Download unit 8 – compound stoichiometry

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Ch.7 – Percent Composition, Empirical Formulas, & Molecular Formulas
I. Introduction:
With this chapter we begin the study of stoichiometry (from the Greek words “stoicheion” which means element and “metron”
which means to measure). Stoichiometry involves chemical reaction calculations. There are 2 types of stoichiometry problems:
1.
Problems that require a balanced chemical equation are known as REACTION STOICHIOMETRY
2.
Problems that do not need a balanced equation but do require a correctly written formula are known as COMPOUND
STOICHIOMETRY
II. Problem Types that you will have to be able to recognize and calculate:
1.
FORMULA or MOLECULAR MASS –
These are two different phrases that really mean the same thing:
Formula mass is used for ionic compounds since they do not exist as molecules but rather as crystals
Molecular mass is used for molecular compounds that exist as discrete molecules.
Steps:
a) Write a correct formula for the compound.
b) Round the average atomic mass (given on the Periodic Table) of each element to 2 decimal places.
Chlorine – use 35.45 amu (do not round to 35)
Copper – use 63.55 amu (do not round to 64)
c) Multiply the rounded average atomic mass of each element by the number of atoms of that element present
(subscripts).
d) Add the total rounded atomic masses of all elements in the compound and label with amu or grams.
Example 8-1: Find the formula mass of calcium phosphate.
Example 8-2: Find the formula mass of ammonium sulfate.
Example 8-3: Find the molecular mass of dichlorine heptoxide.
Hydrates: ionic compounds that TRAP water molecules in their crystal structures. You must know how to name and
recognize hydrates. A hydrate will be written as a correct ionic formula followed by a dot (which looks like a multiplication
dot) and then some number of molecules of water. i.e. CuSO 4 · 5 H2O
Formula Mass of a hydrate:
a) add all the average atomic masses of the ionic compound together
b) add the mass of “x” number of water molecules to it. The dot DOES NOT MEAN multiplication.
Formula Mass of copper (II) sulfate pentahydrate
Cu
63.55 x 1 = 63.45 amu
S
32.07 x 1 = 32.07
O
16
x 4 = 64
H2O
18.02 x 5 = 90.10
249.62 amu.
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Example 8.4: Find the formula mass of barium chlorate hexahydrate.
2.
MOLE PROBLEMS –
a) The term mole is a unit used in chemistry. Mole is a “counting” number.
b) In the American system of measures, we know that a “dozen” is a unit which always means 12 of, no matter whether
we are measuring eggs or pencils or gnats.
c) The mole is a unit that means the formula/molecular mass of a substance expressed in grams.
Example 8-5: What if you have 200 grams of calcium carbonate, how many moles is this?
Example 8-6: How many moles are there in 28.7 grams of lithium nitrate?
3.
AVOGADRO’S NUMBER PROBLEMS –
a) Avogadro’s number (6.022 X 1023) is the number of atoms in one mole of a pure monatomic element…..OR
b) Avogadro’s number is also the number of molecules in one mole of a compound or a diatomic element.
There are 65.38 grams in 1 mole of pure zinc metal. Therefore, there are 6.022 X 1023 atoms of zinc in 65.38 grams (or 1
mole) of zinc.
There are 18.02 grams of water in 1 mole of water. Therefore, there are 6.022 X 10 23 molecules of water in 18.02 grams
(or 1 mole) of water.
There are 32 grams of oxygen in 1 mole of oxygen gas. Therefore, there are 6.022 X 10 23 diatomic oxygen molecules in
32 grams of oxygen gas or 1 mole of oxygen gas. However, there are 2 (6.022 X 10 23) or 1.2044 X 1024 atoms in 32
grams of oxygen gas.
The re a r e 6 .0 2 2 x 1 0 2 3 p a rt ic le s i n 1 m o le o f a nyt h ing . 1 m o le o f p en nie s i s 6 .0 2 2 x 1 0 2 3 p en nie s.
Example 8-11: How many atoms are present in a pure iron nail that weighs 13.2 grams?
Example 8-12: How many moles of carbon dioxide are present in 4.55 X 10 24 molecules of CO2?
Representative particles: a very important concept that you MUST get straight NOW.
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1.
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4.
Element (except diatomics) = ATOM smallest particle of an element which will retain the properties of that
element.
Diatomic elements = MOLECULE
Molecular (2 non-metals) compound = MOLECULE
Ionic compound (metal with nonmetal or polyatomic ion) = FORMULA UNIT
Example 8-13: How many formula units are there in 7.55 grams of sodium sulfate?
Example 8-14: How many carbon atoms are there in 16 grams of glucose?
Example 8-15: How many nitrogen atoms are there in 45.8 grams of ammonium nitride?
4.
PERCENTAGE COMPOSITION PROBLEMS
Just like any percentage problems you are comparing the part to the whole. For example, if you have 5 pieces of fruit in a
dish and 2 of the pieces of fruit are bananas, what percentage of the fruit in the dish is bananas?
In chemistry, however, percentage composition is based on mass, not on numbers of atoms present. We are going to
find the percentage BY MASS of certain elements in a compound by comparing the mass of the element to the mass of
the entire compound.
STEPS: To find percentage composition:
1) write a correct formula for the compound in the question
2) find the formula or molecular mass of the entire compound
3) divide the total mass of the element whose percentage you are looking for by the total mass of the entire compound. Multiply
the result by 100 to convert it to a percentage
4) since there are no significant digits given in the problem except masses which have already been rounded, express your
answer to 1 decimal place and add the % sign as the label
Example 8-16: What is the complete percentage composition (by mass) of potassium dichromate?
Example 8-17: What is the % of calcium in calcium phosphide?
Example 8-18: What is the % of oxygen in barium chlorate hexahydrate?
REMEMBER THAT A % IS AN EXCELLENT CONVERSION FACTOR SINCE IT REPRESENTS A COMPARISON OF UNITS TO
100 UNITS. The following problems can be worked using dimensional analysis or by finding the % of a particular element present
and then multiplying iti times the mass of the compound given.
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Example 8-19: How many grams of pure magnesium could be recovered from the decomposition of 49.4 grams of
magnesium fluoride?
Example 8-20: How many grams of calcium are present in 156.8 grams of chalk (calcium carbonate)?
These problems are also percentage composition problems even though they might not sound like it.
Example 8-21: The Calculator Team needs money for another bus to go to the Austin contest since so many students have
signed up to go. The current price of pure gold is $392/Troy ounce. We have some old gold (III) chloride in
the storeroom which is not being used for anything. How many grams of gold (III) chloride must we
decompose and sell the pure gold out of it in order to raise the $600 necessary for a second bus?
Example 8-22: The officers of the Math-Science Team have met and decided that, as a fund-raiser, they would spend $100 to
buy 1.0 pound of silver nitrate. They plan to decompose the silver nitrate and sell the pure silver for
$7.81/Troy ounce. How much of a profit will they make on this fund-raiser?
5) EMPIRICAL FORMULA PROBLEMS –
 An empirical formula is one that is written in its very lowest terms (cannot be reduced any further).
 it does not necessarily give the exact number of atoms of everything present, but rather the RATIO of the atoms
present
 CH2O is an empirical formula that tells us that there are twice as many hydrogen atoms present as there are carbon
atoms or oxygen atoms.
 C8H12O6 is a molecular formula (not empirical) which could be reduced to lower terms, but IT ISN’T. It tells us
that there are 6 atoms of carbon present, 12 atoms of hydrogen present, and 6 atoms of oxygen present – it gives the
numbers of atoms of each and not the simplest ratio between the atoms
To solve empirical formula problems, you must be given either the grams of each element present or the % of each
element present.
NOTE: formulas are typically written from least electronegative to most electronegative. METALS FIRST!
STEPS to solve empirical formula problems:
1. % to mass – the easiest way to convert % to mass is to assume 100 grams. (42% of 100 g = 42 g!)
2. mass to mole – USE MOLAR MASS! Be sure to leave your answer in at least 4 significant figures to maintain
your accuracy.
3. Divide by the smallest: We want to set up a ratio with the smallest number in the ratio being one.
4. Multiply ‘til whole: If all of the results in #2 are whole numbers, simply use these whole numbers as subscripts
and write the empirical formula. If all of the results in #2 are NOT whole numbers (i.e. some of them are
decimal fraction mixed numbers), you must multiply ALL of the results in #2 by some number which will make
everything a whole number. If the decimal portion is:
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.2 or .8 then multiply by 5
.25 or .75 then multiply by 4
.33 or .67 then multiply by 3
.5 then multiply by 2
Example 8-23: A compound is found to contain 34.39% zinc, 14.82% nitrogen and 50.79% oxygen. What is the empirical
formula of the compound?
Example 8-24: A 200 gram sample of compound which contains only carbon, hydrogen and oxygen is found to contain 94.74
grams of carbon, 21.05 grams of hydrogen and 84.21 grams of oxygen. What is the empirical formula of the
compound?
Example 8-25: An organic compound is analyzed and found to contain 44.226% carbon, 6.388% hydrogen, 23.587% oxygen,
and 25.799% nitrogen. What is the empirical formula for this compound?
6) MOLECULAR FORMULA PROBLEMS –
1. Molecular formulas are those formulas that show the actual number of every different atom present (not just the ratio
between the atoms like the empirical formula does); molecular formulas could be reduced, but the ARE NOT.
2. Hint: Before you can work a molecular formula problem, you must first be given 2 things:
a. the molecular mass of the compound
b. the empirical formula of the compound or the means to find the empirical formula.
3. How to find the molecular formula:
a. Find the molecular mass of the empirical formula
b. Divide the given molecular mass of the entire molecule by the molecular mass of the empirical formula (EFM);
The result should be a whole number or very, very close to a whole number. DO NOT MULTIPLY ‘TIL
WHOLE! You should have very close to a whole number, or you messed up somewhere along the way. This
number indicates the number of empirical formula units (EFU) in the molecular formula.
c. Use this whole number result to multiply all of the subscripts of the empirical formula by and write the
molecular formula.
# EFU =
MM
EFM
Example 8-26: If the empirical formula for a carbohydrate is CH2O and its molecular mass is known to be 240 g/mole, what is
the molecular formula for the carbohydrate?
Example 8-27: If the empirical formula for a hydrocarbon is known to contain 92.3% carbon and its molecular mass is known
to be 78 g/mole, what is the molecular formula for the compound?
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7.
HYDRATE PROBLEMS –
a) Hydrates are compounds which trap water in their crystal structure
b) Hint: Before you can work a hydrate problem you must be sure that you know 2 things:
i) The mass of the anhydrous (dry; water already driven off; dehydrated) compound
ii) Mass of the water which was trapped (and driven off with heat)
STEPS to solve hydrate problems:
1) Divide the mass (or the percentage) of the DRY compound by the formula mass of the ionic compound only (not
the formula mass of the entire hydrate) and leave your answer in at least 4 significant figures.
2) Divide the mass (or percentage) of the water by the molecular mass of water, which is always 18 g/mole, and
leave your answer in at least 4 significant figures.
3) Compare the values you obtained in #1 and #2 by selecting the smaller of the two and divide both of them by it.
Your result here should be very, very close to a whole number and this is the number of molecule of water that
compound traps.
4) Write the formula of the ionic compound, followed by a dot, and then write the number of water molecules you
just calculated as the correct formula of the hydrate.
Example 8-28: A chemist heats a 300 gram sample of hydrated iron (III) nitrate until all of the water of hydration is driven
off. The anhydrous (dry) compound is found to weigh 179.7 grams. What is the correct name of the hydrate?
Example 8-29: A hydrate of iron (II) sulfite is known to contain 44.26% water. What is the correct formula of the hydrate?
Example 8-30: When 500 grams of hydrated copper (II) sulfate were heated strongly in a crucible, 180.35 grams of water
were released and the anhydrous cupric sulfate remained in the crucible. What is the correct formula for the
hydrate?