplest formula not the actual formula of a
compound? At first two phosphorus compounds were thought to have formulas
of P2O3 and P2O5, but it was later discovered that the real formulas were P4O6
and P4O10. Many organic compounds have multiples of the simplest formula for
their real formulas. Ethane, C2H6, has an empirical formula of CH3. There
is no compound with the actual formula of CH3.
How can we tell what the
real formula for a compound is?
If the molecular weight as well as the
percent composition by weight of a compound is known, the actual formula can
be found.
Example 3: A hydrocarbon compound with a calculated molecular
weight of 78.8 was found to consist of 92.3% carbon
of
7.7% hydrogen.
Calculate the actual formula of
the compound.
Solution: To solve this, initially ignore the 78.8 and use the
percentage data as before. Calculate the empirical
formula.
The empirical formula should be C1H1. The real formula
be a multiple of C1H1 so the actual molecular weight
must
multiple of the weight of C1H1.
The weight of C1H1 is approximately 13.
get 6.06. This number is close enough to 6 by
The actual molecular formula is
C6H6. This compound is called benzene.
must
be
Divide 78.8 by
significant
*Just a word of warning. 6 C1H1 is not the same as C6H6
also represents something different. Sometimes it
is very difficult to know whether the correct solution
C6H6 or has units such as Ca(OH)2 included in it.
6 CH is never correct.
Exercises
1.
The analysis of a compound shows that the compound contains
53.33% O, and 20.00% Mg. Calculate its simplest
formula.
2.
13 to
digits.
(CH)6
is
26.67%
S,
A sample of an oxide of iron contains 27.59% O and 72.45% Fe.
Calculate its empirical formula.
3.
Upon analysis, 51.0 grams of aluminum oxide are found to
27.0 grams of aluminum. Calculate the simplest
formula of
oxide.
4.
a
contain
the aluminum
A compound has the following composition by weight:
C, 57.1%
H, 4.8%
O, 38.!%
a. If the atomic weights of carbon, hydrogen, and oxygen are
respectively 12.0, 1.01, and 16.0, calculate the simplest
formula for
this compound.
b. If the true molecular weight is shown to be 126, what must
be the true molecular formula?
Summary for Chapter 1:
Many new ideas that are extremely important have been covered in this
chapter. We saw that although we cannot weigh atoms and molecules directly,
we can obtain their weights in comparison to each other.
These relative
weights (both atomic and molecular or formula) hold between atoms and
molecules whether we are talking about one or ten million
molecules.
This enables us to compare the weights of the same
atoms in common units.
One gram of hydrogen atoms, 12 grams
atoms, and 16 grams of oxygen atoms should contain the same number
atoms or
number of
of carbon
of atoms.
It has been experimentally determined that 0.012 kg of carbon
12
23
contains Avogadro's Number (6.02 x 10 ) of carbon atoms. The mole (mol) is
6.02 x 1023 of anything. The weights on the periodic chart can represent the
weights of the individual atoms in amu's or the weights of moles of atoms in
grams.
Weights of molecules can be calculated by adding up the atomic
weights. The weights of moles of molecules can be found by the same method.
It is possible to interconvert numbers of atoms, molecules, moles, and
grams using moles as the key. The number of moles is really the number of
particles we are dealing with.
6.02 x 1023
M.W.
(atoms)<------>Molecules<------------->MOLES<---->grams
The law of definite composition has long been known.
It would be
impossible to do much chemistry if the formulas of compounds kept changing.
Finding empirical or simplest formulas depended upon having constant
composition as well as having subscripts that could represent atoms in one
case and moles of atoms in another. This was one important use of mole that
we saw in this chapter. In the next chapter we will look at how it is used
in chemical reactions.
To be sure that you really understand the material in this chapter, you
need to do some review exercises.
When you were learning to spell in
elementary school, you learned the list of words but were rather upset when
your mother gave them to you out of order.
As you know now, you do not
spell in that order in real life.
In chemistry you do not have problems
presented to you in a definite order either, so you need to recognize what
the problems are all about to do them better. Try to work as many of these
as you can without looking back through this chapter.
Chapter Review Problems
1. How much does 5.22 x 1023 atoms of nitrogen weigh?
2. How many molecules are present in 48.5 g of HBrO3?
3. Calculate the mass in grams of 1.5 moles Ca(OH)2.
4. Calculate the number of moles in 52.5 g of Zn(HCO3)2.
5. Calculate the number of molecules in 52.5 g of Zn(HCO3)2.
6.
Calculate the number of moles of atoms in 52.5 g of Zn(HCO3)2 7.
Calculate the percent by weight of hydrogen in Zn(HCO3)2.
8. Calculate the total number of atoms in 52.5 g of Zn(HCO3)2.
9. Calculate the number of zinc atoms present in 52.5 g of
Zn(HCO3)2.
10. Find the percentage of nitrogen in ammonium nitrate, NH4NO3.
11. A photography developing compound was found to contain 36.5%
sodium, 25.4% sulfur, and 38.1% oxygen. Calculate its simplest
formula.
12. Analysis of a compound gives 90.6% Pb and 9.4% O. What is the
empirical formula of this compound?
13. Calculate the mass of the metal ion in 50.0 g of MgS.
14. Calculate the moles in 1.505 x 1023 molecules of NaOH.
15. Calculate the grams in 1.505 x 1023 molecules of NaOH.
16. What is the molecular weight of CO2?
17. If the molecular mass of an oxide of nitrogen is 108, and 4.02g
of N combine with 11.48 g of O, what is the molecular formula
of this compound?
18. What is the weight of one mole of CO2?
19. How many atoms does one molecule of H2O contain?
20. How many moles of atoms does one mole of H2O molecules contain?
21. What is the mass in grams of an atom of carbon?
22. Define mole.
23. Which is larger, Avogadro's Number or the number of molecules
g of CO2?
24. Calculate the mass in grams of one mole of Na2SO4 10 H2O.
25. Blue CuSO4 5 H2O and NaSO4 10 H2O are known as hydrates. What
are they called when they no longer contain H2O molecules?
in 66
Chemistry
Chapter 1
Alternate Exercises
Show the calculation setups and answers for all problems.
Find the formula weight (or molecular weight) of:
1. Phosphoric acid, H3PO4
2. Sodium bisulfate, NaHSO4
3. Bismuth(III) nitrate, Bi(NO3)
4. A sample of mercury(II) iodide, HgI2, weighs 9.42 grams.
How many moles are in this sample?
5. What is the weight of 0.l20 mole of ammonium carbonate,
(NH4) CO3?
6. How many molecules are contained in 3.65 moles of chlorine,
Cl2?
Calculate the percent composition by weight of barium sulfate,
7. Ba
8. S
9. O
10.
BaSO4
How many moles are present in 8.42 grams of sodium chlorate,
NaClO3?
A sample of oxygen gas, O2, weighs 30.0 grams. How many molecules and how
many atoms are present in this sample?
11. Molecules?
12. Atoms?
*13. A mixture of sand and salt is found to be 42 percent NaCl by weight.
How many moles of NaCl are in 36 grams of this
mixture?
14. What is the weight of 8.5 x 1022 molecules of carbon dioxide, CO2?
15. A sample of the mineral crysoberyl contains 2.25 g Be, 13.49g
of
Al, and 4.00 g O. What is its empirical formula?
16. What is the empirical formula of a compound that is 47.7% K, 13.2%
B,
and 39.1% O?
17. If the compound in problem 16 has a molecular weight of 246, what
is
its molecular formula?
*18. Pewter is 85% tin and 15% antimony. How many moles of antimony
are
contained in 50 g of pewter?
*19
What mass of silicon is combined with 6.40 g of oxygen in the
compound SiO2?
*20. If one atom of an element has a mass of 3.330 x 10-22g, what
is
that
element's atomic weight?
Name______________________
Date______________________
Chemistry Quiz
Chapter 1 - Section 3
"Percent Composition and Empirical Formulas"
Show your set-up on any calculation problem.
Chart and a calculator.
You may use a Periodic
1.
Five samples of mercury(II) oxide (HgO) were found to contain
mercury and 7.4% oxygen. How does this illustrate the Law
of
Composition?
by
weight
of
aluminum
in
92.6%
Definite
2.
Find the percent composition
hydroxide {Al (OH)3}.
aluminum
3.
Find the percent composition by weight of oxygen in sulfuric acid
(H2SO4).
4.
Identify each of
formula, or both.
a. HO
b. P4O10
c. H2O
d. C12H22O11
e. N2O4
the
following
as
simplest
formula,
molecular
5.
A compound is found by analysis to contain 75.0% carbon and
hydrogen. What is the empirical formula?
25.0%
6.
A compound is found by analysis to contain 19.57 g iron and
oxygen. What is the simplest formula?
8.42
g
7.
The analysis of a gas reveals that it consists of 92.3% carbon
and
7.7% hydrogen. Its molecular weight is 26. What is the
molecular
formula?
Name____________________
Date____________________
Chemistry Quiz
Chapter 1 - Section 3
"Percent Composition and Empirical Formulas"
Show the set-up in each calculation problem to receive credit.
each answer. You may use a Periodic Chart and a calculator.
1.
Identify each of
formula or both.
the
following
as
simplest
formula,
Label
molecular
a.
b.
c.
d.
e.
C12H22O11
C2H6
H2O
CO
HO
2.
The molecular weight of hemoglobin is 68,000. Each molecule of
hemoglobin contains 4 atoms of iron. What is the percent
by
weight
iron in hemoglobin.
3.
78.
Benzene has an empirical formula of CH and a molecular weight
What is the molecular formula for benzene?
of
of
4.
Determine the simplest formula for a compound with 36.5 percent sodium,
0.8 percent hydrogen, and 24.6 percent
phosphorus,
and
38.1
percent
oxygen.
5.
Cyclopropane contains only two elements, carbon and hydrogen.
sample is found to contain 3.00 grams of carbon and 0.250
grams
hydrogen, what is the simplest formula?
6.
If the molecular weight of cyclopropane is found to be 39,
its molecular formula?
If
what
7.
Every sample of pure limestone is found to contain 40.0 percent
calcium, 12.0 percent carbon, and 48.0 percent oxygen. Explain
why
would not expect these values to change from sample to sample.
ANSWERS TO EXERCISES
a
of
is
one
Chapter 1 - Section 1
1.
HNO3
MW = 63.01
The units and either amu's or nothing
2.
NaOH
MW = 40.00
3.
C12H22O
MW = 342.29
4.
C2H5OH
MW = 46.06
5.
CCl4
MW = 153.82
6.
Ca3(PO4)2
MW = 310.19
7.
Na2CO3 10H2O
MW = 286.13
8.
Cl2 MW = 70.91
9.
The term formula weight is a more correct term for compounds that occur
as crystal lattices.
10. The units for relative weight of one atom of gold could be
amu's,
au's, or nothing. The mass of one atom of gold in grams will be calculated
later.
11. This means that copper is 63.54 times heavier than lithium.
6.44
12. According to the electrolysis problem, we can divide the weight of
sulfur by the weight of oxygen. 50lb sulfur = 1 sulfur
50lb oxygen
1 oxygen
The unit of pounds divides our.
This means that the relative weight of
sulfur is equal to the relative weight of oxygen. Sulfur weighs 32.06 and
oxygen weighs 16.00. In SO2, there are two oxygens which would weigh 32.00
together. In SO3, the three oxygens would weigh 48.00. SO2 is the compound
with sulfur and oxygen weighing the same so SO2 is the answer.
11
Section 2
page 12
1. 3.50 moles H2 x 6.02 x 1023 molecules = 2.11 x 1024H2 molecules
1 mole H2
2. 3.50 moles H atoms x 6.02 x 1023 H atoms = 2.11 x 1024 H atoms
1 mole H atoms
23
3. 2.50 x 10 Na atoms x 1 mole Na atoms = 0.415 mole of Na atoms
6.02 x 1023 Na atoms
4. 2.10 x 1021 molecules CH4 x
1 mole CH4___________ =
6.02 x 1023 molecules CH4
5.
2.10 x 1025
.00349 moles CH4 molecules
3.49 x 10-3 moles CH4
molecules CCl4 x 1 mole CCl4
=
6.02 x 1023 molecules CCl4
34.9 moles CCl4 molecules
Page 14
1. 1 mole NaCl has a mass of 58.44 g
2. 18.01 or 18.01 amu but not 18.01 grams. This is for moles
only.
3. AW of I = 126.90
4. 1 mole of F = 19.00g
5.
No.
The weight of a mole of sulfur dioxide is 64.06 grams but
molecular weight has units of amu or nothing. Molecular weight
is a
companion, not an actual weight.
Page 15
1.
48.0g 0 atoms x 1 mole 0 atoms = 3.00 mole 0 atoms
16.0g 0 atoms
2. 0.200g H2SO4 x 1 mole H2SO4 = 0.00204 mole H2SO4
98.08g H2SO4
3. 2.50 moles of N2 x 28.01g N2 = 70.03g N2
1 mole N2
My calculation may round off numbers differently than yours.
Do not
be concerned if your last digit and mine do not agree.
4. 10.0 mole K Cl x 74.56g KCl x 1 lb
= 1.64 lb KCl
1 mole KCl
454g
This last step is a little different.
If you do not know the
English to metric conversions, look them up in your text when
you need
them.
5. One mole of a compound weighs a certain number of grams. It
can
be expressed as ? grams which can equal ? grams.
? moles
1 mole
73.5 grams = 30.0 grams. The mass of one mole is identical
2.45 mole
1 mole
to the molecular weight in amu's.
The answer is 30.0 amu.
Page 16
1.
1.20 x 1021 CH4 molecules x 1 mole CH4___ x
6.02 x 1023 CH4
16.04g CH4 =
1 mole CH4
0.320g CH4
2.
12.0g H2 x 1 mole H2 x 6.02x1023 H2 = 3.58 x 1023 H2 molecules
2.02g H2
1 mole H2
3.
1.50x1023 KCl molecules x 1 mole KCl
x 74.55g KCl x 1000mg
23
6.02x10 KCl molecules
1 mole KCl
1g
= 1.86 x 104mg KCl
This involves a metric-to-metric conversion.
4.
60.0 lb H2SO4 x 454g x 1 mole H2SO4 x 6.02 x 1023 H2SO4 molecules=
1 lb
98.08g H2SO4
1 mole H2SO4
1.67 x 1026 H2SO4 molecules
5.
1 molecule H2O x 1 mole H2O
x 18.01g =
2.99x10-23g H2O
6.02 x 1023 molecules H2O 1 mole H2O
One molecule of H2O is not very large, is it?
Page 17
1.
One molecule of cortisone has 28 H atoms.
2.
50.0g CO2 x 1 mole CO2 x 6.02 x 1023 CO2 molecules x 2 0 atoms
44.0lg CO2
1 mole CO2
1 CO2 molecule
= 1.37 x 1024 atoms of O
3.
One molecule of HNO3 has 3 atoms of O
2.03 x 1024 O atoms x 1 HNO3 molecules
1 mole HNO3 x
3 atoms of O
6.02 x 1023 molecules HNO3
63.01g HNO3
1 mole HNO3
This was a different problem.
4.
= 70.8g HNO3
2.50 moles H2SO4 x 6.02x1023 H2SO4 molecules x 2 atoms H
1 mole H2SO4
1 molecule H2SO4
= 3.01 x 1024 atoms H
Did you get confused and try to use grams?
5.
Read each problem and think.
7.35 x 1022 H atoms x 1 H2SO4 molecules x 1 mole H2SO4
2 H atoms
6.02 x 1023 molecules
x
1000 millimole H2SO4 =
1 mole H2SO4
61.0 millimoles H2SO4
This is also a difficult problem. The problem, itself, did not state
that the millimole were to be of H SO . That was my mistake and I hope it
did not confuse you too much.
Page 19 - top of page
1.
2.
3.
4.
1
1
1
1
H2SO4
mole
mole
mole
molecule contains 7 atoms
H2SO4 contains 6.02 x 1023 molecules H2SO4
H2SO4 contains 6.02 x 1023 x 7 = 4.21 x 1024 atoms
H2SO4 contains 6.02 x 1023 x 2 = 1.20 x 1024 H atoms
5.
6.
1 mole H2SO4 contains 6.02 x 1023 x 1 = 6.02 x 1023 S atoms
1 mole H2SO4 contains
2
moles of H atoms
P 19 - Exercise
1.
MW of sucrose = 342.20
MW of sodium chloride = 58.44
342.30g of C12H22O and 58.44g of NaCl are both one mole.
would have 6.02 x 1023 molecules.
Each
11
2.
6.02 x 1023 molecules
3.
1 mole
4.
1 mole C12H22O11 x 6.02 x 1023 Molecules C12H22O11 x 11 O atoms
1 mole C12H22O11
1 mole C12H22O11
=
5.
There are 11 moles of oxygen atoms.
6.
11 O atoms x 16.00g = 176.00g
atoms
176.00g
x
100
=
342.20g
6.62 x 1023 O atoms
51.43% O
Do not worry about
You will look at this in the next
section.
7.
6.02 x 10
8.
1 mole of NaCl
9.
1 mole of chloride ions
10.
22.99g x 100 = 39.34% Na
58.44g
this
molecules
Do not worry about this problem yet.
Section 3
P.24
1.
MW of C2H5Cl = 64.51
24.02g C x 100 = 37.23% C
64.51g C2H5CL
2.
MW of Al2(SO3)3 = 294.15
3.
MW of CH3COOH = 60.05
4.
0.350g of metal
96.19g S
x 100 = 32.70% S
294.15g Al2(SO3)3
24.02g C
x 100 = 40.00% C
60.05g CH3COOH
0.350g metal x 100 = 57.9% metal
problem.
+0.255g of oxygen
0.605g compound
.605g of metal oxide
The method for find percent by weight is still identical.
5.
CuSO4 5H2O
MW = 339.74
180.14g H2O x 100 = 53.02% H2O
339.74g CuSO4 5H2O
P. 29 Exercises
1.
Assume 100g of compound is present
26.67g S x 1 mole S atoms = 0.8319 moles
32.06g
Hg should be
53.33g O x 1 mole O atoms = 3.333 moles O
16.00g
20.00g Mg x 1 mole Mg
24.31 g
There is a
a mistake in
the problem.
Mg.
= 0.8227 moles Mg
Mg .8227
S 0.8319
O 3.333
<-- 1 st in order of
character
Mg .8227 S 0.8319 O 3.333 --> Mg S 1.011 O 4.051 --> MgSO4
.8227
.8227
.8227
2.
Assume 100g of compound - 27.59g O and 72.41g Fe
27.59g O x 1 mole O = 1.724 mole O atoms
16.00g O
72.41g Fe x 1 mole Fe
= 1.296 mole Fe atoms
55.85g Fe
Fe 1.296 O 1.724 --> Fe 1.296 O 1.724 --> Fe1 O 1.330 -->
Fe3O4
3. 51.0g of aluminum oxide - 27.0g aluminum = 24.0g oxygen
27.0g x 1 mole Al = 1.00 mole 24.0g O atoms x 1 mole O atoms=
26.98g Al
Al
16.0g
1.50 mole O
Al1O 1.50 --> Al2O3
4.
(a)
Assume 100g of compound
57.1g C x 1 mole C atoms = 4.76 mole C
12.0gC
4.8g H x 1mole H atoms = 4.75 mole H
1.01g H
38.1g O x 1 mole O atoms - 2.38 mole O
16.0gO
C 4.76 H 4.75 O 2.38 --> C 4.76 H 4.75 O 2.38 --> C2H2O1
2.38
2.38
2.38
metallic
(b)
C2H2O1 's weight adds up to 42. The real weight is 126.
126 = 3 so 3 units of C2H2O1 are needed.
42
The true formula is C6H6O3. {Note: 3C2H2O1 is wrong.}
Page 31 - Review problems
1.
5.22 x 1023 N atoms x 1 mole N atoms x 14.01g N = 12.1gN atoms
6.02 x 1023 atoms N
1 mole N atoms
2.
48.5g HBrO3 x 1 mole HBrO3 x 6.02 x 1023HBrO3 molecules =
128.92g HBrO3 1 mole HBrO3
2.26 x 1023HBrO3
3.
1.5 moles Ca(OH)2 x 74.09g Ca(OH)2 = 111g but with only 2 sig.
1 mole Ca(OH)2
figs. The answer must be
110g Ca(OH)2
4.
52.5g Zn(HCO3)2 x 1 mole Zn(HCO3)2 = 0.280 mole Zn(HCO3)2
187.41g Zn(HCO3)2
5.
From #4
0.280 moles Zn(HCO3)2 x 6.02 x 1023 molecules =
1 mole Zn(HCO3)2
1.69 x 1023 Zn(HCO3)2
molecules
6.
0.280 moles Zn(HCO3)2 x 11 moles atoms = 3.08 moles of atoms
1 mole Zn(HCO3)2
1 molecule of Zn(HCO3)2 contains 11 atoms
1 mole of Zn (HCO3)2 molecules contains 11 moles of atoms
This is a new type of problem but uses old ideas. There are
other ways of solving it also.
7.
2.02g H
x 100 = 1.08%H
187.41g Zn(HCO3)2
8.
52.5gZn(HCO312 x 1 mole Zn(HCO3)2x6.02x1023Zn(HCO3)2 moleculesx
187.41gZn(HCO3)2
1 moleZn(HCO3)2
11 atoms
= 1.86 x 1024 atoms
1 molecule Zn(HCO3)2
We could have started the answer to #6 also.
3.08 moles of atoms x 6.02 x 1023 atoms = 1.85 x 1024 atoms
1 mole atoms
Notice the difference of 1 in the final digit.
digit is always uncertain and can vary.
9.
The last sig.
52.5gZn(HCO3)2 x 1 mole Zn(HCO3)2x6.02x1023 Zn(HCO3)2moleculesx
187.41gZn(HCO3)2
1 mole Zn(HCO3)2
1 atom Zn
1 molecule Zn(HCO3)2 =
1.69 x 1023 Zn atoms
10.
28.01g N
80.04g NH4NO3
11.
Assume 100g of compound
36.5g Na x 1 mole Na atoms = 1.59 mole Na atoms
22.99g Na
25.4g S x 1 mole S atoms = 0.792 mole S atoms
32.06g S
38.1g O x 1 mole O atoms = 2.38 mole O atoms
16.00g O
Na2SO3
12.
<---
Na2.01 S1 O3.01
Assume 100g
90.6g Pb x 1 mole Pb atoms = 0.437 mole Pb
207.19g
9.4g O x 1 mole O atoms = 0.59 mole O
16.00gO
Pb3O4
13.
x 100 = 35.0% N
<-- multiply x 3 <--- Pb1O1.35
Calculate the percent of metal first in MgS.
24.31gMg x 100 = 43.13%
43.13% of 50.0g of MgS must be
5
56.37gMg
Mg.
43.13gMg x 50.0g MgS = 21.6g Mg
100g MgS
14.
1.505 x 1023 moleculse NaOH x 1 mole NaOH x 40.00g NaOH =
6.02 x 1023 molecules 1 mole NaOH
10.0g NaOH
Of course, you could have begun with the answer to #14.
16.
44.01 amu's - Did you say grams?
Review that vocabulary.
17. This is a different problem to read. Molecular mass is the
molecules weight. The problem is an empirical formula problem.
4.02gN x 1 mole N atoms = 0.287 mole N atoms
14.01gN
same
as
11.48gO x 1 mole O atoms = 0.718 mole O atoms
16.00 O
N .287 O .718 --> N1O2.5 --> multiply by 2 --> N2O5
--N2O5 has a weighted 108 so the actual formula is N2O5.
18.
44.01g
19.
3 atoms - This problem asked for atoms in a molecule, not
20.
3 moles of atoms - See your teacher if 19 and 20 confuse you.
21.
1 C atom x 1 mole C atoms x 12.01g C atoms = 2.00x10-23 g C
6.02x1023 atoms of C
1 mole C atoms
This is backward from many of the problems that we have done.
22.
A mole is 6.02x1023 of anything.
A mole, by deifintion, is a number and not a weight.
23.
Avogadro's Number is 6.02 x 1023 or 1 mole.
6 6g CO2 x 1 mole
= 1.50 mole (or 9.03x1023 molecules)
44gCO2
S or the number of molecules in 66g of CO2 is larger.
24.
322.18g
mole.
25. They are called anhydrons copper (II) sulfate and anhydrous sodium
sulfate instead of copper (II) sulfate pentahydrate (or copper (II) sulfate
5 hydrate) and sodium sulfate decalydrate.
Anhydrous
means
"without
water."
CHEMICAL FORMULAS
AND
NOMENCLATURE
CHEMICAL FORMULAS
and
NOMENCLATURE
One of the basic skills that must be developed by beginning cheistry
students is that of writing and naming formulas.
Many of the laboratory
reagent bottles will have either names or formulas of compounds while the
laboratory experiments will list the compound in the other mode.
Just as
you learned teh alphabet to put words together, you must first be able to
recognize symbols before you can learn formulas. Learning to write formulas
is much easier than learning to spell English words, however, since the
rules are straight forward. The mastery of formula writing as the mastery
of spelling requires practice. This is one of the few sections of chemistry
where moemorization is a must so clear the cobwebs out of your memory banks.
Objective:
1.
When given the name of an ion group, write its formula.
2.
When given the formula of an ion group, write its name.
3.
When given a Periodic Chart, identify the most common
number of representative group elemental atoms that
can
metal-nonmetal compounds.
4.
When given the formula of a compound, write its name.
5.
When given the name of a compound, write its formula.
combining
be
used
6.
When presented with a list of problems to name or write formulas,
identify and correctly perform the naming or formula
writing
nonmetal-nonmetal compounds.
7.
Define the words ion, anion, and cation as they were used in this
section.
in
for
Formulas for compounds were not standarized until the middle to late
nineteenth century.
This is not say that water was H 2O,
H3O, and H3O2
before that but that man had to discover that water was H 2O. The formulas
of compounds are invariant.
Water is H2O.
It just took a long time to
discover that.
Before John Dalton's model of the atom in 1802 to 1808, many scientists
thought that matter was infinitely divisible.
There was no smallest
particle such as the atom. Dalton's model stated that there was a smallest
particle for each element, and the particles of one type of element were
similar in weight to each other but different from the particles of other
elements.
Dalton also made an initial chart of atomic weights.
In 1828
Berzelius published a more accurate table of atomic weights. Later, other
scientists found even more accuarte values.
Another contribution of Berzelius was that of a standard symbol and
nomenclature system. The ancient alchemists used assorted symbols such as
moon slivers, triangles, and circles to represent elements. Dalton proposed
that circles with various shadings and other delineations be adopted.
Berzellius chose the first letter of the name of the element where possible
or the first and second letter of the elemental name for the symbol. The
symbol often represented the Latin name of the element rather than the
English name.
An example is that of iron and ferrum, Fe.
Gold has the
symbol Au from "aurum", silver, Ag form "argentum", and lead, Pb from
"plumbum".
Once the system for symbols and formulas were standarized and tables of
atomic weights were established, it became a matter of time before chemists
succeeded in deducing formulas for compounds.
They did this by either
forming or breaking apart the compound to see how much of each element it
contained.
Since it was (and is) not possible to see the atoms, it was
necessary to use indirest means to discover weight relationships. You will
look at weight relationships in the next unit.
As formulas became to be known, the idea of combining number or
combining power was advanced.
In metal-nonmetal compounds, it can be
observed that elements have certain tendencies for combination.
With the
exception of metal to hydrogen combinations called hydrides, hydrogen can be
thought of as a metal. It will combine to form H 2O, HCl, H2S, HBr, and HF.
Sodium can form NaCl, Na2S, NaBr, and NaF.
Sodium and hydrogen have
combining numbers of one.
Magnesium forms compounds of MgO, MgCl2, MgBr2,
and MgI2. Its combining number is two as are those of beryllium, calcium,
strontium, barium and radium.
In metal-nonmetal compounds, chlorine,
bromine, fluorine, and iodine all have combining numbers of one with oxygen
and sulfur having two.
By looking at a Periodic Chart, the most common
combining numbers of many elements can be deduced.
The combining numbers of middle and the bottom groups cannot be deduced in
this manner but must be learned.
Binary metal-nonmetal compounds are formed by making the combining
numbers of the metal and nonmetal equal.
In magnesium chloride, the
magnesium has a combining number of two and the chlorine has a combining
number of one.
It is necessary to have two chlorines to add up to the
magnesium combining number. The formula has one magnesium and two chlorines
or MgCl2. The idea of combining numbers is sufficient to form most binary
compounds. You may have, however, seen bottles in the laboratory with more
than two elements in the formula. Common battery acid, or sulfuric acid, is
H2SO4. How do we obtain a formula such as this?
The ancient Greeks had known that amber when rubbed attracted light
objects just as bits of paper will be attracted to a comb after it has been
put through dry hair in the winter. A French chemist, du Fay, discovered in
1733 that there were two kinds of electrical charge. One could be put on
glass and the other on amber. Particles containing one of the charges would
attract particles containing the other charge, but particles of like charges
repelled each other.
In the 1740's, Benjamin Franklin suggested that on
electrical charge resuleted from a substance which contained more than a
greater amount of electrical fluid; and the other charge resulted from a
substance which contained a lesser amount of electrical fluid.
He called
the glass charge positive and the amber charge negative.
Volta, Carlisle, Nicholson, and Davy worked with chemistry and
electricity in the late 1700's and early 1800's. Volta found that a current
could be produced by two different metals, brine soaked blotting paper in
between, and wire connecting them. This voltaic pile (battery) enabled the
other men to use electrolysis as a means of separating certain compounds.
Davy, in particular, observed that certain substances such as hydrogen and
metals would go to the negative pole and oxygen would go to the positive
pole.
An atom of sodium is electrically neutral.
Sodium metal reacts
readily, vigorously, and sometimes violently with water. When Davy isolated
the metal sodium at the negative electrode, he observed this phenomena. The
particles of sodium that was attracted to the negative electrode had to be
positively charged.
It did not react with the water.
This particle is
called a sodium ion. An ion is a charged particle of an atom.
If a salt water solution is used in this electrolysis, chlorine as well
as sodium will be present since ordinary table salt is sodium chloride. The
chlorine will be formed as diatomic molecules in normal conditions and is a
poisonous pale green gas.
Chlorine is electrically neutral.
The particle that migrates to the
positive electrode must be negatively charged.
This charged particle is
called a chloride ion. Table salt is made up of sodium ions and chloride
ions.
You would definitely not want to eat sodium atoms or chlorine
molecules but eat their ions everyday.
Ions, or charged particles, are
chemically different from their original species.
How does this help us to deduce a formula such as H2SO4? Some groups of
atoms tend to combine together to form charged species that act as if they
were single charged atoms. Ions are charged atoms or groups of atoms. It
is know that sodium and chlorine both have combining numbers of one but that
their ions have opposite charges which cause them to attract. Na1+ and Cl1can form NaCl.
The hydrogen ion has a combining number of one and is
positively charged, H+, except when it combines with metals. The one for a
combining number is sometimes omitted but any other number must be included.
The sulfate group is SO42-. This means that it is negatively charged with a
combining number of two. The two ions attract each other because they have
opposite charges. It requires two hydrogens to equal the combining number
of one sulfate so the formula is H2SO4.
The next page has a chart of common ions and ion groups. It will be
necessary to learn these. In this class we will use the modern notation of
iron (III) for Fe3+ and iron (II) for Fe2+
instead of the older ferric and
ferrous. You may occasionally find these older names so be aware that they
do exist.
Rules for writing and naming formulas follow the ion chart.
Compounds can also be made from nonmetals combining with other nonmetals.
These compounds have an entirely different set of rules involved with their
formulas and names. This will follow the other set of rules.
TABLE OF COMMON IONS AND THEIR CHARGES
_________________________________________________________________NAME
SYMBOL
CHARGE
NAME
SYMBOL
CHARGE
-----------------------------------------------------------------Aluminum
Al3+
+3
Lead (II)
Pb2+
+2
-----------------------------------------------------------------Ammonium
NH4 2+ +1
-----------------------------------------------------------------Barium
Ba2+
+2
Magnesium
Mg2+
+2
-----------------------------------------------------------------Calcium
Ca3+
+2
Mercury (I) Hg2 +2
-----------------------------------------------------------------Chromiuim
(III) Cr
+3
mercurous
----------------------------------------------------------------chromic
2+
Mercury (II) Hg
+2
-----------------------------------------------------------------Cobalt (II)
Co2+
+2
mercuric
----------------------------------------------------------------Nickel (II) Ni2+ +2
-----------------------------------------------------------------Copper (I)
Cu+
+1
----------------------------------------------------------------cuprous
Potassium
K+
+1
-----------------------------------------------------------------Copper (II)
Cu2+
+2
Silver
Ag+
+1
----------------------------------------------------------------cupric
Sodium
Na2+
+1
-----------------------------------------------------------------Hydronium
H3O
+1
Tin (II)
Sn
+2
----------------------------------------------------------------stannous
-----------------------------------------------------------------Iron
(II)
Fe2+
+2
Tin (IV)
Sn4+
+4
----------------------------------------------------------------ferrous
stannic
-----------------------------------------------------------------Iron (III)
Fe3+
+3
Zinc
Zn2+
+2
----------------------------------------------------------------Ferric
--------------------------------------------------------------------------------------------------------------------------------Acetate
C2H3O2
-1
Hydroxide
OH-1
-----------------------------------------------------------------Arsenate
AsO43-3
Hypochlorite C10- -1
-----------------------------------------------------------------Bromide
Br-1
Iodide
I-1
-----------------------------------------------------------------Carbonate
CO32-2
Nitrate
NO3-1
-----------------------------------------------------------------Chlorate
C103-1
Nitrate
NO2-1
-----------------------------------------------------------------Chloride
Cl-1
Oxalate
C2O42-2
-----------------------------------------------------------------Chromate
CrO42- -2
Oxide O2-2
-----------------------------------------------------------------Cyanide
CN-1
Permanganate MnO4
-1
-----------------------------------------------------------------Dichromate
Cr2O72-2
Peroxide
O22-2
-----------------------------------------------------------------Dihydrogen
Phospate
PO43-3
----------------------------------------------------------------phosphate H2PO4-1
Sulfate
SO42-2
-----------------------------------------------------------------Fluoride
F-1
Sulfide
S2-2
-----------------------------------------------------------------Hydride
H-1
Sulfite
SO32-2
-----------------------------------------------------------------Hydrogen
carbonate HCO3- -1
-----------------------------------------------------------------Hydrogen
sulfate HSO4- -1
-----------------------------------------------------------------
FORMULA WRITING
for
METAL-NONMETAL COMPOUNDS
1.
All compounds are electrically neutral and have balanced
numbers.
combining
2.
An atom is electrically neutral; an ion is a charged species.
molecule is electrically neutral; an ion group is a charged species.
3.
A negative ion is called an anion; a positive ion is a cation.
4.
A cation is always written first in a formula.
In table salt, sodium is first.
In ammonium chloride, the ammonium ion is first.
5.
The total combining number of the cation species must equal the
combining number of the anion species.
In sodium chloride, Na+ and Cl- ----> Na Cl.
In magnesium chloride, Mg2+ and Cl- requires two Cl- to
equal one Mg2+ -----> MgCl2.
The one is understood.
In magnesium oxide, Mg2+ and O2-
total
---> MgO.
6.
When the cation or anion is an ion group, parenthesis are used
more than one group in needed.
In magnesium sulfate, Mg2+ and SO42-----> MgSO4.
Only one sulfate ion is needed.
In sodium sulfate, Na+ and SO42Two sodium ions are needed.
A
-----> Na2SO4.
In calcium hydroxide, Ca2+ and OH- ----> Ca(OH)2.
Two
when
hydroxides are needed. This is written Ca(OH)2 and not
CaO2H2.
It is really the calcium ion with two hydroxide
ion groups attached.
In magnesium nitrate, Mg2+ and NO3-
---> Mg(NO3)2.
Exercises:
Write the formula for each of the following compounds:
1.
2.
3.
4.
5.
Calcium phosphate
Sodium oxide
Potassium chloride
Sodium carbonate
Calcium sulfate
9.
6. Lithium carbonate
7. Sodiuim hydrogen carbonate
8. Sodium sulfite
Sodium sulfide
10. Ammonium sulfate
7.
Metals with combining numbers that vary are written
combining number is given in the name.
In iron (III) chrloride, Fe3+ and Cl- ----> FeCl3.
In iron (II) bromide, Fe2+ and Br-
so
that
the
----> FeBr2.
Exercises
Write the formula for each of the following compounds:
1.
2.
3.
Tin (IV) oxide
Tin (II) fluoride
Lead (II) chloride
4.
5.
6.
Copper (II) sulfate
Copper (I) sulfate
Iron (II) hydroxide
NOMENCLATURE RULES
for
METAL-NONMETAL COMPOUNDS
1.
The compound has a cation (+) and an anion (-).
2.
The cation is listed first in a compound.
3.
In most cases the cation has no prefix.
It nevr has a suffix
(unless the old system for multivalent compounds is used, I.e.
ferric for Fe3+.)
4.
Some cations have more than one common combining number and this
number must be identified by the use of Roman numerals in parentheses after
the elemental name.
IUPAC (modern) rules: Fe3+ ----> iron (III)
Fe2+ ----> iron (II)
3+
Old system
Fe
----> ferric
Fe2+ ----> ferrous
The combining number must be deduced from the formula of a
compound.
FeS would be iron (II) sulfide since sulfur in the ion form is S2-.
Fe2(SO4)3 would be iron (III) sulfate since the sulfate ion is
SO
and there are three of them to make a total combining
number of six.
Since there are two iron ions present, each
iron must have a combining
number of three to also make a
total of six.
24
Exercises
What is the combining number of the metal in each of the
1.
2.
3.
4.
5.
5.
FePO4
Fe3(PO4)2
SnO2
SnO
PbCl2
6.
7.
8.
9.
10.
following?
HgCl2
CuSO4
CuCl2
CuI
SnF2
All anions have suffixes. The elemental ion will end in -ide.
O, oxygen ---> ion is oxide, O2Cl, chlorine ---> ion is choride ClN, nitrogen ---> ion is nitride, N3-
Exercises
Name the anions of the following atoms:
1.
2.
3.
bromine
phosphorus
sulfur
6.
4. iodine
5. selenium
hydrogen (as an
anion)
The anion groups have suffixes that are already incorporated into their
names. This has nothing to do with the number of oxygens attached (except
more oxygens
ate, fewer
ite) or
the combining number and must be
memorized.
sulfate SO42nitrate NO32Sulfite SO3
nitrate NO26.
An aqueous (aq) solution of a compound of hydrogen and an
form an acid. The same substance in gaseous (g)
form
may
considered to be an acid.
anion
not
may
be
The hydrogen is shortened to hydro- and the anion ending is changed
from -ide to -ic. Acid is added to the name.
HCl (g) is hydrogen chloride. HCl (aq) is hydrocloric
acid.
HF(g) is hydrogen fluoride.
HF(aq) is hydrofluoric
acid.
An aqueous solution of a compound of hydrogen and an anion
group
may
also form an acid. The hydrogen part of the name is dropped entirely.
-ate suffix becomes an -ic while the
-ite suffix becomes an -ous.
HNO3(g) is hydrogen nitrate. HNO3(aq) is nitric acid.
H2SO3(g) is hydrogen sulfite. H2SO3(aq) is sulfurous
acid.
The
Exercises
Name the following:
1.
HBr(g)
6.
2.
H2SO4(aq)
3.
HI(aq)
4.
H3PO4(aq)
5.
HC2H3O2(aq)
H2CO3(aq)
7.
8.
HCN(aq)
9.
10.
HCl(aq)
HCl(g)
H2CrO4(aq)
7.
For acids and salts which have several anion groups of a
nonmetal
and oxygen, a special nomenclature is used. A prefix
may be necessary.
ClO4perchlorate ion
ClO3chlorate ion
ClO2chlorite ion
ClO
hypochlorite ion
Examples:
KBrO3
Potassium bromate
HClO2(aq)
Chlorous acid
Exercises
Name the following:
1.
NaClO
4.
HClO4(aq)
2.
HClO(aq)
5.
HClO3(aq)
3.
NaClO4
6.
KClO2
BINARY NONMETAL-NONMETAL COMPOUNDS
1.
The first species in the name or formula is always (almost)
left or below in the Periodic Chart.
to
the
In a compound of sulfur and oxygen, sulfur is first.
In a compound of nitrogen and oxygen, nitrogen is first.
2.
The first species will have no suffix. It may have a prefix if
than one atom is present.
In CO, the first species is carbon-.
In N2O4, the first species in dinitrogen-.
3.
The second species will have a number prefix and an -ide
In CO, the second species is -monoxide.
In N2O4, the second species is tetroxide.
more
ending.
4.
The prefixes will indicate the number of atoms of that element
present. They are mono, di, tri, tetra, penta, hexa, hepta, octo, and
so on. The last letter may be omitted to facilitate
naming.
In CO, the second species is -monoxide rather than
monoxide.
In N2O4, the second species is -tetroxide rather than
tetraoxide.
5.
The rules for metal-nonmetal compounds do not hold here. Ions
are
not formed and combining numbers can vary. The formula is
given
by
the
name.
Carbon dioxide ---> CO2
This must have only one carbon (no prefix) and two oxygens (di
prefix). The name can be taken directly from the formula
according
to the above rules.
SO3 ----> sulfur trioxide
This has only one sulfur so no prefix is needed.
It has three
oxygens so a tri- prefix is needed as well as an -ide suffix.
Exercises
Name or write the formula for each of the following:
1.
CO2
4.
dichlorine monoxide
2.
sulfur dioxide
5.
P4O10
3.
carbon tetrachloride
6.
NO2
Exercises
In order to master a skill such as formula writing and naming,
it is
necessary to practice. The following exercises are
divided into groups.
Do the first group, check your answers, and review any types that you
missed. Do the same with the next group.
Be sure that your spelling is
correct.
1.
2.
3.
4.
5.
Potassium bromide
SF6
Ag3PO4
Chromium (III) iodate
SnCl4
1.
2.
3.
4.
5.
CsClO3
Phosphorus pentafluoride
Zinc hydrogen carbonate
Aluminum sulfate
Pb(MnO4)2
1.
2.
3.
Iron (III) chloride
KClO3
Ba(HSO4)2
chromate
5. Hg2(NO3)2
1.
2.
3.
4.
5.
MgC2O4
Beryllium nitrite
Aluminum sulfide
Al2(SO3)2
nitrogen tri-iodide
1.
2.
3.
4.
5.
Barium hypobromite
CaH2
Arsenic (III) sulfate
Bismuth (III) oxide
PbCO3
8.
9.
6.
9.
6. Calcium nitrate
7. Boron nitride
Zinc cyanide
Sn3(PO4)2
10. Al(ClO4)3
(NH4)3N
7. Lead (II) sulfide
8. Bromic acid
Cd(IO3)2
10. Cl2O
6. Phosphorus trichloride
7. Cu2CO3
8. Calcium potassium
phosphate
4. Aluminum
9.
HClO
10.
6.
Cl2O7
8.
KMnO4
7. Aluminum acetate
Ammonium nitrite
9. CoCl3
10. N2O5
8.
6. AuHSO3
7. Cesium hypoiodite
Hg(CN)2
9. Co(MnO4)2
10. IO5
Exercises
If astatine (At) is similar in properties to chlorine (Cl) and
gallium (Ga) is similar to aluminum (Al), write the formulas for
the
following
compounds.
1. potassium astatate
2. barium astatide
3. gallium sulfate
4. hydrastatic acid
5. gallium cyanide
6. gallium hypoastatite
If selenium (Se) is similar in properties to sulfur (S) and francium
(Fr) is similiar to sodium (Na), write the formulas
for
the
following
compounds:
1. zinc selenide
2. francium phosphate
3. cobalt (II) selenite
4. selenium dioxide
5. francium selenate
6. selenium hexafluoride
7. francium hydride
Answers to Chemical Formulas and Nomenclature
P.8 Exercises
1.
2.
3.
4.
5.
P.9
1.
2.
3.
P.9
1.
2.
3.
4.
5.
Ca3(PO4)2
Na2O
KCl
Na2CO3
CaSO4
Exercises - top
SnO2
SuF2
PbCl2
Fe(III)
Fe (II)
Sn (IV)
Sn (II)
Pb (II)
3.
4.
CuSO4
Cu2SO4
Fe(OH)2
6.
7.
8.
9.
10.
Hg
Cu
Cu
Cu
Su
(II)
(II)
(II)
(I)
(II)
Exercises - top
bromide
phosphide
sulfide
P. 10
1.
2.
4.
5.
6.
Exercises - bottom
P. 10
1.
2.
3.
9.
6. Li2CO3
7. NaHCO3
8. Na2SO3
Na2S
10. (NH4)2SO4
4.
5.
6.
iodide
selenide
hydride
Exercises - bottom
hydrogen bromide gas
sulfuric acid - dilate
(conc. sulfuric acid is 95%
sulfuric acid. It is a liquid) 9.
hydroiodic acid
phosphoric acid
6. carbonic acid
7. hydrochloric acid
8. hydrocyanic acid
hydrogenchloride gas
10. chromic acid
5.
acetic acid
Exercises p. 11
1.
2.
3.
sodium hypochlorite
hypochlorous acid
sodium perchlorate
4.
5.
6.
perchloric acid
chloric acid
potassium chlorite
4.
5.
Cl2O
P4O10 - tetraphosphorus
decoxide
(phosphorus (V) oxide)
nitrogen dioxide
(nitrogen (IV) oxide
Exercises p. 12
1.
2.
3.
carbon dioxide
Carbon (IV) oxide
SO2
CCl4
6.
Exercies p. 13
1.
2.
3.
4.
5.
KBr
Sulfur hexafluoride
(sulfur (VI) fluoride
silver phosphate
Cr I3
Tin (IV) chloride
1.
2.
3.
4.
5.
Cesium chlorate
PF5
Zn(HCO3)2
Al2(SO4)3
lead (II) permangamate
1.
2.
3.
FeCl3
potassium chlorate
barium hydrogen sulfate
barium bisulfate
Al2(CrO3)3
mercury (I) nitrate
4.
5.
8.
9.
6. Ca(NO3)2
7. BN
Zn(CN)2
Sn (II) phosphate
10. aluminum perchlorate
6.
7.
8.
9.
10.
9.
ammonium nitride
PbS
HBrO3
Cadmium (II) iodate
dichlorine monoxide
6. PCl3
7. Copper (I) carbonate
8. CaKPO4
hypochlorous acid
10. dichlorine heptoxide
Exercises p. 13 continued
1.
2.
3.
4.
5.
magnesium oxalate
Be3N2
Al2S3
aluminum sulfite
NI3
9.
6. potassium permanganate
7. Al(C2H3O2)3
8. NH4NO2
cobalt (III) chloride
10. dinitrogen pentoxide
1.
Ba (BrO)2
2.
3.
4.
calcium hydride
As2(SO4)3
Bi2O3
5.
lead (II) carbonate
6. gold (I) hydrogen
sulfite gold (I)
brsulfite
7. Cs IO
8. mercury (II) cyanide
9. cobalt (II) per
manganate
10. iodine pentoxide
Exercises p. 14
1.
2.
3.
4.
5.
6.
KAt
BaAt2
Ga2(SO4)3
HAt
Ga(CN)3
Ga (AtO)3
1.
2.
3.
4.
5.
6.
7.
ZnSe
Fr3PO4
Co SeO3
Se O2
Fr2SeO4
SeF6
FrH
CHAPTER 2 -- STOICHIOMETRY
SECTION 1 -- CHEMICAL REACTIONS
Chemistry evolved from alchemy in which the aim was tranmutation of
metals to gold. The early alchemists did make some important observations
about chemical reactions as they tried to convert lead to gold but these
were usually not quantitative in nature.
One question about chemical
reactions is what happens if something is mixed with something else.
A
modern industrially important question is how much of a chemical is needed
to put into a reaction to get the desired amount of product. This chapter
will attempt to look at these questions although they involve many more
factors than can be addressed here. We will begin by investigating chemical
reactions.
Objectives:
1.
Given a chemical equation, form pictorial representations of the
reactants and products.
2.
Define chemical equation, products, and reactants.
3.
Illustration of the Law of Conservation of Matter using a
reaction.
4.
Given a formula equation, balance it.
5.
Given a word
balance it.
equation,
transform
it
to
a
formula
chemical
equation
and
6.
Given a list of equations of the five given types, classify
to type.
according
7.
Given a partial single replacement, metathesis, combustion
complete and balance the equation.
reaction,
8.
Given a solubility chart, determine if a metathesis reaction will
occur.
CHAPTER 2
SECTION 1
Modern chemisty, as compared to alchemist chemistry, began in 1860.
The period from 1750 to 1860 was the time of changeover.
The ideas of
Antoine Lavoiser (1743 - 1794), a wealthy French scientist, helped to guide
chemistry in the right direction. Lavoisier kept careful records (or rather
had his wife do so) of his experiments and extensively used the balances to
weigh both reactants and products.
Although the idea of conservation of
matter had been observed earlier by Joseph Black, Henry Cavandick, and a
Russian, Mikhail Lamonosov, it was up to Lavoisier to publicize the idea.
The law of Conservation of Mass of Conservation
of Matter states that the weight of the products
in a chemical reaction should be equal to the
weight of the reactants.
The writing of modern chemical equations had to follow the introduction
and popularization of several theories.
In the period from 1802 to 1808,
John Dalton, a poor English school teacher, formulated a very important
theory of the atom.
In it he stated that all matter is composed of tiny
particles called atoms, atoms of one element have the same weight, atoms are
indivisible, and chemical changes are changes in the combinations of atoms
with each other.
Although some ancient Greeks such as Leucippets and his
student Democritus theorized that there was a smallest particle of nature
such as an atom, it remained just a philosophical entity with the idea that
matter was infinitely divisible being more accepted.
The idea that there
are finite particles of definite masses for the elements paved the way for
later ideas. We often accept this idea and the important idea that chemical
changes are changes in the combinations of atoms with one another without
second thought.
Another aid to the writing of equations was the
standardization of chemical symbols. Many new
elements were discovered after the 1750 time
period
mentioned above. The symbols used for
the
elements
were
left over from the alchemist
days. The Wizard of Id's hat,
in the comics,
contains many of those symbols. Some of the
chemistry texts in your classroom should have other
examples.
These were cumbersome to remember and to use.
Dalton made an
attempt at a more modern system of symbol writing by using circles with
various dots and lines to indicate the different elements but this system
was also tedious to use.
Berzellius (1779-1848) proposed in 1813 that
letters and subscripts (raised numbers) be used to indicate compounds.
Water would be H2O.
With the exception of subscripts instead of
superscripts, this is the system used today.
What else had to preceed equation-writing?
It was necessary to know
the formula of a compound first.
Some compounds such as water, H 2O, and
hydrogen peroxide, H2O2 consist of the same type elements. Dalton observed
that when two elements combine to form more than one compound, the masses of
one element that combine with a given mass of the other element are in the
ratio of small whole numbers.
This is known as the Law of Multiple
Proportions.
In the water and hydrogen peroxide case set the mass of
hydrogen at 2.0 grams.
The mass of oxygen in water would then be 16.0
grams. The mass of oxygen in hydrogen peroxide would be 32.0 grams. The
ratio of the mass of oxygen in water to mass of oxygen in hydrogen
Hydrogen (set)
Oxygen (found)
Ratio of O masses
-----------------------------------------------------------------water
2.0g
16.0g
1
hydrogen peroxide
2.0g
32.0g
2
peroxide would be one to two. According to Dalton's ideas, the formula for
water should then be HO and the formula for hydrogen peroxide should be HO 2.
The true formulas were deduced later when better analytical methods were
developed and more accuarate atomic weights were found.
Gay-Lussac's Law of Combining Volumes both confirmed the Dalton model
of the atom and illustrated numerical observations that are very close to
our modern equations. He observed that two volumes of hydrogen gas and one
volume of oxygen would produce two
2 vol.
hydrogen
+
1 vol. oxygen gives 2 vol water vaper
volumes of water vaper.
Dalton disputed this observation by saying that it was the same as two atoms
of hydrogen and one atom of oxygen giving two molecules
of water (HO). For this to happen, it would be necessary to cut an atom in
half.
Even if he had known that water was really H2O, only one molecule
could have been predicted.
Gay-Lussac also observed that one volume of
hydrogen and one of chlorine would give two volumes of hydrogen chloride
gas. Three volumes of hydrogen and one volume of nitrogen gs would yield
two volumes of ammonia gas (NH3). Look at the pictures below.
Gay-Lussac's observations:
one vol. hydrogen + one vol. chlorine gives two vol. hydrogen chloride
three vol. hydrogen + one vol. nitrogen gives two vol. ammonia
Dalton's view
one atom hydrogen + one atom chlorine
three atoms hydrogen
two molecules hydrogen chloride
one atom nitrogen gives two molecules ammonia
It remained fo rthe Italian, Amedo Avogadro (1776-1856) to publish a
paper in 1811 to explain the problem in Dalton's idea. He stated that the
gases were diatomic in nature. Diatomic means that hydrogen gas occurs not
as the atomic hydrogen but as a molecule with two hydrogen atoms chemically
combined. Those common elements that exist naturally as diatomic molecules
are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine.
All but the last two are gases.
The pictorial explanation for the three reactions above would be:
two molecules of
hydrogen
+
one molecule
of oxygen
gives
two molecules of water
one molecule of
+ one molcule of gives
two molecules of
chlorine
hydrogen chloride
------------
-------------
hydrogen
----------------
Draw in the pictures for the third reaction yourself and write the statement
for what is happening as in the first two examples.
We would write these reactions as:
2H2
+
O2
H2
+
Cl2
-----> 2HCl
---N2
---->
---H2
+
----->
2H2O
---NH3
The statements above are symbolic representations, of what happens in a
chemical reaction. They are called chemical equations. The substances on
the left side of the equation are called reactants and those on the right
side are called products. According to the Law of Conversation of Matter,
we should have the same amount of oxygen and hydrogen on each side of the
equation. Nothing can be lost or added during a chemical reaction. The law
can also be called the Law of Conservation of Atoms. This means that the
right and left side should have the same number of atoms.
How would you go about making the number of atoms the same on each side
of the ammonia equation above? This is called balancing the equation.
------H2
+
------N2
------>
----NH3
What we cannot do is change any subscript.
The formula for ammonia is NH3
and nothing else.
N2H6 and NH3 are very different formulas.
From the
preceeding discussion we now know that hydrogen and nitrogen occur as
diatomics.
The only part that appears to be adjustable is the number of
molecules that we bring to the reaction and the number that we get out.
When balancing an equation, adjust only the coefficients.
The method we
will use here is called "balancing by inspection" or "trial and error".
Begin by counting the number of atoms on each side.
----H2
H's
+
-
----N2
2
--------->
----NH3
H's
-
3
N's
-
2
N's
-
1
Next change the coefficients from the understood 1 to whatever is
needed. In this case, adjust the product nitrogen (which, of course, also
adjusts the product hydrogen).
______H2
+
______N2
Recount the atoms:
--------->
H's - 2
N's - 2
__2___NH3
H's -6
N's - 2
Readjust the coefficients if needed.
3
H2
+
Count atoms:
N2
------->
H's - 6
N's - 2
The equation is balanced:
2
NH3
H's - 6
N's - 2
3H2
+ N2
------> 2NH3
Try the next example yourself.
Ca(OH)2
+
H3PO4
-------> Ca3(PO4)2
+
H2O
There are some hints that make working with an equation such as this a
little easier.
In equations such as this, it is better to save the
hydrogens and oxygens for last.
Adjust the calcium and phosphorus first.
In some cases it is easier if you rewrite water as NOH to show that it can
represent a hydroxide ion and a hydrogen ion. If the ion groups such as OHand PO43- do not break up the reaction, it is possible to count groups. The
answer to the equation is
3Ca(OH)2
+ 2H3PO4
--------> Ca3(PO4)2
+ 6H2O
After the experiment you will have plenty of practice with balancing
equations.
Experiment:
Balancing Chemical Equations
In chapter one we looked at the electrolysis
of water apparatus. Here we will use it again
along with molecular models to aid in balancing
equations according to the Law of Conservation of
Atoms.
If your lab has several electrolysis
apparatuses, you will need to set up the
apparatus. If there is only one, skip parts
1, 2, and 3.
1.
Set up the electrolysis apparatus. The apparatus should be connected
to some D-C 6 volt source. If a rectifier is used,
be sure that it is
off before you connect the apparatus. If
batteries are used, include a
knife switch and keep it open until you are ready to begin.
2.
Slowly, with stirring, add 10 ml of concentrated sulfuric acid ` to
600 ml of water in a 1000-ml breaker. Open the stopcocks
to
the
collecting tubes. Add the water-acid mixture to the
apparatus
with
a
small excess collects in the reservoir.
3.
Close the switch and allow the electrolysis to proceed until one of the
tubes is two-thirds full of gas. Record both
quantitative
and
qualitative observations.
Observations:
4.
Collect a test tube full of gas from the tube with the larger
volume by holding an inverted test tube over the stopcock,
opening
the stopcock, and then closing the stopcock. Keep the test tube inverted.
Hold a blazing splint to the mouth of the
test tube. This is a test for
which gas? Collect the second
gas in the same manner but thrust a
glowing splint into this test tube. For which gas is this a test?
Observations:
5. If it is marked on the rectifier or battery, follown the connecting
wires to identify which is the positive electrode and
which
is
negative. Which gas collected at each
electrode?
Reverse the electrical connections and close the switch.
is collecting at each electrode here.
6.
Write a word equation to describe what is happening.
Which
the
gas
7.
Write a formula equation (not balanced) to describe what is
8.
Use the molcular models available to represent the above
Construct any additional models needed to conform to
the
Conservation of Atoms.
happening.
equation.
Law
of
Draw pictures of the models or use circle models.
9.
Write the complete balanced equation.
Questions:
1.
How does the volume of the hydrogen produced compare to the
the oxygen produced?
2.
At which electrode does the hydrogen collect?
3.
Does the oxygen always collect at a particular electrode?
volume
of
4.
What relationship do you find between the relative number of molecules
represented by the balanced formula equation and the
experimentally
measured volumes of the gases produced?
5.
The relative volumes of gases collected in the tubes are not exactly
two to one, although exactly twice the volume of one
gas
is
set
free
compared to the other gas. How do you account
for the difference?
(Hint: Why is it possible to collect the gases in this manner
(displacement of water)?)
Now it is time for a little practice work.
As with the skill of formula
writing, the only way in which you become a good equation balancer is
practice. Remember that practice makes perfect:
1. _____BaCl2
+ _____(NH4)2 CO3
-----> _____BaCO3 + ____NH4Cl
2. _____KClO3
-----> _____KCl + _____O2
3. _____Al(OH)3
+ ___NaOH ----> ___Na AlO2
4. _____Fe(OH)3
+ ___H2SO4
+ _____H2O
____> ___Fe2(SO4)3
+ ___H2O
5. _____Na + ___H2O ----> ___NaOH + ___H2
6. _____Mg + ____N2
----> _____Mg3N2
7. _____Mg + ____O2
-----> ___MgO
8. _____AgNO3
+ ___CuCl2
9. _____C2H6O + ___O2
10. _____FeCl2
11. _____HNO3
-----> ___AgCl + ___Cu(NO3)2
---> ___CO2
+ ___Na3PO4
+ ___H2O
----> ___Fe3(PO4)2
+ ___Sn ----> ___SnO2
12. ____NH4OH + ___AgNO3
+ ___NO2
----> ___NH4NO3
+ ____NaCl
+ ___H2O
+ ___Ag2O + ___H2O
13. ____NaOH + ___NH4Cl -----> ___NaCl + ___NH3
14. ____NaOH + ___AgNO3
15. ____CuSO4
----> ___NaNO3
5H2O ----> ____CuSO4
+ ___H2
+ ___Ag2O + ___H2O
+ ___H2O
Formula Writing and Equations
Now that you are an expert in balancing equations and should have
already been an expert in formula writing, you are going to be a double
expert.
You will get some practice in writing and balancing equations.
Look at the following example:
Borium chloride + ammonium carbonate -----> borium carbonate +
ammonium chloride
You must write each formula first. Much of what I will write down here
should be done in your head. To write down the formulas, think of the ion
symbols and their combining numbers.
Ba2+ Cl1- + NH41+
CO32-
Ba Cl2 + (NH4)2
CO3
----> Ba2+
CO32- + NH41+
Cl1-
---> BaCO3 + NH4Cl
After you write the formulas, you may not change the subscripts when
balancing the equation.
BaCl2
+ (NH4)2
CO3
Balance the equation next.
---> BaCO3
+ 2 NH4Cl
If you encounter an equation which apparently cannot be balanced, check
your formula writing for mistakes first.
In the following exercises, write and balance the equations for the
first ten. Do not forget those elements that are diatomic. In the second
ten, write the names of the elements and compounds represented by the
formulas.
Exercises:
1.
Sodium chloride + silver nitrate ---> silver chloride + sodium
nitrate
2.
Zinc + copper (II) sulfate ---> zinc sulfate + copper
3.
Silver nitrate + copper ---> copper (II) nitrate + silver
4.
Barium chloride + sodium sulfate ---> sodium chloride + barium
sulfate
5.
Zinc chloride + ammonium sulfide ---> zinc sulfide + ammonium
chloride
6.
Nickel + hydrochloric acid ---> nickel (II) chloride + hydrogen
7.
Calcium carbonate ---> calcium oxide + carbon dioxide
8.
Iron (III) chloride + sodium carbonate ---> sodium chloride +
iron (III) carbonate
9.
Calcium carbonate + hydrochloric acid ---> calcium chloride +
water + carbon dioxide
10.
Copper + sulfuric ace ---> copper (II) sulfate + water +
sulfur dioxide
11.
MgBr2
12.
Ca(OH)2
13.
Nc + 2HCl ---> NiCl2
14.
Zn + Pb(C2H3O2)2
15.
NH4NO2
+ Cl2
---> MgCl2
+ CO2
---> N2
+ Br2
---> CaCO3
+ H2O
+ H2
---> Pb + Zn(C2H3O2)2
+ 2H2O
16.
2AgNO3
+ Cu ---> Cu(NO3)2 + 2 Ag
17.
Fe S + 2 H Cl ---> H2 S + Fe Cl2
18.
CaCO3
19.
2 Al(OH)3 + 3 H2SO4
20.
Ca(OH)2
+ 2 HCl ---> CaCl2 + H2O + CO2
+ (NH4)2
---> Al2
SO4
(SO4)3
---> CaSO4
+ 6 H2O
+ 2NH3
+ 2H2O
Types of Chemical Reactions
If sodium chloride and silver nitrate solutions are combined, what are
the products?
Now that you are doubly expert on writing formulas and balancing
equations, you are going to be triply expert. This new skill does not refer
to being a hamburger stand expert but being able to write and balance
equations as in the question above.
Many, but not all, chemical reactions can be divided into categories.
These categories make it possible for us to write equations.
In some
reactions, however, the only way we know what will happen is to do the
reaction and analyze the products.
Here are five types of chemical
reactions and examples of each.
1.
Combustion Reactions
When a hydrocarbon (hydrogen, carbon compound) is burned,
the products are carbon dioxide and water.
Example:
CH4 + 2O2 ---> CO2 + 2H2O
Some other organic compounds such as ethanol, C2H5OH,
also form carbon dioxide upon combustion.
Example:
C2H5OH + 3O2 ---> 2CO2 + 3H2O
2.
Combination Reactions
When two or more substances (elements or compounds)
combine to
form one substance, the reaction is called a combination reaction.
Example:
2H2 + O2
---> H2O
S
+ O2
---> SO2
2Na + Cl2
---> 2NaCl
When it rains over an area heavily polluted with sulfur dioxide, a
combination reaction often happens.
Example:
H2O + SO2
---> H2SO3
3.
Decomposition Reactions:
When one substance breaks up to form two or more
reaction is called decomposition.
substances, the
The electrolysis of water gives a good example.
Example:
2H2O ---> 2H2
+ O2
When baking soda (sodium bicarbonate) is heated, it
breaks up
into three compounds: sodium carbonate (which must
be
neutralized
or
the food will taste bitter), water, and carbon dioxide (which makes a cake
rise).
Example:
2Na HCO3
---> Na2 CO3
+ H2O + CO2
When Epson Salts, a hydrate of magnesium sulfate, are
the water comes off.
Example:
MgSO4
4.
7H2O ---> MgSO4
heated,
+ 7H2O
Single Replacement Reaction
When one element replaces another element in a compound,
the
reaction is called a single replacement, or displacement
reaction.
Example:
Fe + CuSO4 ---> FeSO4
+ Cu
2K + 2H2O ---> 2KOH + H2
Cl2 + 2NaBr ---> 2NaCl + Br2
In the first of the two experiments accompanying Types of Chemical
Reactions, you will observe and write many single replacement reactions.
Just because you can write a reactoin on paper does not mean that it will
happen. Some elements are not capable of replacing others as you will see.
Perform the experiment Single Replacement Reactions before proceeding
to Metathesis Reactions.
Experiment:
Single Replacement Reactions
Some elements are so reactive that they are never found in the free or
uncombined state.
Other elements are nearly inert.
The reactivity of an
element is related to its tendency to lose or gain elections.
A list or
series can be made which indicates which elements are capable of replacing
other elements from their compounds. In the general example: A + BC ---> B
+ AC.
A is the more active element since it replaces element B from the
compound BC.
In a second general reaction X + YZ ---> Y + XZ whre --->
means the reaction does not go, element Y is more active than element X so
element X cannot replace it.
Procedure:
1.
In some cases you will notice no immediate reaction.
test tube and observe it again after ten minutes.
Set
Evidence of a reaction will be either evolution of a gas or
appearance of a deposit on the surface of the metal strip.
2.
Label six clean test tubes numbers 1 through 6 and place them
rack. Add the following to the test tubes:
Tube 1 =
M silver
Tube 2 =
nitrate
Tube 3 =
Tube 4 =
Tube 5 =
acid
Tube 6 =
acid
aside
the
the
in
a
Copper strip, about 1 X 2 cm, and about 4 ml of 0.1
nitrate.
Lead strip and about 4 ml of 0.1 M copper (II)
Zinc strip and about 4 ml of 0.1 M lead (II) nitrate
Zinc strip and about 4 ml of 0.1 M magnesium sulfate
Copper strip and about 4 ml of dilute (3M) sulfuric
Zinc strip and about 4 ml of dilute (3M) sulfuric
Record your observations on the data sheet.
_________________________________________________________________Evidence of
Reaction
Equation (to be completed)
_________________________________________________________________Describe
any evidence of reaction
Write "No reaction," if no
if no reaction was observed, write
reaction was observed
"None"
_________________________________________________________________1.
Cu + Ag NO3 --->
__________________________________________________________________
2.
Pb + Cu(NO3)2
--->
_________________________________________________________________3.
Zn + Pb(NO3)2
--->
_________________________________________________________________4.
Zn + MgSO4 ---->
_________________________________________________________________5.
Cu + A2SO4 ---->
_________________________________________________________________6.
Zn + H2SO4 --->
_________________________________________________________________
Questions:
1.
Complete the following table by writing the symbols of
elements whose reactivities are being compared in each test:
the
two
Tube Number
1
2
3
4
5
6
_________________________________________________________________
Greater Activity
_________________________________________________________________Lesser
Activity
_________________________________________________________________
2.
Arrange Pb, Mg, and Zn in order of their activities, listing the
active first.
(1)_______
(2)_______
(3)_______
most
3.
Arrange Cu, Ag, and Zn in order of their activities, listing the
active first.
most
(1)_______
(2)_______
(3)_______
4.
Arrange Mg, H and Ag in order of their activities, listing the
active first.
most
(1)_______
(2)_______
(3)_______
5.
Arrange all five of the metals (excluding hydrogen) in an
series, listing the most active first.
activity
(1)_______
(2)_______
(3)_______
(4)_______
(5)_______
6.
On the basis of the reactions observed in the six test tubes,
explain why the position of hydrogen cannot be fixed exactly with
respect to all of the other elements listed in the activity
series
in
Question 5.
7.
What additional reaction(s) would be needed to establish the exact
position of hydrogen in the activity series of the elements
listed
Question 5?
8.
On the basis of the evidence developed in this experiment:
(a) Would silver react with dilute sulfuric acid? Why or why
not?
(b) Would magnesiuim react with dilute sulfuric acid?
why not?
5.
in
Why or
Metathesis Reactions
When two positive ions exchange negative ions (or partners),
metathesis or double replacement reaction has
happened.
a
As above, it is often possible to write reactions that are
not
observed. Most metathesis reactions are done in
aqueous
solution
(the
solid dissolved in water). If the possible products of the reaction are
also soluble in water,
there iis no driving force to make the reaction
happen. If
one of the possible products is insoluble or only slightly
soluble in water, the solid product is observed.
+
Ax
(mg)
Bz
(aq)
If the possible
products AZ and
AX
+
X- + B + Z-.
BZ in water are really
BX are soluble
in water, the
ions have no
driving force to
combine.
(Note: the subscripts (s) for solid, (l) for liquid, (g) for
and (aq) for aqueous are often found in reactions to
indicate
states.)
A
+
gas,
physical
If AZ or BX is
either insoluble
or slightly
soluble in water,
it will form a
solid called a
precipitate.
The B+ and X- ions will
still be in solution.
To write these reactions you must have a solubility
One follows this discussion.
chart.
Examples:
AgNO3 + H Cl ----> AgCl
+ HNO3
Ni(NO3)2 + 2NaOH -----> Ni(OH)2
+ 2NaNO3
The arrow going down indicates a precipitate.
If a gaseous product is formed during a reaction such as
this, the reaction can also be written.
Example:
Mg CO3
+ 2HCl ---> MgCl2
+ H2O + CO2
and CO2.
You would have predicted that the product was MgCl2 +
H2CO3, H2CO3, carbonic acid, will decompose to form H2O
Other examples are in the second experiment.
A Neutrolization reaction is a type of metathesis
reaction. A salt and water are formed as well as heat
energy.
Example:
HCL + NaOH ---> NaCl + H2O
2HCL + Mg(OH)2 ---> MgCl2 + 2H2O
Solubility Rules
1.
Compounds of the Group IA metals and NH4+ are soluble.
2.
Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl,
Hg2
Cl2, and PbCl2.
3.
(PbCl2 is soluble in hot water.)
Compounds containing the NO3- group are all soluble.
4.
Compound containing SO4 2- are soluble, except BaSO4, SrSO4 ,
CaSO4 and Ag2SO4 are only slightly soluble.
and PbSO4.
5.
Binary compounds of S2- are insoluble except fo rthe sulfide
metals of Group IA, IIA, and NH4+.
of
6.
Compounds containing the OH- group are insoluble except those
Group IA and Sr, Ba, and Ra.
7.
Compounds of metals containing CO32-, PO43-, and SO32- are
except those of Group IA and NH4+.
the
of
insoluble
Insoluble = less than lg of salt 100gH2O dissolves
Soluble = dissolves in water to a greater extent than above.
Use this chart in the next experiment on Double Replacement Reactions.
This experiment will give you plenty of practice on writing reactions. It
is not necessary to memorize the above chart for work in this chapter. The
lab portion of the experiment will take one period, but the questions and
problems will take longer.
Experiment:
Double Replacement Reactions
A general metathesis or double replacement reaction follows:
AB + CO -----> AD + CD
In solution we really have ions of A+, B-, C+, and D- as is seen in
these diagrams.
AB(aq)
+
CD
------->
AD + CB
no reaction
1.
If the four ions can exist mutually in solution, they do not recombine
to form AD + CB. An example of this case is
NaCl + KNO3
-----> KCl + NaNO3
According to the solubility rules the products are both soluble
so
no
evidence of a reaction is present.
NaCl + KNO3
2.
This would
best be written as
----->
A second situation involves the formation of a solid or precipitate.
NaCl + AgNO3
-----> NaNO3 + AgCl
According to the solubility rules NaNO3 is soluble and will
stay
solution. AgCl, however, is insoluble and will form
a precipitate.
is indicated by a downward arrow
.
NaCl + AgNO3
----> NaNO3
in
This
+ AgCl
3.
A third situation involves the formation of a gaseous product.
An
unstable product is formed which will decompse to form water and a gas.
Na2CO3
+ 2HCl ---> 2NaCl + H2CO3
unstable
H2CO3 will decompose to form H2O and CO2.
correctly written as:
Na2CO3
The entire reaction
is
+ 2 H Cl ----> 2 NaCl + H2O + CO2
Other examples of metathesis reaction products that decompose
are
H2SO3 (H2O and SO2 ) and NH4OH(H2O and NH3 ). The upward
arrow indicates
the formation of a gas.
4.
The fourth type involves the formation of heat in a
reaction.
neutralization
NaOH + HCl ------> NaCl + H2O
Water is formed in the reaction so its ions, H+ and OH-, are no longer
free in the solution. The formation of any slightly
ionized
compound
(water, H2O; acetic acid, HC2H3O2; oxalic acid, H2C2O4; and phosphoric acid,
H3PO4) causes the reaction
to occur and liberate heat.
Double Replacement Reactions will occur if at least one
following classes of substances is formed by the reaction:
1.
2.
3.
Procedure:
A precipitate
A gas
A slightly ionized compound
of
the
Measure out 3 ml of distilled water into a test tube.
Observe this
level. This is the approximate level to which each solution should be added
to obtain 3 ml.
It is not necessary to measure each volume accurately.
Record your observations at the time of mixing.
If there is no visible
reaction, feel the test tube to determine if heat has been evolved.
Complete and balance the equation using
for gases and
for precipitates
where appropriate.
If there is no evidence of a reaction, write "No
reaction".
Follow the following instructions and be sure to use the proper
concentrations of solutions. In 1, for example, 3 ml 0.1M sodium chloride
and 3 ml 0.1M potassium nitrate solutions are mixed together in a test tube.
1.
Mix 0.1M sodium chloride and 0.1M potassium nitrate solutions.
2.
Mix 0.1M sodium chloride and 0.1M silver nitrate solutions.
3.
Mix 0.1M sodium carbonate and conc. (12M) hydrochloric acid solutions.
4.
Mix 10 percent sodium hydroxide and dil. (6M) hydrochloric
acid
solutions.
5.
Mix 0.1M barium chloride and dil. (3M) sulfuric acid
solutions.
6.
Mix dil. (6M) ammonium hydroxide an dil (3M) sulfuric acid
solutions.
7.
Mix 0.1M copper (II) sulfate and 0.1M zinc nitrate solutions.
8.
Mix 0.1M sodium carbonate and 0.1M calcium chloride solutions.
9.
Mix 0.1M copper (II) sulfate and 0.1M ammonium chloride solutions.
10. Mix 10 percent sodium hydroxide and dil. (6M) nitric acid
solutions.
11. Mix 0.1M iron (III) chloride and dil. (6M) ammonium hydroxide
solutions.
12. Add lg of solid sodium sulfite to 3 ml of water and shake to dissolve.
Add about 1 ml of conc. (12M) hydrochloric acid
solution, dropwise, using
a medicine dropper.
Data for Double Replacement Reactions
_________________________________________________________________ Evidence
of Reaction
Equation
-----------------------------------------------------------------1.
NaCl + KNO3
------>
-----------------------------------------------------------------2.
NaCl + AgNO3 ------>
-----------------------------------------------------------------3.
Na2CO3 + HCl ------>
-----------------------------------------------------------------4.
NaOH + HCl ------>
-----------------------------------------------------------------5.
BaCl2 + H2SO4 ------>
-----------------------------------------------------------------6.
NH4OH + H2SO4 ------>
-----------------------------------------------------------------7.
CuSO4 + Zn(NO3)2 ----->
-----------------------------------------------------------------8.
Na2CO3 + CuCl2 ------>
-----------------------------------------------------------------9.
CuSO4 + NH4Cl ------>
-----------------------------------------------------------------10.
NaOH + HNO3 ------>
-----------------------------------------------------------------11.
FeCl3 + NH4OH ------>
-----------------------------------------------------------------12.
Na2SO3 + HCl ------>
----------------------------------------------------------------Questions and Problems
1.
The formation of three classes of substances caused double
replacement reactions to occur in this experiment. Name them.
a.
b.
c.
2.
Write the equation for the decomposition of carboric acid.
3.
Using the three criteria for double replacement reactions,
together
with the solubility chart of page 59, predict whether
a double replacement
reaction will occur in each example
below.
If reactin will occur,
complete and balance the equation, using arrows to indicate gases and
precipitates. If you believe no reaction will occur, write "no reaction" at
the right side of the equation.
a.
b.
c.
d.
e.
f.
g.
h.
i.
NH4OH + HCl ---->
Na2S + CuSO4 ---->
NaC2H3O2 + HCl ---->
NaOH + NH4Cl ---->
(NH4)2 SO4 + KNO3 ---->
K2CrO4 + Pb(NO3)2 ---->
K2CO3 + HNO3 ---->
BiCl3 + KOH ---->
NaC2H3O2 + CuSO4 ---->
These two experiments should have given you plenty of practice in
writing single and double replacement reactions. Can you pick out types of
reactins when you see them?
Go back to the twenty exercises starting on
page 50. Classify each as to type of reaction. Most are single and double
replacements but there are a couple of others also so review all five types.
For 1-5, balance the following equations:
1.
KNO3 ----> KNO2 + O2
2.
Fe + H2O ----> Fe3O4 + H2
3.
CO2 + NaOH ---> NaHCO3
4.
Zn + H2SO4 ---> ZnSO4 + H2
5.
Fe Cl2 + Na3PO4 ---> Fe3(PO4)2 + Na Cl
6.
Write a balanced equation for the formation of magnesium
nitride
from its elements.
7.
Write a balanced equation for the decomposition of mercury
(II) oxide
to form free elements.
8.
Write a balanced equation for the combustion of propane
(C3H8).
For 9-18, write a balanced equation and indicate the reaction type.
Use
for gases and
for precipitates where appropriate.
If no reaction is
predicted, write the formulas for the reactants only.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Magnesium bromide + chlorine ---> magnesium chloride + bromine
Sodium + water ---> sodium hydroxide + hydrogen gas
Potassium nitrate ---> potassium nitrite + oxygen gas
Zinc + hydrochloric acid --->
Calcium oxide + hydrochloric acid --->
Aluminum nitrate + ammonium hydroxide --->
Potassium chlorate ---> potassium chloride + oxygen gas
Zinc chloride + ammonium sulfide --->
Mercury (II) sulfate + ammonium sulfide --->
Silver nitrate + potassium chloride --->
CHAPTER 2 -- STOICHIOMETRY
SECTION 2 --- MOLES AND REACTIONS
By now you should be perfect on writing chemical equations since you
have had plenty of practice. Are you wondering how all of this relates to
what you studied in Chapter 1? That is precisely what we will learn in this
section. This is a very useful area of quantitative chemistry. As moles
provided the key for converting between grams and molecules, moles will be
the key in working with relationships within balanced chemical equations.
After spending all of that time on the previous section, you may feel a
little rusty on moles and need to review.
Objectives:
1.
Given a formula equation, translate it into an English
both moles and molecules.
2.
sentence
Given a chemical equation, state the chemical equivalence
using
relations.
3.
Given a chemical equation and the moles of one substance,
the moles of another substance within that equation.
calculate
4.
Given a chemical equation and the moles of one substance,
the grams of another substance within that equation.
calculate
5.
Given a chemical equation and the grams of one substance,
the grams of another substance within that equation.
calculate
The first question in Chapter 1 was about the meaning of the formula
H2O. The answer at the beginnig of the chapter was that
H2O represented one molecule of water which consisted of two atoms of
hydrogen and one atom of oxygen. At the end of the chapter we said that H 2O
not only represented the above but also one mole of water molecule which
consisted of two moles of hydrogen atoms and one mole of oxygen atoms.
How would you read the balanced equation 2H2 + O2 ---> 2H2O?
This equation could be read in two equally correct ways. It could be two
molecules of hydrogen gas mixed with one molecule of oxygen gas reacts to
give two molecules of water. Pictorially, this is represented by:
2 molecules H2 + 1 molecule O2
---> 2 molecules
H2O.
The equation
could also be read: two moles of hydrogen gas added to one mole of oxygen
gas yields two moles of water. Pictorially, this would take quite a long
time to draw since each one mole, as you remember, is 6.02 x 1023 molecules.
Even when we want to talk about moles, our pictorial representations will
have to be of molecules. You will find it helpful to make models for these
reactions. There will be models provided for your individual use.
Look at the equation 2H2 + O2 ---> 2H2O again. This balanced equation
represents the mole ratio by which the reactants combine to form the
product. What would happen if you combined three moles of hydrogen with one
mole of oxygen? Look at the following models:
3 H2
+
1 O2
---> ?
water (H2O)
Water is still H2O. In this case you should have 2H2O formed with a hydrogen
mole (or molecule in the picture) left over. Hydrogen is in excess. The
reaction will still occur but the ratio of reactants to products will still
be 2:1:2.
What would be produced if you combined four moles of hydrogen with one
mole of oxygen?
+
---->
?
4 H2
+
102
---->
?
This is really the same question as before. Hydrogen is again in excess.
Two moles of water should be formed with two moles of hydrogen left over.
What would be produced if four moles of hydrogen were combines with two
moles of oxygen?
4 H2
+
--->
+
--->
?
H2O
If you used the models, you could see that there would be enough atoms
present to form four water molecules. Notice that 4H2 + 202 ---> 4H2O is
really a multiple of the original equation.
What would be produced if one mole of hydrogen were combined with a
half mole of oxygen?
Did you answer that one
mole of water formed or that it was not possible since not enough was
present? You were correct if the first was your answer. One to one-half to
one is still the same combining ratio as two to one to two. Remember that
we are working with moles in this problem. Although we do not talk about
half of a molecule, there is no reason not to talk about half a mole or 3.01
x 1023 molecules reacting.
What would be produced if three moles of hydrogen were combined with
excess oxygen?
Did you say three moles of water
were produced or did you say two moles of water with one mole of hydrogen
left over? Look at the following picture:
3 H2
+
xs O2
---> ?
H2O
(xs means excess)
There is no reason that all three moles of hydrogen in our problem should
not react. Three moles of water would be produced. Notice that as long as
one of the reacting reagents is in excess it is only necessary to specify
the amount of the other reactant.
If five moles of water were produced, how many moles of oxygen must
have racted to form the water? This is the reverse of the problem above.
Since the balance equation is still 2H2 + O2 ---> 2H2O, the same reacting
ratios hold. You may be able to work this in your head and obtain 2.5 moles
of oxygen for the answer.
Most problems in the lab do not work out as nicely. The dimensional
analysis method of working problems comes in handy here. According to the
equation,m it takes two moles of hydrogen and one mole of oxygen to form two
moles of water. Although two moles of hydrogen are not equal to one mole of
oxygen or two moles of water, in this balanced equation they can be said to
be "chemically equivalent". This could be indicated as 2 mole H2
1 mole
O2
2 mole H2O with the symbol
indicating chemical equivalence. These
can then be used as unit factors as
2 moles H2
1 mole O2
or
1 mole O2 or 1 mole O2 and so on.
2 moles H2
2 mole H2O
In a
problem, the amount of oxygen needed to form five moles of water could be
set up like this:
5 mole H2O x 1 mole O2 = 2.5 mole O2) .
2 mole H2O
The "mole H2O" units divide
out and leave "mole O2" for the units. Do not forget to include all of the
necessary units.
Try one more problem before we look at the exercises. Use dimensional
analysis to help you.
Example:
Oxygen gas is prepared by heating KClO3 in the presence of a catalyst.
KCl is the other product in the reaction.
If 0.0244 moles of KClO3
decompose completely, how many moles of oxygen would form?
This problem has a different wrinkle. You must first write and balance
the equation. The only reactant is KClO3 and the two products are KCl and
O2. Were you able to deduce that from the statement? Did you remember that
oxygen was diatonic?
Learning to read what the question asks is an
important part of working problems.
2 K Cl O3
---> 2 K Cl + 3O2
Now that we have the equation and have balanced it, we must look at
what we have and want.
2 K Cl O3
0.0244 moles
------> 2 K Cl + 3 O2
? moles
We hav 0.0244 moles of potassium chlorate and want the number of moles
of oxygen.
It is often helpful to underline the known and unknown.
The
equation says that two moles of K Cl O3 will form three moles of O2. As was
said before, it is possible to have more or fewer than two moles of K Cl O 3.
The coefficients do not indicate the absolute numbers that must be present.
0.0244 moles K Cl O3
x 3 mole O2
2 mole K Cl O3
+ 0.0366 mole O2 formed
0.0366 mole O2 would be formed.
To check your understanding, work the following exercises.
on your own paper and show all of your work.
Work these
Exercises:
1.
If 0.0 3195 mole of copper reacts with sulfur to form copper (II)
sulfide, how many moles of sulfur would be needed to
completely
react
with the copper? How many moles of copper
(II) sulfide would be formed?
2.
If 0.0178 moles of magnesium are burned in air to form magnesium
oxide, how many moles of oxygen would be needed?
How
many
moles
magnesium oxide would be formed?
of
3.
In the reaction 2 Fe (s) + 2 H2O(1) + O2(g) ---> 2 Fe(OH)2(s)
how
many moles of Fe(OH)2 will form when 0.25 mole O2 reacts
with iron?
*4. According to the equation in question 3, how many moles of
would be formed if 2.5 mole Fe, 5.2 mole H2O, and 1.5
mole
combined? Why?
Fe(OH)2
O2
were
*5. In the equation A l (OH)3 + H2SO4 ---> A l2(SO4)3 + H2O, how many moles
of sulfuric acid (H2SO4) are needed to ract with
0.250 mole A l(OH)3?
The next pages contain an experiment on mole ratios.
Experiment:
Mole Ratios and Chemical Reactions
In this experiment you will investigate the relationships between moles
of reacting chemical species.
1.
Weigh a clean, dry 250 ml Erlenmeyer flask to the
nearest 0.0lg. Keep the flask on the balance pan.
________
2.
Add between 15.0 and 16.0g CuSO4 5H2O to the flask.
(Adjust the balance so that it weighs between 15.0g
and 16.0g more than the flask. Add the salt until
the pan begins to drop. Stop adding the salt and
adjust the balance to the nearest 0.0lg)
Weight of CuSO4 5H20 and flask
________
3.
Add 200 ml of distilled water to the flask to
dissolve the salt. Swirl the flask to help in
dissolving. It may be necessary to gently heat
the flask.
4.
Weigh out a 3.0g tuft of steel wool to the nearest
0.0lg. The steel wool may be placed directly on
the balance pan but be sure to blow off any small
loose steel particles before recording the final
weight.
Weight of steel wool (Fe)
________
moles of Fe
________
5.
Add the steel wool to the flask. Swirl the contents
of the flask intermittently for 5-10 minutes or until
the steel wool has disappeared.
6.
Let the solid copper particles which have formed settle
to the bottom. Carefully decant, or pour off the liquid
as is shown in the picture to the right. Wash the residue
with 10 ml of distilled
water, and decant again.
Repeat the washing procedure
at least three times.
Prepare a filtering setup by folding
a piece of filter paper and fitting
it into a funnel. Moisten the paper
with distilled water.
Add 10 ml of distilled
water to the flask
containing the washed
residue. Swirl the
contents of the flask
and filter the
contents. The tip of
the
funnel should
touch the collection
beaker so that a
steady filtrate stream
can run down the side.
Wash any residue in the flask into
the filter paper. Rotate the flask
to remove any residue from it.
Carefully remove the filter paper and
contents. Place it on a watch glass
and place in a drying oven at no more
than 100 C over night. If any of the
paper accidently tore off, include this
on the watch glass. Weigh the dried
paper and contents to the nearest 0.0lg.
Weight of filter paper and copper
________
7.
Weigh ten sheets of filter paper. To find the average
weight of one sheet, divide the weight of the ten by
ten.
_________
8.
Weight of copper formed during the reaction
Moles of copper formed
9.
Calculate the ratio of moles of iron (in the steel wool) to
the moles of copper involved in the reaction.
_________
_________
Questions and Problems
1.
On the basis of this experiment, write the equation for the reaction
between iron in steel wool and on aqueous solution of
CuSO4 5H2O.
2.
Was all of the copper ion in CuSO4
copper? Explain your answer.
5H2O converted to metallic
3.
How many atoms of iron and how many atoms of copper were
involved in your experiment?
4.
Though it is predominantly iron, steel wool does contain a
small percentage of other elements. How does this reflect
upon your experimental results?
5.
If you had evaporated the filtrate (liquid) of step 6 to
dryness, what would you expect the composition of the residue
to have been?
Since you have had practice on the mole type problems, look at the
following problem. Try this on your own before you look at the solution.
Example:
How many grams of calcium acetate can be produced from 1.50 moles of
acetic acid (H C2H3O2) when it reacts with limestone according to the
following equation?
CaCO3
+ 2HC2H3O2
----> Ca(C2H3O2)2
+ CO2
+ H2O
You know how to do this problem if the answer requests moles.
out in moles.
1.50 moles HC2H3O2
x 1 mole Ca(C2H3O2)2 = 0.750 mole Ca(C2H3O2)
2 moles HC2H3O2
Work it
The answer, however, should be in grams. In Chapter 1, you learned how
to change moles to grams by using the weight of one mole.
0.750 mole Ca(C2H3O2)2
x 158.17gCa(C2H3O2)2 = 118.63g
1 mole Ca(C2H3O2)2
or
119g Ca(C2H3O2)2
This problem could be set up as a continuous dimensional analysis
problem.
1.50 moles HC2H3O2 x 1 mole Ca(C2H3O2)2 x 158.17gCa(C2H3O2)
2 moles H(2H3O2
1 mole Ca(C2H3O2)2
= 119gCa(C2H3O2)2
Example:
How many grams of copper are needed to replace silver in 0.0235 moles
of silver nitrate dissolved in water? One of the products is copper (II)
nitrate.
What answer did you get?
This problem is similar to the last
experiment.
The reactin is of a type that you did earlier.
What is the
name of this type of reaction?
The first step
is to write to the balanced chemical equation for the single replacement
reaction.
Cu
?
+
2 AgNO3
0.0235 moles
---> Cu(NO3)2
+ 2Ag
It is set up just like the previous problem.
0.0235 moles of AgNO3
x 1 mole Cu
2 mole AgNO3
x 63.54gCu
1 mole Cu
= 0.747g Cu
Did you have trouble reading the problem? The first part asks, "How
many grams of copper. . ." so the unknown must be copper. The next part is,
". . .are needed to replace silver in 0.0235 moles of silver nitrate
dissolved in water?"
Reading this correctly requires that you recognize
what happens in a single displacement reaction. The silver must be one of
the products.
Try these problems on your own.
Exercises:
1.
Calculate the number of grams of oxygen required to burn 2.40
of C2H6.
mole
2.
How many grams of K Cl O3, would be produced from the reaction
3.55 mole of Cl2 with KOH, according to the following
of
balanced equation?
3 Cl2
+ 6 KOH ---> 5 K Cl + K Cl O3
+ 3 H 2O
Look at the following problem.
How many grams of silver would a silver nitrate solution? The
equation is
Cu
175g
+ 2 AgNO3
---> Cu(NO3)2
+ 2 Ag
? g
Do you recognize that this problem is the reverse of an earlier
example?
Attempt to solve it on your own and then read the following
discussion. At this point you should recall the information found from the
balanced chemical equation.
The coefficients in front of the formulas
represent molecules or moles.
Either way, they represent the number of
particles that combine in a chemical reaction, not the weights of particles.
Using pictorial representations we have
Cu
2 Ag NO3
---> Cu(NO3)2
+
2Ag
To work a problem of the type above, a good method is to change the
weight of what we know to moles (number of particles). Since the equation
tells us how the particles will numerically combine, we can use this
information.
In a continuous dimensional analysis calculation:
175g of Cu x 1 mole Cu
63.54gCu
x 2 moles Ag x 196.97g Ag = 1085g Ag
1 mole Cu 1 mole Ag = 1090g Ag
or by sig. liq.
This corresponds to 2.75 moles
This is the mole conversion factor
and gives the moles of silver or
5.51 moles
This is the final conversion of moles of
silver to grams of
silver
The word "stoichiometry" was used in this section.
It means that
mathematics of chemistry and entails the type of problems that we have been
doing. These are very useful for real laboratory situations as you will see
in the next experiment.
Before you perform the experiment, look at the
following example.
Example:
In the decomposition of 3.00 grams of potassium chlorate, how many
grams of oxygen will be formed?
How many molecules of oxygen?
Work out the first part yourself. If you are not sure of the equation,
look back in this section since we have already looked at an example using
this reaction. I get 1.17 grams of oxygen for my answer.
The answer to the second part of the question can found in two ways.
You could begin with the 3.00 grams of potassium chlorate or begin with 1.17
grams of oxygen. The two set-ups follow:
3.00g K ClO3 x 1 mole K ClO3
122.55g K ClO3
x 3 mole O2
2 mole K ClO3
x 6.02x1023 molecules
1 mole O2
2.20 x 1022 molecules O2
1.17g O2 x 1 mole O2 x 6.02 x 1023 molecules O2 = 2.20x1022moleculesO2
32.0g O2
1 mole O2
Experiment:
Synthesis of Acetylsalicylic Acid
Purpose: To use stoichiometry problems in an actual laboratory
situation.
Theory:
By knowing the formulas for the reactants and products in
a chemical reaction, it is possible to compute the
theoretical
yield of acetylaslicylic acid or aspirin.
this involves the use of
chemical equations and mole
relations.
The theoretical yield is
what would ideally
be expected for the product. In many reactions a
lesser
yield than expected is obtained. Some of the reasons
involve the purification procedures. Other reasons,
inherent
within the system itself, will be converted next
semester.
A percentage yield may be computed by
Actual Yield in grams
x 100 = % yield
Theoretical Yield in grams
Chemicals needed:
Salicylic acid
Acetic anhydride
Conc. surfuric acid
0.5 M acetic acid (cold)
Caution: Acetic anhydride fumes irritate the eyes and
nasal membranes. Both it and the conc.
sulfuric
acid will cause skin burns. Perform
the experiment under
a hood or in an area with
plenty of ventilation.
Equipment needed:
Test tube
Water bath (400 ml beaker will work)
Ice bath (Ice and water in a beaker)
Bunsen Burner
Suction flask
Vaccum tubing
Means of suction (your teacher will instruct
you on this.)
Procedure: Come to class prepared. Although the experiment is
not
difficult, it may be necessary to wait for equipment. The
early bird gets
the Buchner funnel. As with any experiment, keep
your
goggles
on
the
entire time.
Weigh to the nearest 0.0lg a 3.01g or 3.0g sample of salicylic
acid.
Place the solid in your largest test tube. Working
under
the
hood,
carefully add 4.5 ml of acetic anhydride,
followed by 15 drops of conc
H2SO4.
Place the test tube in a water bath maintained at a
temperature of
60 C over a low Bunsen burner flame. The
temperature is maintained at
60 C + 5 by alternately placing
and removing the burner as needed. Heat
the system, with
intermittent shaking, for 20 minutes.
After the heating cycle has ended, cool the contents of the test tube
in an ice bath.
Add 5 ml of distilled water to the tube and stir the
mixture for 5 minutes while it is kept cold. Be careful not to punch a hole
in the bottom of the test tube while stirring. Cut a piece of filter paper
to fit the bottom of the Buchner funnel and filter the crystals with the
funnel and a suction flask.
Buchner funnel
to vacuum pump or
asirator
filtratim flask
Wash the residue with three 5-ml portions of ice cold 0.5
M
acetic acid. Wash with one portion, filter; wash with a
second, filter;
wash a third time, and filter the third portion.
Remove the filter paper
and crystals. Place on a
watch glass and allow to dry overnight in the
drying oven.
Fill out the data chart, calculate the theoretical yield
from
the equation, and calculate the percentage yield of the synthesis.
Acetic anhydride
H3CCO- O - OCCH3 + C6H4
H SO4
COOH OH -------> C6H4COOHOOCCH3 + H3 C-COOH
Added in xs
Data: 1. Weight of salicylic acid
=
g
2.
Weight of dried aspirin and filter paper =
g
3.
Weight of filter paper
=
g
4.
Weight of aspirin
=
g
5.
Theoretical yield (from equation)
show calculations:
=
g
6.
Percentage yield of aspirin
=
g
To
finish up this section, you need to try some exercises.
1.
How many grams of diphosphorus dioxide (P4O10) must be reacted
water to produce 4.00g of phosphoric acid (H3PO4)?
P4 O10
+ H2O ----> H3PO4
with
2.
How many moles of nitrogen dioxide may theoretically be produced
when
copper reacts completely with 21.0 grams of
concentrated nitric acid?
Cu + HNO3 -----> NO2
+ CuO + H2O
3.
How many moles of barium hydroxide can be neutralized by the reaction
with 145 grams of sulfuric acid?
Ba(OH)2 + H2SO4 ---> BaSO4 + H2O
4.
How many grams of zinc chloride can be prepared by reacting
zinc withexcess hydrochloric acid?
10.0g
5.
If excess sulfuric acid reacts with 30.0g of sodium chloride,
many grams of hydrogen chloride are produced?
6.
Calculate the number of grams of lead (II) chloride produced by
reacting 0.400 mole of chloride ions with excess lead (II)
ions.
2+
Pb
+ 2Cl- ---> Pb Cl2
The following is a balanced equation for the combustion of
of
how
the
organic substance butadiene in oxygen:
2 C4H6(g)
+ 11 O2(g)
----> 8 CO2(g)
+
6 H2O(g)
7.
What weight of CO2 will be produced by the combustion of 40.0
of C4H6?
grams
8.
How many molecules of CO2
40.0 grams of C4H6?
will be produced by the combustion
of
*9. How many atoms of hydrogen will be produced by the combustion
40.0 grams of C4H6?
of
10. How many moles of O2
H2O by this reaction?
must be consumed to produce 54.0 grams
of
11. How many molecules of O2 must be consumed to produce 54.0
H2O by this reaction?
grams
12.
Write the following equation as an English sentence in two
2H2O + 2 NaCl ----> 2H2 + Cl2 + Cl2 + 2Na + O2
ways.
13.
What is the Law of Conservation of Matter:
What is conserved in the equation in 12?
of
A compound hs the following composition by weight.
C, 57.1%
H, 4.8%
O, 38.1%
14.
Calculate the simplest formula for this compound.
15. If the molecular weight is shown to be 126, what then must be
true molecular formula?
DO NOT WRITE ON THIS SHEET
Activity Series
K Most Active
Ca
Na
Mg
Al
Zn
Fe
Ni
Sn
Pb
H2
the
Cu
Ag
Au
Least Active
Solubility Rules
1.
Compounds of the Group IA metals and NH4+ are soluble.
2.
Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl,
Hg2Cl2, and PbCl2. (PbCl2 is soluble in hot water.)
3.
Compounds containing SO42- are soluble, except BaSO4, SrSO4 ,
CaSO4 and Ag2SO4 are only slightly soluble.
4.
and PbSO4.
Compounds containing NO3- group are all soluble.
5.
Binary compounds of S2- are insoluble except for the sulfides
the metals of Group IA, IIA, and NH4+.
of
6.
Compounds containing the OH- group are insoluble except those
Group IA and Sr, Ba, and Ra.
of
7.
Compounds of metals containing CO32-, PO43-, and SO32- are
except for those of Groups IA and NH +.
insoluble
CHAPTER 2 - STOICHIOMETRY
SECTION 3 - LIMITING REAGENTS
In many of the exercises you were given the weight of only one
substance.
Often you had to assume that you had an excess of any other
needed reagents.
How do you determine which is the limiting reagent (the
one that makes the difference) and which is in excess of you are given the
mass of both? This section will address that problem.
Objectives
1.
Define limiting reagent.
2.
Given the amounts of the reactants in a reaction, determine
limiting reagent.
the
3.
Given the amounts of the reactants in a reaction, determine
of excess reagent left after the reaction is completed.
the amount
You will perform a short experiment to begin this section.
EXPERIMENT:
You will use exactly 10.0 ml of 6 M HCl.
Place two pieces of zinc
(each approximately 2 cm x 1 cm, strip form) in evaporating with the 10.0 ml
of acid. In a second dish place 10 pieces of zinc. Wait until the reaction
has stopped and the questions below.
Questions:
1.
When two strips of zinc are used, which reagent, the acid or the
is in excess? Why?
Test the solutions in each dish with litmus paper.
litmus turns red. In base red litmus turns blue.)
(In acid blue
Does this observation agree with your answer to the first
2.
zinc,
question?
When 10 strips of zinc are used, which reagent is the limiting
reagent? Why?
Again test the solution with litmus and compare to your answer
to 2.
3.
If more acid were used with the two strips, would you expect more
product?
If more acid were used with the ten strips, would you expect more
product?
4.
Write a balanced chemical equation for this reaction.
In the first part of Section 2 in this chapter, we looked at the
formation of water reaction:
2H2
+
O2
=
2H2O
One of the accompanying questions asked how many moles of water could
be formed from four moles of hydrogen and one mole of oxygen.
By using
pictorial representations we can again see what is occuring.
We begin with four water molecules (since I did not feel like
drawing out 4 x 6.02 x 1023
molecules) and one oxygen
molecule.
Each water molecule consists of
two hydrogen atoms and one
oxygen atom.
The two oxygen atoms can "grab"
two hydrogen atoms each. Two
hydrogen molecules remain unreacted.
Four moles of hydrogen and one mole of oxygen can form only two moles of
water with two moles of hydrogen unreacted.
In the above example oxygen is called the limiting reagent. All of the
oxygen is completely consumed during the reaction and the amount of oxygen
originally present will determine how much product is formed. Hydrogens was
present is excess of the amount needed for complete reaction of the oxygen.
Look at the following exercises:
1.
A mother has four children and three lollipops. Would the
or the lollipops be the limiting "reagent" in this case?
children
2. There
are seven lab benches which can have four students apiece in the chemistry
room. There are also twenty four
student desks in the room.
Which of
these is the limiting
factor on possible student enrollment per section
in that
classroom.
3.
How much H2SO4 can be prepared from 5.0 moles of SO2, 1.0 mole
of
O2, and an unlimited quantity of H2O according to the
equation
2SO2 + O2 + H2O = 2H2SO4?
4.
What would be the limiting reagent, according to the equation
2Fe + 2H2O + O2 = 2Fe(OH)2 ,
if 2.5 mole Fe, 5.2 mole H2O, and 1.5 mole O2 were put together
way that the reaction above could take place?
How did you do on the last two exercises?
in
a
Essentially they were the
same as the water example. If you had trouble, reread the material on water
and then ask your teacher for help.
We have used the idea of limiting
reagent in some earlier laboratory experiments.
In the experiment where
hydrated copper (II) sulfate was mixed with steel wool, which reagent, the
steel wool or the copper (II) sulfate, was used in the calculation? Why was
that used instead of the other reagent? Would it have made any difference
in your answer? Which was the limiting reagent in the aspirin experiment?
Look at the following example:
If 32.0 g of NaOH reacts with 75.0 g of H2SO4, which one of the reagents
will be in excess?
There are many ways in which this problem can be solved.
however, requires that a balanced equation be written. Why?
2NaOH
+
H2SO4
= Na2SO4
+
Each way,
2H2O
Once again, what does this equation say?
(Besides - Nothing, an
equation can't talk!)
The equation shows that two moles (or particles or
molecules) of sodium hydroxide react with one mole of sulfuric acid to form
one mole of sodium sulfate and two moles of water. The example is given in
grams and not moles.
32.0 g NaOH x 1 mole NaOH
40.0 g NaOH
75.0 g H2SO4
= 0.800 mole NaOH
x 1 mole H2SO4 = 0.765 mole H2SO4
98.1 g H2SO4
There are fewer moles of H2SO4 so does this mean that NaOH is in excess
with H2SO4 limiting? If you answered yes, you forgot the stoichiometry or
combining mole ratio of the equation. If you answered nothing, you really
are not doing a very good job of reading this. According to the mole ratio,
two moles of sodium hydroxide must have one mole of sulfuric acid.
0.800 mole NaOH x 1 mole H2SO4 = 0.400 mole H2SO4
2 mole NaOH
0.800 mole of NaOH needs only 0.400 mole of H 2SO4 so we have plenty of
H2SO4 present for all of the NaOH to react.
The sodium hydroxide is the
limiting reagent and the acid is present in excess.
If you had chose to work with sulfuric acid first, you could have
deduced that it was present in excess by a similiar mole ratio.
0.765 mole H2SO4
x 2 mole NaOH
1 mole H2SO4
= 1.53 mole NaOH
Here, there is not enough NaOH present to react with all of the sulfuric
acid so the acid must be in excess with the base (NaOH) limiting.
A third method involves using both reagents as limiting to calculate
the amount of product.
32.0 g NaOH x 1 mole NaOH
40.0 g NaOH
75.0 g H2SO4
x 2 mole H2O x 18.0 g H2O = 14.4 g H2O
2 mole NaOH 1 mole H2O
x 1 mole H2SO4 x 2 mole H2O x 18.0 g H2O = 27.6 g H2O
98.1 g H2SO4
1 mole H2SO4 1 mole H2O
In this case the reagent that produced the least amount of product is
limiting and the other reagent is in excess.
Try the following exercises:
1.
When solutions of lead (II) nitrate, Pb(NO3)2, and sodium
NaCl, are mixed, the following reaction occurs:
Pb(NO3)2
+ 2NaCl = PbCl2
chloride,
+ 2NaNO3
How many grams of PbCl2 can be prepared from 1.00 g of
Pb(NO3)2 and 1.00 g of NaCl?
2.
Which is th limiting reagent if 5.00 g og copper metal reacts
with
a
solution
containing
20.0
g
of
AgNO3?
3.
If 20.0 g of KOH reacts with 15.0 g of (NH4)2 SO4, calculate the
following: a) the moles of K2SO4 produced and b) the grams
of
NH3
produced.
4.
It is necessary to prepare the maximum possible amount of
magnesium
acetate by a reaction involving 15.0 g of iron(III)
acetate with either
10.0 g of MgC2O4 or 15.0 g of MgSO4. Both
reactions produce the desired
magnesium acetate and a precipitate.
Which reaction will give the most
Mg(C2H3O2)2
and how many grams will be produced?
5.
A 36.0 g sample of calcium hydroxide, Ca(OH)2, is allowed to react with
a 54.0 g sample of phosphoric acid, H3PO4. How
many grams of calcium
phosphate could be produced? If 45.2 g of
calcium
phosphate
could
be
produced in an actual run of
this reaction, what is the percent yield?
(Refer back to the aspirin
experiment
for
percent
yield.)
These exercises really require that you understand moles and chemical
reactions. They are also meant to be somewhat challenging. The exercises
that follow are review exercises for the entire chapter.
Chapter Review Exercise:
1. At elevated temperatures the compound NF3 decomposes to N2 and
Find the amounts of N2 and F2
formed by the decomposition
exactly 3 mole of NF3, b) 0.268 mole of NF3.
F2 .
of: a)
2.
Direct reaction of P with Cl2 can form PCl5 under
conditions. Find the mass of Cl2 required to form 2.70 mole of
suitable
PCl5
3. Find the mass of AsCl3
with arsenic.
formed by the reaction of 0.133 mole of
Cl2
4.
Carbon
disulfide,
CS2,
is
a
very
flammable
substance
that
reacts with O2 to form carbon dioxide and sulfur dioxide.
Find
the
mass of O2 that is required to react with 9.34 g of CS2.
Find the mass
of carbon dioxide and of sulfur dioxide formed in
this reaction.
5.
A certain oxide of lead is converted by H 2 to lead
water. One mole of the oxide reacts with 8.1 g of H 2 and forms
of lead. Find the formula of the oxide.
6. Quantities of 11.1 g of H2 and 33.3 g of Cl2 are mixed.
mass of hydrogen chloride that forms.
metal and
622 g
Find
the
7.
The reaction of carbon with calcium oxide produces carbon
monoxide and calcium carbide, CaC2.
A total of 2.45 g of CaC2
is
isolated from the reaction of 5.00 g of calcium oxide with
2.50 g of
carbon. Find the percent yield of the CaC2.
8.
The oxide of an unknown element is believed to have the formula
XO2.
Heating
a
25.2
g
sample
of
the
compound
decomposes
it
completely to form 3.2 g of O2.
Find the atomic weight of X,
the
unknown element.
Alternate Problems
1.
An important step in the manufacture of nitric acid is the
reaction of ammonia with O2 to form NO and water. Find the
amount of
O2 required to react completely with 76.0 g of
ammonia, NH 3. Find
the mass of water and of NO produced.
2. The Cl2 formed by the decomposition of 1.30 mole of PCl3 is used
to
convert carbon to CCl4. Find the amount of CCl4 formed.
(2PCl3 = 2P +
3Cl2, C + 2Cl2 = CCl4)
3.
An important reaction that takes place in a blast
during the production of iron is the formation of iron metal
furnance
and
carbon dioxide from Fe2O3 and carbon momoxide. Find the
mass of Fe 2O3
required to form 910 kg of iron. Find the amount
of carbon dioxide
that forms in this process.
4. Incomplete combustion of hydrocarbons is an important cause of
air
pollution because it produces carbon monoxide. Find the
difference
between the mass of O2 required to convert one mole
of octane, C 6H18,
completely to CO and the mass of O2 required
to convert one mole of
octane completely to CO2.
5.
Reaction of tungsten with Cl2 forms WCl6.
Find the mass of the
unreacted starting material when 12.6 g of tungsten is treated
with
13.6 g of Cl2 and the reaction takes place. Find the mass
of WCl 2 that
forms.
6.
When HgO is heated, it decomposes to mercury and O 2.
Careful
technique makes it possible to isolate 62.7 g of mercury from
the
decomposition of 75.8 g of the oxide. Find the percent
yield of the
reaction.
7. Find the mass of sodium chloride and the mass of water needed
prepare 125 g of an 18.0% sodium chloride solution by mass.
to
8. A mixture of 50.0 g of S and 100 g of Cl 2 reacts completely to form S2Cl2
and SCl2 and no other products. Find the mass of S2Cl2 formed.
DO NOT WRITE ON THIS SHEET
Activity Series
K Most Active
Ca
Na
Mg
Al
Zn
Fe
Ni
Sn
Pb
H
Cu
Ag
Au Least Active
Solubility Rules
1.
Compounds of the Group IA metals and NH4+ are soluble.
2.
Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl,
Hg2Cl2,
and
PbCl2.
(PbCl2
is
soluble
in
hot
water.)
23.
Compounds containing SO4 are soluble, except BaSO4, SrSO4
and
PbSO4.
CaSO4
and
Ag2SO4
are
only
slightly
soluble.
4.
Compounds containing NO3- group are all soluble.
5.
Binary compounds of S2- are insoluble except for the sulfides of
the
metals
of
Group
IA,
IIA,
and
NH
.
6.
Compounds containing the OH- group are insoluble except those
of
Group
IA
and
Sr,
Ba,
and
Ra.
7.
Compounds of metals containing CO32-, PO42-, and SO32- are insoluble
except for those of Groups IA and NH4+.
CHAPTER 2
Section 1
Page 42
+
------->
These molecules of hydrogen react with one molecule of nitrogen to form two
molecules of ammonia.
3H2
+ N2
---> 2NH3
Page 44
3Ca(OH)2
+ 2H3PO4
---> Ca3(PO4)2
+ 6H2O
Page 48
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
BaCl2 + (NH4)2CO3 ----> BaCO3 + 2NH4Cl
2KClO3 ---> 2KCl + 3O2
Al(OH)3 + NaOH ---> NaAlO2 + 2H2O
2Fe(OH)3 + 3H2SO4 ---> Fe2(SO4)3 + 6H2O
2Na + 2H2O ---> 2NaOH + H2
3Mg + N2 ---> Mg3N2
2Mg + O2 ---> 2MgO
2AgNO3 + CuCl2 ---> 2 AgCl + Cu(NO3)2
C2H6O + 3 O2 ---> 2CO2 + 3 H2O
3FeCl2 + 2Na3PO4 ---> Fe3(PO4)2 + 6NaCl
4HNO3 + 1Sn ---> 1SnO2 + 4NO2 + 4H2O
2NH4OH + 2AgNO3 ---> 2NH4NO3 + Ag2O + H2O
13.
14.
15.
NaOH + NH4Cl ---> NaCl + NH3 + H2O
2NaOH + 2AgNO3 ---> 2NaNO3 + Ag2O + H2O
CuSO4
5H2O ---> CuSO4 + 5H2O
Page 50
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
NaCl + AgNO3 ---> AgCl + NaNO3
Zn + CuSO4
---> ZnSO4 + Cu
2AgNO3 + Cu ---> Cu(NO3)2 + 2Ag
BaCl2 + Na2SO4 ---> 2 NaCl + BaSO4
ZnCl2 + (NH4)2S ---> ZnS + 2 NH4Cl
Ni + 2HCl ---> NiCl2 + H2
CaCO3 ---> CaO + CO2
2FeCl3 + 3Na2CO3 ---> 6NaCl + Fe2(CO3)3
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Cu + 2H2SO4 ---> CuSO4 + 2H2O + 2SO2
Magnesium bromide + chlorine ---> magnesium chloride + bromine
Calcium hydroxide + carbon dioxide ---> calcium carbonate + water
Nickel + hydrochloric acid ---> nickel (II) chloride + hydrogen
Zinc + lead(II) acetate ---> lead + zinc acetate
Ammonium nitrite ---> nitrogen + water
Silver nitrate + copper ---> copper(II) nitrate + silver
Iron(II) sulfide + hydrochloric acid ---> hydrogen sulfide +
iron (II) chloride
Calcium carbonate + hydrochloric acid ---> calcium chloride +
water + carbon dioxide
Aluminum hydroxide + sulfuric acid ---> aluminum sulfate +
water
Calcium hydroxide + ammonium sulfate ---> calcium sulfate +
ammonia + water
Page 65
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
2KNO3 ---> 2KNO2 + O2
3Fe + 4H2O ---> Fe3O4 + 4H2
CO2 + NaOH ---> NaHCO3
Zn + H2SO4 ---> ZnSO4 + H2
3FeCl2 + 2Na3PO4 ---> 6Fe3(PO4)2 + NaCl
3Mg + N2 ---> Mg3N2
2MgO ---> 2Hg + O2
C3H8 + 5O2 ---> 3CO2 + 4H2O
MgBr2 + Cl2 ---> HCl2 + Br2(e)
2Na + 2H2O ---> 2NaOH + H2
2KNO3 ---> 2KNO2 + O2
Zn + 2HCl --->ZnCl2 + H2
CaO + 2HCl --->
No reaction
Al(NO3)3 + 3NH4OH ---> Al(OH)3 + 3 NH4NO3
15.
16.
17.
18.
2KClO3 ---> 2KCl + 3O2
ZnCl2 + (NH4)2S ---> ZnS + 2NH4Cl
HgSO4 + 2NH4NO3 ---> No reaction
AgNO3 + KCl ---> AgCl + KNO3
-Section 2-
Page 72 - Exercises
1.
Cu + S --->
0.03195 mole?
Cu S
?
0.03195 mole Cu x 1 mole S = 0.03195 mole S
1 mole Cu
0.03195 mole Cu X
2.
2Mg +
0.0178 moles
1 mole CuS = 0.03195 mole CuS
1 mole Cu
O2
---> 2MgO
? moles
? moles
0.0178 moles Mg x 1 mole O2
2 mole Mg
= .00890 mole O2
0.0178 mole Mg x 2 mole MgO = 0.0178 mole MgO
2 mole Mg
3.
2 Fe(s)
+ 2H2O
0.25 mole O2
+ O2(g)
---> 2 Fe(OH)2(s)
0.25 mole
? moles
x 2 mole Fe(OH)2 = 0.50 mole Fe(OH)2
1 mole O2
4.
2 Fe(s)
+
2 H2O(l)
+ O2(g)
--->
2 Fe(OH)2(s)
2.5 mole Fe
5.2 mole
1.5 mole
? moles
2.5 mole Fe would form 2.5 mole Fe(OH)2 if plenty of the other
two were present.
5.2 mole H2O would form 5.2 mole Fe(OH)2 if plenty of the
other two
were present.
1.5 mole O 2 would
form 3.0 mole Fe(OH)2 if plenty of the other two
were
present.
2.5 mole Fe will produce the least amount of product or 2.5 mole
Fe(OH)2.
There would be plenty of H2O and O2 present for
all of the Fe to react
but enough Fe for either of the other
two to be used up completely. The
amount of Fe present is what
limits
this
reaction.
5.
2Al (OH)3 + 3H2SO4 ---> Al2(SO4)3 = 6H2O
0.250 moles
? moles
0.250 mole Al(OH)3
x 3 mole H2SO4 = 0.375 moles H2SO4
2 mole Al(OH)3
Page 77
1.
2C2H6
+ 7O2 ---> 4CO2
2.40 mole
?g
2.40 mole C2H6
2.
X 7 mole O2 x 32.00g O2
2 mole C2H6 1 mole O2
3 Cl2
+ 6 KOH
3.55 mole
3.55 mole Cl2
+ 6H2O
= 269g O2
---> 5 KCl = KClO3 + 3H2O
? g
x 1 mole KClO3 x 122.56g KClO3 = 145g KClO3
3 mole Cl2
1 mole KClO3
Page 83
1.
P4O10
?g
+ 6H2O ---> 4H3PO4
4.00g
4.00g H PO
2.
x 1 mole P4O10 x 283.88g P4O10 = 284g P4O10
4 mole H3PO4
1 mole P4O10
Cu + 2HNO3 ---> 2NO2 + CuO + H2O
21.0g
? g
21.0g HNO3 x 1 mole HNO3 x 2moles NO2 x 46.01gNo2 = 14.2gNO2
68.01gHNO3
2 moles HNO3 1 mole NO2
3.
Ba(OH)2 + H2SO4 ---> BaSO4
?moles
145g
+ 2H2O
145g H2SO4 x 1 mole H2SO4 x 1 mole Ba(OH)2 x 171.35gBa(OH)2 =
98.08g H2SO4
1 mole H2SO4
1 mole Ba(OH)2
253 Ba(OH)2
4.
Zn
+ 2 HCl ---> ZnCl2 + H2
10.0g
?g
10.0g Zn x 1 mole Zn
x 1 mole ZnCl2 x 136.28gZnCl2 =
65.37gZn
1 mole Zn
1 mole ZnCl2
20.8gZnCl2
5.
2NaCl + H2SO4 ---> HCl + Na2SO4
30.0g
?g
30.0gNaCl x 1 mole NaCl x 1 mole HCl x 36.469HCl = 9.36gHCl
58.44g
2 moleNaCl
1 mole HCl
6.
Pb2+
+ 2 Cl0.400 mole
---> PbCl2
? g
0.400 mole Cl- x 1 mole PbCl2 x 278.109PbCl2 = 55.6g PbCl2
2 mole Cl1 mole PbCl2
7.
2 C4H6
(g)
+ 11 O2(g) ---> 8 CO2(g)
+ 6 H2O(9)
40.0g x 1 mole C4H6
x 8
54.099C4H6
2
8.
40.0g C4H6 x 1 mole C4H6
x
1
23
6.02x10 molecules
CO2
=
9.
40.09C4H6 x 1 mole C4H6 x 6 moles
54.09C4H6
2moles C4H6
40.0g
mole CO2 x 44.019CO2 = 130.189CO2
mole
C4H6
1
mole
CO2
mole
C4H6
x
8mole
CO2
x
24
1.78
x
10
molecules
CO2
H2O x 6.02 x 1023molecules H2O
1 mole H2O
x 2 atoms H
=
2.68
x
1024
H
atoms
10. 54.0g H2O x 1 mole H2O x 11 moles O2 x 32.00g O2 = 5.50 molesO2
18.019H2O 6 molles H2O 1 mole O2
11.
Use # 10 to make this simpler
5.50 moles O2 x 6.02 x 1023 molecules O2 = 3.31 x 1024
moleculesO2
12.
Two molecules of water react with two molecules of sodium
chloride to form two molecules of hydrogen gas and one molecule
of
chlorine gas (oops, another mistake. My typist
was not a chemist.)
and
two molecules of sodium metal and one
molecule
of
oxygen.
Two moles of water react with two moles of sodium chloride to form two
moles of hydrogen, one mole of chlorine gas, two moles of sodium metal
atoms,
and
one
mole
of
oxygen.
13. In a chemical reaction, matter can be neither created nor
destoyed.
These are the same number of atoms on both sides of
the equation.
The
mass is the same on both sides of the
equation.
14. Assume 100 grams
57.19 C x
1mole C = 4.75 moles C
12.01gC
4.8g H x 1 mole H
1.01 g H
= 4.75 moles H
2.38
2.38
38.1 g O x 1 mole O = 2.38 mole O
16.00g O
15.
C2H2O
C 4.75 H 4.75 O2.38
C
4.75
2.38
H
C2H2O
adds up to 42
126 = 3
There are 3 -C2H2O units so the true molecular
4.75
O
2.38
42
you already?
here.
formula is C6H6O3. Does this problem look like one
It should. I thought that you needed a little
review
Answers to Exercises
page 87
1.
2.
The lollipops are the limiting reagent
reasons.
7 lab benches x 4 students = 28 students
1 bench
for,
24 desks
hopefully,
obvious
benches
desks
The limit for enrollment would be the number of desks.
3.
2 SO2 + O2 + H2O ---> 2H2SO4
Given:
5.0 moles 1.0 mole x S
? moles
5.0 moles SO2 x 2 moles H2SO4 = 5.0 moles H2SO4
2 moles SO2
1.0 mole O2
the
4.
x 2 moles H2SO4 = 2.0 mole H2SO4
1 mole O2
There is only enough O2 present to form 2.0 mole H2SO4 so O2
limiting
2Fe + 2H2O + O2 ---> 2Fe(OH)2
Given: 2-5 mole 5.2 mole 1.5 mole
must
be
reagent.
In words:
2.5 mole of Fe would produce 2.5 moles Fe(OH)2 and require 2.5
mole
H2O and 1.25 mole of O2 to do so. There is excess of H2O
and O2. Fe is
the
limiting
reagent.
-2
7.87 x 10 mole Cu x 2 mole AgNO3 = 0.157 mole AgNO3
1 mole Cu
Thre are only 0.118 moles of AgNO present, however, so Cu
must
be
present
in
excess
with
AgNO3
limiting.
3.
2KOH + (NH4)2SO4 ---> K2SO4 + 2NH3 + 2H2O
This was probably a difficult equation for you.
This type was
covered in the double replacement type of reaction earlier.
20.0g KOH x 1 mole KOH = 0.356 mole KOH
56.11g KOH
15.0g (NH4)2SO4 x 1 mole (NH4)2SO4 = 0.114 mole (NH4)2SO4
132.14
0.356 mole KOH x 1 mole (NH4)2SO4 = 0.178 mole (NH4)2SO4
2 mole KOH
We do not have enough ammonium sulfate for all the KOH to
ammonium
sulfate
is
a)
0.114 mole (NH4)2SO4 x 1 mole K2SO4 = 0.114 mole (NH4)2SO4
1 mole (NH4)2SO4
b)
react
so
limiting.
0.114 mole (NH4)2SO4 x 2 mole NH3 x 17.03g NH3 = 3.88g NH3
mole (NH4)2SO4 1 mole NH3
If all of MgSO4 is to be used up, 0.0831 mole of Fe(C2H3O2)3
is needed. How much do we have? only 0.644 mole. Here
,
Fe
(C2H3O2)3
is
the
limiting
reagent.
.0644 mole Fe (C2H3O2)3 x 6 mole Mg(C2H3O2)2x 142.20g Mg(C2H3O2)2 =
4 mole Fe(C2H3O2)3
1 mole
13.7g Mg(C2H3O2)2
Reaction 2 will produce the greatest yield.
5.
3Ca(OH)2 + 2H3PO4
36.0g 54.0g
---> Ca3(PO4)2
?
+ 6H2O
Part 1 of this problem is to find the limiting reagent.
Mw's - Ca(OH)2
74.10g/mole
H3PO4
98.00g/mole
Ca3(PO4)2 310.19 g/mole
36.0g
74.10g
Cu(OH)2
x
1
mole
Ca(OH)2
=
0.486
mole
Ca(OH)2
54.0g H3PO4 x 1 mole H3PO4 = 0.551 mole H3PO4
98.00g
.486 mole Ca(OH)2 x 2 mole H3PO4 = 0.324 mole of H3PO4
3 mole Ca(OH)2
Do we have enough H3PO4 for all of the Ca(OH)2 to react?
so Ca(OH)2 is the limiting reagent.
Chapter Review Exercises
Page 91
1.
2NF3
--->
N2
a) (exactly) 3 mole
+ 3F2
?g
?g
Yes, we have plenty
3 mole NF3 x 1 mole N2 x 28.01g N2 = 42.02g N2
2 mole NF3 1 mole N2
3 mole NF3 x 3 moles F2 x 38.00g F=2 = 171.0g F2
2 mole NF3
1 mole F2
b)
2 NF3
---> N2
0.268 mole NF3
+ 3F2
x 1 mole N2 x 28.01gN2 = 3.75g N2
2 mole NF3 1 mole N2
.268 mole NF3 x 3 mole Fe
2 mole NF3
2.
2p + 5Cl2
?g
x 38.00gFe = 15.3g F2
1 mole F2
---> 2 PCl5 (or you can use P4)
2.70 mole
2.70 mole PCl5 x 5 mole Cl2 x 70.91g Cl2 = 478.62g =479gCl2
2 mole PCl3
1 mole Cl2
3.
2 A s + 3Cl2 ---> 2AsCl3
0.113 mole ?
0.113 mole x Cl2 2 mole AsCl3 x 181.28gAsCl3 13.66g = 13.7AgCl3
3 mole Cl2
1 mole AsCl3
6.
H2 + Cl2 ---> 2HCl
11.1g 33.3g ---> ?g
Cl2 is so much heavier than H2 that I would guess 33.3g Cl2
is the limiting reagent, but we should check it out anyway.
11.1g H2 x 1 mole H2 = 5.51 mole H2; 33.3gCl2 x 1 mole Cl2
2.02
70.91g Cl2
=0.470 mole Cl2
Since H2 and Cl2 react in a 1:1 mole ratio, Cl2 is definitely the
limiting reagent
0.470 mole Cl2 x 2 mole HCl x 36.45g HCl = 34.3g HCl
1 mole Cl2
1 mole HCl
7.
3C +
2.50g
CaO ---> CO + CaC2
5.00g
2.45g <--- actual find theoretical
*
Find the moles of C and CaO
2.50gC x 1mole = .208 mole C ; 5.00gCaO x 1 moleCaO =
12.01g
56.08
0.0892 mole CaO
*
Let's assume CaO is limiting and see if there is enough C
0.0892 mole CaO x 3 mole C = 0.2675 mole C
1 mole CuO
*
We do not have 0.2675 mole of C. C must be limiting.
mole of C.
0.208 mole C x 1 mole CaC2 x 64.10gCaC2 = 4.44g CaC2
3 mole C
1 mole CaC2
*
present
We do have 0.208
Find the percent yield
% yield = Actual yield x 100 = 2.45g x 100 = 55.2%
Theoretical yield
4.44g
Chapter 2 - Alternate Problems
1.
4NH3 + 5O2
76.0g ?g
---> 4NO + 6H2O
?g
?g
76.0gNH3 x 1 mole NH3 X 5 MOLE O2 X 32.00gO2 = 178.51G = 179g O2
17.03g NH3
4NH3
1 mole O2
76.0gNH3
x 1 mole NH3 x 4 mole NO x 30.01gNO = 134g NO
17.03gNH3
4 mole NH3 1 mole NO
We could just subtrac 134g NO from (76.0 + 179 O2) to find the
of H2O (Law of Conservation of Mass) but I'll work it out
anyway.
mass
76.0gNH3 x 1 mole NH3 x 6 mole H2O x 18.02gH2O = 121gH2O
17.03gNH3 4 mole NH3
1 mole H2O
2.
2 PCl3 ---> 2 P + 3Cl2 ; C + 2Cl2 ---> CCl4
1.30 mole
1.30 mole PCl3 x 3 mole Cl2 = 1.95 mole Cl2 formed
2 mole PCl3
This 1.95 mole Cl2 can then be used in the second equation as
Don't try to adjust the moles.
C
+
2Cl2 ---> CCl4
1.95 mole
?g
1.95 mole Cl2
x 1 mole CCl4 x 153.82g CCl4 = 150gCCl4
2 mole Cl2
1 mole CCl4
is.
3.
Fe2O3
?g
+ 3CO
---> 2Fe + 3CO2
910kg ?g
910kgFe x 1 mole Fe x 2 mole Fe2O3 x 159.70gFe2O3= 5200KgFe2O3
55.85gFe
1 mole Fe
1 mole Fe2O3
It looks as if either might be limiting.
the moles of the other.
12.6gW x 1 mole W
183.85gW
Try one and compare
with
x 3 mole Cl2 = 0.206 mole Cl2
1 mole W
We do not have quite 0.206 moles of Cl
so W is in excess.
* To answer the first part of the problem we must find the
of W that reacted:
grams
0.192 mole Cl2 x 1 mole W x 183.85gW = 11.8g W
3 mole Cl2 1 mole W
*
To find the mass of unreacted W, subtract:
12.6g W - 11.8gW = 0.8gW left over
*
Now we need to find the mass of WCl6 that forms.
Either work the stoichiometry or add 11.8gW to 13.6gCl2=
25.4g
0.192 mole Cl2 x 1 mole WCl6 x 396.57gWCl6 = 25.4gWCl6
3 mole Cl2
1 mole WCl6
6.
2HgO ---> 2Hg + O2
75.8g 62.7g
75.8g HgO x 1 mole HgO x 2 mole Hg x 200.59gHg = 70.2gHg <-2 mole HgO 1 mole Hg
216.59gHgO
% yield = actual
x 100 62.7gHg x 100 = 89.3%
theoretical
70.2gHg
Answers to Exercises
Page 87
1.
The lollipops
reasons.
are
the
limiting
reagent
2.
7 lab benches x 4 students = 28 students
1 bench
24 desks
for,
hopefully,
benches
desks
obvious
The limit for enrollment would be the number of desks.
3.
2 SO2 + O2 + H2O --> 2H2SO4
Given:
5.0 moles 1.0 mole xs
? moles
5.0 moles SO2 x 2 moles H2SO4 = 5.0 moles H2SO4
2 moles SO2
1.0 mole O2 x 2 mole H2SO4 = 2.0 mole H2SO4
1 mole O2
There is only enough O2 present to form 2.0 mole H2SO4 so O2
the limiting reagent.
4.
2Fe + 2H2O + O2 ---> 2Fe(OH)2
Given:
2-5 mole 5.2 mole
must
be
1.5 mole
In words:
2.5 mole of Fe would produce 2.5 mole Fe(OH)2 and require 2.5
H2O and 1.25 mole of O2 to do so. There is excess of
H2O and O2. Fe is the limiting reagent.
mole
Answers to Exercises, page 90:
1.
Pb(NO3)2 + 2NaCl ---> PbCl2 + 2NaNO3
1.00g1.00g ?g
Which is the limiting reagent?
1.00g x 1 mole Pb(NO3)2 = 3.02 x 10-3 mole Pb(NO3)2
331.2Og
1.00g x 1 mole NaCl = 1.71 x 10-2 mole NaCl
58.44
3.02 x 10-3 Pb(NO3)2
x 2 mole NaCl = 6.04 x 10-3 mole NaCl
1 mole Pb(NO3)2
Since 6.04 x 10-3 mole of NaCl is needed to react with 3.02 x 10-3
mole
Pb(NO3)2 and there is 1.71 x 10-2 mole of NaCl
present, the NaCl must be
present in excess with Pb(NO3)2 the limiting reagent.
3.02 x 10-3 mole Pb(NO3)2 x 1 mole PbCl2 x 2.78.10gPbCl2 =
1 mole Pb(NO3)2 1 mole PbCl2
0.840g of
Pb(NO3)2
There are other ways (even easier) of working this problem.
2.
Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2
5.00g
20.0g
5.00g x 1 mole Cu
63.54gCu
= 7.87 x 10-2 mole Cu
20.0g x 1 mole AgNO3 = 0.118 mole AgNO3
169.88g AgNO3
7.87 x 10-2 mole Cu x 2 mole AgNO3 = 0.157 mole AgNO3
1 mole Cu
There are only 0.118 mole of AgNO3
present in excess with AgNO3 limiting.
3.
2KOH + (NH4)2SO4
---> K2SO4 + 2NH3
present, however, so Cu
must
be
+ 2H2O
This was probably a difficult equation for you.
This type was
covered in the double replacement type of reaction earlier.
20.0gKOH x 1 mole KOH = 0.356 mole KOH
56.11gKOH
15.0g(NH4)2SO4
x 1 mole (NH4)2SO4 = 0.114 mole (NH4)2SO4
132.14
0.356 mole KOH x 1 mole (NH4)2SO4 = 0.178 mole (NH4)2SO4
2 mole KOH
We do not have enough ammonium sulfate for all the KOH to
ammonium sulfate is limiting.
a)
0.114 mole (NH4)2SO4 x 1 mole K2SO4 = 0.114 mole (NH4)2SO4
1 mole (NH4)2SO4
b)
0.114 mole (NH4)2SO4 x 2 mole NH3
x 17.03gNH3 = 3.88gNH3
mole (NH4)2SO4 1 mole NH3
4.
Reaction 1:
4Fe(C2H3O2)3 + 6MgC2O4 ---> 6Mg(C2H3O2)2 + 2Fe2(C2O4)3
15.0g
10.0g
?
MW of Fe(C2H3O2)3 = 232.99g/mole
MgC2O4
= 112.33g/mole
15.0gFe(C2H3O2)3 x 1 mole
= 6.44 x 10=-2 mole Fe(C2H3O2)3
232.99g
10.0g MgC2O4
x 1 mole = 8.90 x 10-2 mole MgC2O4
112.33g
react
so
By stoichiometry
8.90 x 10-2 mole MgC2O4 x 4 mole Fe(C2H3O2)3 = 5.93 x 10-2 mole
6 mole MgC2O4 of Fe(C2H3O2)3
We have enough Fe(C2H3O2)3
MgC2O4 is limiting.
for all of MgC2O4 to react so
8.90 x 10-2 mole MgC2O4 x 6 mole Mg(C2H3O2)2 x 14240g mg(C2H3O2)2
6 mole MgC2O4 1 mole Mg (C2H3O2)2
12.7g Mg(C2H3O2)2
Reaction 2:
4Fe(C2H3O2)3 + 6MgSO4 ---> 6Mg(C2H3O2)3
15.0g
10.0g
+ 2Fe2(SO4)3
MW of MgSO4
= 120.38
15.0g MgSO4
x 1 mole MgSO4 = .125 mole MgSO4
120.38gMgSO4
.125 mole MgSO4 x 4 mole Fe(C2H3O2)3 .0831 mole of Fe(C2H3O2)3
6 mole MgSO4
4. cont'd
If all of MgSO4 is to be used up, 0.0831 mole of Fe(C2H3O2)3 is needed.
How much do we have? - only .0644 mole. Here
, Fe(C 2H3O2)3 is the limiting
reagent.
.0644mole Fe(C2H3O2)3 x 6 mole Mg(C2H3O2)2 x 142.20g Mg(C2H3O2)2 =
4 mole Fe(C2H3O2)3
1 mole
13.7g MG(C2H3O2)2
Reaction 2 will produce the greatest yield.
5.
3Ca(OH)2 + 2H3PO4
36.0g54.0g
?
---> Ca3(PO4)2
+ 6H2O
Part 1 of this problem is to find the limiting reagent.
MW's - Ca(OH)2
74.10g/mole
H3PO4
98.00g/mole
Ca3(PO4)2 310.19g/mole
36.0gCu(OH)2
54.0g H3PO4
x 1 mole Ca(OH)2 = 0.486 mole Ca(OH)2
74.10g
x 1 mole H3PO4 = 0.551 mole H3PO4
98.00g
.486 mole Ca(OH)2 x 2 mole H3PO4 = 0.324 mole of H3PO4
3 mole Ca(OH)2
Do we have enough H3PO4 for all of the Ca(OH)2 to react?
Yes, we have plenty so Ca(OH)2 is the limiting reagent.
5.
Part 2 of this problem is to calculate the theoretical yield.
Start with the limiting reagent.
0.486 mole Ca(OH)2 x 1 mole Ca3(PO4)2
3 mole Ca(OH)2
x 310.19g = 50.3g
1 mole Ca3(PO4)2
The theortical yield (maximum possible) is 50.3g of Ca3(PO4)2
Part 3 of this problem is to calculate the percent yield.
Percent yield = Actual yield
x 100
Theoretical yield
= 45.2g
50.3g
x 100
= 89.9%
Chapter Review Exercises
Page 91
1.
2NF3 ---> N2 + 3F2
a) (exactly) 3 mole
?g
?g
3 mole NF3 x 1 mole N2 x 28.01gN2 = 42.02gN2
2 mole NF3 1 mole N2
3 mole NF3
b)
x 3 mole F2 x 38.00gF2 = 171.0gF2
2 mole NF3 1 mole F2
2NF3
---> N2 + 3F2
0.268 mole
?g
?g
.268 mole NF3 x 1 mole N2 x 28.01gN2 = 3.75gN2
2 mole NF3 1 mole N2
.268 mole NF3
x 3 mole Fe x 38.00gFe = 15.3gF2
2 mole NF3
1 mole F2
2.
2P + 5Cl2
---> 2PCl5
(or you can use P4)
?g
2.70 mole
2.70 mole PCl5 x 5 mole Cl2 x 70.91gCl2 = 478.62g = 479gCl2
2 mole PCl3 1 mole Cl2
3.
2As + 3Cl2 ---> 2AsCl3
0.113 mole
?
0.113 mole x Cl2 2 mole AsCl3 x 181.28gAsCl3=13.66g =13.7gAgCl3
3 mole Cl2
1 mole AsCl3
4.
CS2 + 3O2 ---> CO2 + 2SO2
9.34g ?g
9.34gCS2 x 1 mole CS2 x 3mole O2 x 32.00gO2 = 11.78 = 11.8gO2
76.14gCS2 1 mole CS2 1 mole O2
5.
PbyOx + H2 ---> Pb + H2 O
We are trying to find x and y.
1 mole
8.1g 622g
One mole of PbyOx reacts with 8.1g H2 to form 622gPb.
many ways to solve this problem. This is only one way.
Change all grams to moles:
622gPb x 1 mole Pb = 3.00 mole Pb
207.19gPb
8.1gH2 x 1 mole H2 = 4.00 mole H2
2.02gH2
There
are
The Pb caould only have come form PbOx so we must have three mole of Pb
present originally.
The 4.00 mole of H2 must have formed 4.00 mole of H2O (to
balance
the hydrogens). This means that there had to be 4.00
mole
of
O
atoms
present originally.
This gives a formula of Pb3O4.
Check this:
Pb3O4 + 4H2 ---> 3Pb + 4H2O
685.57
8.1g = 622g + 72.06
693
694
694
6.
H2
+
11.1g
Cl2
---> 2HCl
33.3g---> ?g
Cl2 is so much heavier than H2 that I would guess 33.3gCl2 is the
limiting reagent, but we should check it out anyway.
11.1gH2 x 1 mole H2 = 5.51 mole H2; 33.3gCl2 x 1 mole Cl2 =
2.02
70.91gCl2
= 0.470 mole Cl
Since H2 and Cl2 react in a 1:1 mole ratio, Cl2 is definitely the
limiting reagent.
0.470 mole Cl2 x 2 mole HCl x 36.45g HCl = 34.3gHCl
1 mole Cl2
1 mole HCl
7.
3C
+
2.50g
CaO
---> CO + CaC2
5.00g
2.45g <--- actual, find theoretical
* Find the moles of C and CaO
2.50gC x 1 mole = .208 mole C; 5.00gCaO x 1 mole CaO = 0.0892
12.01g
56.08
moleCaO
* Let's assume CaO is limiting and see if there is enough C
present.
0.0892 mole CaO x 3 mole C = 0.2675 mole C
1 mole CaO
8.
*
We do not have 0.2675 mole of C. C must be limiting.
We do have 0.208 mole of C.
0.208 mle C x 1 mole CaC2 x 64.10gCaC2 = 4.44gCaC2
3 mole C
1 mole CaC2
*
Find the percent yield
% yield = Actual yield x 100 = 2.45g x 100 = 55.2%
Theoretical yield
4.44g
XO2
---> x
+ O2
25.2g
3.2g
*
*
How much X do we have? Subtract 25.2g - 3.2g = 22.0g
How many moles of x do we have? According to this we
have 3.2gO2 x 1 mole O2 = .10 mole O2.
32.00gO
This .10 mole O2 must have come from XO2.
have .10 mole XO2 present and then .10 mole x.
We must also
* If .10 mole of X weighs 22.0g, we can easily figure the
weight of one mole.
22.0g of X = 22Og/1 mole
.10 mole
The atomic weight of X must be 220.
Chapter 2 - Alternate Problems
1.
4NH3 + 5O2 ---> 4NO + 6H2O
76.0g ?g
?g
?g
76.0gNH3 x 1 mole NH3 x 5 mole O2 x 32.00gO2 = 178.51g = 179gO2
17.03gNH3
4NH3 1 mole O2
76.0gNH3 x 1 mole NH
_______
17.03gNH3
x 4 mole NO x 30.01gNO = 134gNO
4 mole NH3 1 mole NO
______
______
We could just subtract 134gNO from (76.0 + 179gO2) to find the
of H2O (Law of Conservation of Mass)
76.0gNH3 x 1 mole NH3 x 6 mole H2O x 18.02gH2O = 121gH2O
17.03gNH3
4 mole NH3
1 mole H2 O
2.
mass
2PCl3
---> 2P + 3Cl2 ; C + 2Cl2 ---> CCl4
1.30 mole
1.30 mole PCl3 x 3 mole Cl2 = 1.95 mole Cl2 formed
2 mole PCl3
This 1.95 mole Cl2 can then be used in the second equation as
Don't try to adjust the moles.
C
is.
2Cl2
---> CCl4
1.95 mole
?g
1.95 mole Cl2 x 1 mole CCl=4 x 153.82gCCl4 = 150gCCl4
2 mole Cl2
1 mole CCl4
3.
+
Fe2O3 + 3CO
--->
?g
910kgFe x 1 mole Fe
55.85gFe
2Fe
+ 3CO2
910kg
?g
x 2 mole Fe2O3 x 159.70gFe2O3 =
1 mole Fe
1 mole Fe2O3
5200 kg Fe2O3
I did not change to grams because I wanted to find answer is kg.
could have converted or could have used 1k mole Fe
55.85kg Fe
910kgFe x 1 mole Fe x 3 mole CO2 x 44.01gCO2 = 2150kgCO2
55.85gFe
1 mole Fe
1 mole CO2
4.
2C6H18 + 21O2
1 mole
1 mole C6H18
2C6H18
---> 12CO2
?g
+
18H2O
x 21 mole O2 x 32.00g = 336gO2
2 mole C6H18 1 mole O2
+ 15O2
---> 12CO + 18H2O
You
1 mole
1 mole C6H18
x 15 mole O2 x 32.00gO2 = 240.gO2
2 mole C6H18 1 mole O2
Difference = 336g - 240g = 96g
5.
W
+
12.6g
3Cl2
--->
13.6g
WCl6
This is a limiting reagent - type problem.
12.6gW x 1 mole W = .0685 mole W
183.85gW
13.6g Cl2 x 1 mole Cl2 = 0.192 mole Cl2
70.91
It looks as if either might be limiting. Try one and compare
the moles of the other.
12.6gW x 1 mole W x 3 mole Cl2 = 0.206 mole Cl2
183.85gW
1 mole W
with
We do not have quite 0.206 moles of Cl2 so W is in excess.
* To answer the first part of the problem we must find the
of W that reacted:
0.192 mole Cl2 x 1 mole W x 183.85gW = 11.8gW
3 mole Cl2 1 mole W
*
To find the mass of unreacted W, subtract:
12.6gW - 11.8gW = 0.8g W left over
* Now we need to find the mass of WCl6 that forms.
the stoichiometry or add 11.8gW to 13.6gCl2 = 25.4g
Either
0.192 mole Cl2 x 1 mole WCl6 x 396.57gWCl6 = 25.4gWCl6
3 mole Cl2
1 mole WCl6
6.
2HgO
75.8g
---> 2Hg
62.7g
+
O2
75.8gHgO x 1 mole HgO x 2 mole Hg x 200.59gHg = 70.2gHg
216.59gHgO
2 mole HgO 1 mole Hg
% yield = Actual
x 100 = 62.7gHg x 100 = 89.3%
Theoretical
70.2gHg
7.
grams
This is a difficult kind of problem for you.
An 18.0% NaCl solution by mass is (assumeing 100g of solution)
work
18.0gNaCl
100g solution
What is the 100g of solution?
is made up of 18.0gNaCl and 82g of water.
The solution
We have 18.0gNaCl
82.0gH2O
125g of solution x 18.0gNaCl
= 22.5gNaCl
100g solution
125g solution - 22.5gNaCl = 102.5gH2O
8.
3S
+
50.0g
2Cl2
100g
---> S2Cl2 + SCl2
?g
All of the S and Cl2 have reacted.
either present and use this.
We need to find the moles
of
50.0gS x 1 mole S x 1 mole S2Cl2 x 135.03gS2Cl2 = 70.2gS2Cl2
32.06gS
3 mole S
1 mole S2Cl2
Check this by using Cl2
100gCl2 x 1 mole Cl2 x 1 mole S2Cl2 x 135.03gS2Cl2=95.2gS2Cl2
70.91gCl2
2 mole Cl2
1 mole S2Cl2
-Oops- Despite what the problem says, Cl2
must be present
excess so theoretical mass of S2Cl2 formed must me 70.2gS2Cl2
This laboratory experiment should preceed Chapter 1, Section 3.
class period may be needed to perform this experiment.
in
An entire
Partial Weight Analysis of a Compound
The blue copper (II) sulfate found in the lab is an example of hydrate.
Hydrates have definite amounts of water incorporated periodically within the
crystalline structure. The five moles of water molecules per mole of copper
sulfate can be removed by heating.
This enables you to determine the
percentage of water of hydration present in a sample.
Weigh a clean, dry crucible to the nearest 0.0lg.
Grind up crystalline hydrated copper (II) sulfate
________g
in a mortar. Weigh out between 2.50 and 3.50 grams
of the salt in the crucible. Be sure to clean up
any spilled copper sulfate from the balance pan
before you record the weight.
Weight of sample and crucible.
_________g
Weight of sample, calculated
_________g
4.
Place the crucible and its contents on a pipestem triange set
ring clamped to a ring stand.
Heating the system gently for 3 to 5 minutes to avoid
Continue to heat but more strongly for 10 more
minutes.
observations during this time.
on
a
splattering.
Record
any
Observations:
Obtain a desiccator from the lab or
make one from a large jar. The
bottom of the jar should have 1/2
inch solid anhydrous calcium
chloride.
A
commercial laboratory
desiccator may have the
lid coated
with petroleum jelly so do not set
the lid down flat on the table. The
top should be placed upon the jar as
soon as it is filled and not removed
until necessary. The crucible will
rest on a wire screen or plate placed
above the calcium chloride.
At the end of the 10-minute heating, allow the crucible to cool on the stand
for 2 minutes and then use tongs to transfer it to the desiccator.
5.
After the crucible and contents have cooled to room
temperature
the closed desiccator, weigh the crucible and
solid.
Weight of crucible and residue.
_________g
6.
Calculate the weight of the residue
_________g
7.
Calculate the weight of the water loss by the original sample
_________g
8.
Calculate the percentage of water of hydration in the sample
_________g
9.
Place a few drops of water on the cooled, weighed residue.
Feel the bottom of the crucible before and after doing
Record any observations.
this.
in
Observations:
An entire clas period may be needed to perform this experiment.
Experiment:
The formula of a compound
Each compound has its own distinct formula, as
was just seen in the previous section, has its
own constant composition. In this experiment
you are going to make a compound and then
calculate
its simplest formula.
1. Weigh a clean, dry crucible and record its
weight to 0.0lg
_________
2. Obtain a clean piece of magnesium ribbon
about 35 cm in length. Use a fine piece of
steel wool to clean the ribbon. Cut the
ribbon
into pieces of approximately 1 cm
and add to the
crucible. Weigh the
crucible
and
contents.
_________
3.
Weight of magnesium ribbon by difference.
_________
4.
Calculate the moles of magnesium (M.W. of Mg = 24.31)________
5.
Set up the ring, triangle, and ring
stand as in the previous experiment.
Heat the covered crucible (picture 1)
gently at first and then gradually
increase the heat intensity for 2
minutes.
With your tongs, carefully tilt the
lid (picture 2) and continue strong
heating of the partially uncovered
crucible for an additional 10 minutes.
At the end of this time, allow the
covered crucible and contents to cool.
Check to see that all of the magnesium
has reacted before covering. If there
is still unreacted magnesium, heat
strongly again.
Since the magnesium reacted with both the oxygen and nitrogen
in
the air, it is necessary to convert the mixture of compounds to only the
oxide compound.
Use a medicine dropper to add only enough distilled water to cover the
contents of the crucible Carefully heat the system.
Observe the odor of
any vapor that has been produced. When the water has evaporated, add a
little more and carefully heat
again. Continue to do this until there
is no longer a distinct odor produced upon heating.
When this has happened, continue careful heating until all of
the
water has evaporated. Strongly heat the uncovered
crucible
for
five
more minutes. Allow the crucible and its
product to cool.
Weigh the
crucible and its contents.
Weight of crucible and contents ________
6.
Determine the weight of the magnesium oxide bydifference______
7.
Determine the weight of oxygen that combined with the
weight of magnesium
(You know how much Mg is present in the compound.) _______
8.
Calculate the number of moles of combined oxygen
_______
9.
Determine the simplest formula for magnesium oxide.
(If you are not sure what to do here, read ahead in
section 3.)
_______
original