Download Mass # = Atomic # + # Neutrons

Law of Conservation of Mass:
Mass is neither created nor destroyed in a chemical reaction.
Therefore in a chemical reaction, the Mass of Reactants (Left side of arrow) must
equal the Mass of Products (Right side of arrow).
For the general reaction:
A + B → C + D where A, B, C, D represent chemicals
mA + mB = mC + mD
Example:
Mercury (II) Sulfide
HgS
(reactant)
Makes
Mercury
→
Hg
+
Sulfur
+
S
(products)
If 216.0 g of Mercury (II) sulfide are used, and 200.0 g of Mercury are made, how
many grams of Sulfur must have been made?
216.0 g = 200.0 g + mS
Click to see Solution
mS = 216 g – 200.0 g = 16.0 g of S made
Make special note of all problems! Expect to have to do this on a quiz or test!
Law of Conservation of Mass Practice
N2 (g) + 3 H2 (g)
→
2 NH3 (g)
1) If 28.0 g of N2 react exactly with 6.0 g of H2, how many grams of NH3 are
produced?
N2 (g) + 3 H2 (g)
→
2 NH3 (g)
2) If 53.5 g of NH3 are produced by the reaction of 42.42 g of N2 with excess
H2, how many grams of are H2 consumed?
2 NH3 (g)
→
N2 (g) + 3 H2 (g)
3) If 128.4 g of NH3 are reacted to produce 98.4 g of N2, how many grams of
are H2 produced?
Law of Definite Proportions:
A pure substance always contains the same proportions by mass of each atom
in the substance regardless of where the sample of the substance was found or
how much of the substance was analyzed.
This law really just tells us that if we have a sample of some chemical (like
water for example), then the chemical will have to have the same formula if it
really is the same substance. We know that water has the formula H2O. No
matter where we find “water” it will always have the formula H2O. If we
find something that has the formula H2O2 for example, we will know that it is
not water because its formula is different.
The Law is worded the way it is because we can not “see” molecules to
determine their formulas. Instead, we calculate their formulas based on the
mass percent of each element present in a sample of the substance.
Percent by Mass (or Mass Percent)
Mass of Element
Mass % =
Mass of Compound
x 100
Mass % Example:
Sodium chloride (NaCl) is known to have 39.34% sodium ions (Na+1) and
60.66% chloride ions (Cl-1). If we have a 14.67 g sample of sodium chloride,
how many grams of sodium ion will be present in the sample? How many
grams of chloride ion would be in the sample?
Given: Mass%Na = 39.34%
Mass%Cl = 60.66%
mNaCl = 14.67 g
mNa = ?
mCl = ?
mNa
Mass%Na =
mNaCl
x 100
mNa =
mNa =
Click
(Mass%Na)(mNaCl
) to see Solution
100
(39.34)(14.67 g)
= 5.771 g Na+1
100
Finding the mass of Cl-1 would be similar using: Mass%Cl =
mCl
mNaCl
x 100
Make special note of all problems! Expect to have to do this on a quiz or test!
Mass % Practice
1) If 28.0 g of a sample of compound are found to contain 8.00 g of calcium and
20.00 g of iodine, what is the mass % of calcium in the sample? What is the mass %
of iodine in the sample?
2) If a compound is known to be 22.00% carbon and 78.00% chlorine by mass, how
many grams of carbon would there be in 3.721 g of the compound? How many grams
of chlorine?
Law of Multiple Proportions:
If two or more compounds are composed of the same two elements, then
the mass ratio(s) of one of the elements will always be a ratio of small
whole numbers.
Example: at least three different compounds containing just chromium and chlorine
are known. Data for these three compounds is given in the table below.
Compound Mass of Chromium Mass of Chlorine
A
1.0000 g
1.3635 g
B
1.0000 g
2.0450 g
C
1.0000 g
4.0916 g
Using the data in the table, the law of multiple proportions says that the ratios of the
masses of the chlorine will be small whole numbers.
Ratio of chlorine in B to A is 1.4998 or 1.500 which is 3/2.
Ratio of chlorine in C to B is 2.000 which is 2/1.
Ratio of chlorine in C to A is 3.000 which is 3/1.
What does this do for us in terms of our understanding of chemistry?
Atomic Theories (theories of what atoms are like)
John Dalton proposed a new atomic theory based on the three Laws we have
been discussing and the ideas of Democritus.
See page 64 for a discussion of Daltons ideas.
Dalton’s Atomic Theory (1808 A.D.)-very similar to ideas of Democritus (~400 B.C.)
•
Each element is made up of tiny particles called atoms that can not be broken
down into smaller particles.
•
The atoms of a given element are identical; the atoms of different elements are
different in some fundamental way or ways.
•
Chemical compounds are formed when atoms combine with each other in
simple whole number ratios. A given compound always has the same relative
numbers and types of atoms.
•
Chemical reactions involve reorganization of the atoms - changes in the way
they are bound together. The atoms themselves are not changed in a chemical
reaction.
Was Dalton 100% correct?
Cathode Ray: stream of charged particles produced by an
electric field. The green “glow” in the picture.
The particles move from the cathode (negatively charged
plate) to the anode (positively charged plate).
Cathode “rays” are influenced by both magnetic and electric
fields (electric field shown in the diagram).
Opposite charges are known to attract each other. Since the
cathode “ray” was deflected towards the positively charged plate,
what charge do the particles in the cathode “ray” have?
J. J. Thompson and Millikan demonstrated that the mass of the particles in a
Cathode Ray are much less than the mass of the smallest atom.
How did this change our understanding of Dalton’s theory?
J. J. Thompson proposed the Plum Pudding Model
Plum Pudding Model:
Electrons are evenly distributed
throughout a uniform positive
charge equal to the negative
charge of all of the electrons.
The part with the positive
charge is where most of the
mass of an atom is located.
Ernest Rutherford (1911) “gold foil” experiment
Large, high energy particles (alpha particles, +2 charge) do not always pass
straight through a thin sheet of matter. The alpha particle being deflected or
reflected is like having a 30-06 bullet bounce off of a sheet of tissue paper.
a) Results of golf foil experiment expected if Plum Pudding Model
were true.
b) Results of golf foil experiment explained by new model.
Rutherford proposed a nuclear model of the atom.
The small, dense nucleus contains virtually all the mass of the atom and
all of the positive charge while the negatively charged electrons exist
apart from the nucleus. Rutherford did not know where the electrons
were-they were just outside of the nucleus.
How does this model fit Rutherford’s results?
Nuclear Model of an Atom
Small, very dense nucleus containing
massive protons and neutrons,
surrounded by small rapidly moving
electrons
If all atoms contain the same types of
particles, what makes one atom
hydrogen and another carbon?
The number of protons in the nucleus
determine what element an atom is.
Remember that in normal atoms, the
number of protons is equal to the
number of electrons (they are neutral-a
total charge of zero).
How many times bigger is the atom compared to the nucleus?
If the nucleus is very small compared to the overall size of the atom, then the
atom consists of mostly empty space!
Rutherford called the positive particle in the nucleus a proton.
In 1932, James Chadwick discovered a neutral particle in
the nucleus and called it a neutron.
Summary of Subatomic Particles (mid 1900’s)
Particle
Symbols
Relative Charge
Mass #
Relative Mass
Actual Mass
-1
0
1/1837
9.11X10-31kg
Electron
e-
Proton
p+
1
H
1
+1
1
1
1.67X10-27 kg
Neutron
no
1
n
0
0
1
1
1.68X10-27 kg
1/1837
9.11X10-31kg
0
e
-1
Other Related Particles (mid 1900’s)
Beta
Alpha
e-
0
0
e
b
-1
-1
4
He
a
2
-1
0
+2
4
4
6.64X10-27 kg
Atomic Number:
Atomic number is the number of protons in the nucleus of an
element. Since in neutral atoms the number of protons is equal
to the number of electrons, atomic number also indicates the
number of electrons in a single atom of an element.
The Periodic Table is arranged according to atomic number.
For example, hydrogen is element 1 and has 1 proton in its nucleus.
Similarly, helium is element 2 and has 2 protons in its nucleus.
How many protons does element number 25 contain?
What is the name of this element?
How many electrons does element 25 contain?
What is the charge of one atom of element 25?
Isotopes are atoms with the same number of protons (same element)
but with a different number of neutrons.
Na-23 (11 + 12 = 23)
Na-24 (11 + 13 = 24)
Identifying Specific Isotopes (Nuclides)
Mass Number (total number of protons and neutrons in a nucleus)
A
Z
X
Element Symbol
Atomic Number (total number of protons in a nucleus)
Mass # = Atomic # + # Neutrons
# Neutrons = Mass # - Atomic #
or
# Neutrons = A - Z
Use a periodic table to help you fill in the missing information in
the following tables.
Isotope
Atomic #
Mass #
# Protons
# Neutrons
137
Ba
14
14
207
Isotope
K-42
Atomic #
Mass #
80
# Protons
13
# Neutrons
14
207
82
18
Unified AMU: unified atomic mass unit is a unit of mass based on 12C.
The mass of one atom of 12C = 12 u
so 1 u = 1/12 of the mass of one atom of 12C
On this scale, 1 proton = 1.007276 u
1 neutron = 1.008665 u
1 electron = 0.0005486 u
The Mass of “C” on the periodic table is 12.011 u because
pure carbon is made up of both 12C (12 u exactly) and 13C
(close to 13 u).
The mass of “C” is a weighted average of all isotopes of C!
19
Average Atomic Mass: the weighted average of the atomic masses of all naturally
occurring isotopes of an element. The weighted average is found by multiplying the
natural abundance (%Abund) for a given isotope times the actual mass (m) of that
isotope and then adding all of those masses together.
For example: if an element (X) has three isotopes, the formula for Average atomic
Mass would be:
Average Mass = [(%Abund1)*(m1) + (%Abund2)*(m2) + (%Abund3)*(m3)]
How would the formula change if an element had 5 naturally occurring isotopes?
What would it look like if an element only had 2 naturally occurring isotopes?
20
Find the Average Mass of Magnesium (Mg)
Given:
Data for the known isotopes of Magnesium
Isotope
24
Mg
25
Mg
26
Mg
Abundance
78.99%
Mass
10.00%
23.985Click
u to see Solution
24.986 u
11.01%
25.982 u
Average Mass =
[(%AbundMg-24)*(mMg-24) + (%AbundMg-25)*(mMg-25) + (%AbundMg-26)*(mMg-26)]
= [(0.7899)*(23.985 u) + (0.1000)*(24.986 u) + (0.1101)*(25.982 u)]
= 24.3049697 u
= 24.30 u
21
One Mole is the number of atoms in exactly 12 g of carbon-12.
Avogadro’s Number is the number of particles in exactly one mole and
has a value of 6.022X1023
The definition of “mole” and Avogadro’s number allows us to “count” atoms
by weighing them when we know what the mass of one mole of a substance is.
Molar Mass is the mass in grams of one mole of a pure substance. When the
atomic masses on the periodic table are expressed in grams (instead of “u”),
the atomic mass represents the mass of one mole of that element.
Important Equivalence Statements:
1 mol = 6.022X1023 particles (atoms, molecules, ions, etc…)
1 mol = atomic mass in grams for each element on the periodic table
for example: 24.3050 g Mg = 1 mol Mg and 55.845 g Fe = 1 mol Fe
Remember that Equivalence Statements can be used to convert
from one set of units to another!
When working on problems, we will use “n” to represent the variable of moles
in a problem (just like we may use “l” to represent length in a math problem).
When working on problems, we use “m” to represent the variable of mass in a
problem. Please note that variable and units mean different things.
When working on problems, I will try to use “MW” to represent the variable
of molar mass. The book often uses “M” for molar mass, but later in the year
the “M” will stand for molarity which means something entirely different.
In equation form the relationship between molar mass, moles, and mass is:
m
MW =
n
Therefore we can write an equation for mass:
m = nMW
and we can write an equation for moles:
n=
m
MW
When working on problems, sometimes we will use the equivalence statement
approach and other times we will use the equation approach. They both do the
same math, but they use a slightly different kind of logic to solve.
Mass to Moles and Moles to Mass Example Problems:
What is the mass in grams of 0.0695 moles of silver?
First we will solve this using the equivalence statement method:
Start with what was given and then use the equivalence statement from
the periodic table: 1 mole Ag = 107.9 g Ag
(
)(
0.0695 moles Ag
)
107.9 g Ag
1 mol Ag
Click to see Solution
= 7.50 g Ag
This is a conversion factor based on the
equivalence statement from the periodic
table.
Now we can solve the same problem using the equation method:
Given:
n = 0.0695 mol
MW = 107.9 g/mol (form periodic table)
m = nMW
m = (0.0695 mol Ag)(107.9 g Ag/mol Ag) = 7.50 g Ag
Which method should we use?
Both methods (equation versus equivalence statement) work well, but the
equivalence statement method will work in many instances where we do not
have an obvious equation to use.
Ultimately, you may use any method you want, but you need to learn both
because there are times when really only one of the two methods will work.
Can you decide which method to use with a given problem?
If a reaction is producing water at a rate of 3.5 grams per second, how many
moles of water per hour will be produced.
Use the Equivalence Statement method
Lead has a density of 13.2 g/mL. If you have a sample of lead that has a mass of
276.4 g, what is the volume of the sample?
Use the Equation method
How many moles of copper (Cu) are present in a sample of copper that
has a mass of 127.3g?
First we will solve this using the equivalence statement method:
Start with what was given and then use the equivalence statement from
the periodic table: 1 mole Cu = 63.55 g Cu
(
127.3 g Cu
)(
)
1 mol Cu
63.55 g Cu
Click to see Solution
= 2.003 mol Cu
Now we can solve the same problem using the equation method:
Given:
m = 127.3 g
MW = 63.55 g/mol (form periodic table)
m
MW
127.3 g
=
63.55 g/mol
n=
n =
m
MW
= 2.003 mol Cu
Particles to Moles and Moles to Particles
How many atoms of Zn are contained in 0.0087 moles of Zn?
Given:
0.0087 moles Zn
1 mole Zn = 6.022X1023 atoms Zn
)(
(
0.0087 moles Zn
6.022X1023 atoms Zn
1 mol Zn
)
Click to see21 Solution
= 5.2X10 atoms Zn
How many moles of copper are present if you have a sample
containing 3.8X1021 atoms of copper?
Given:
(
3.8X1021 atoms Cu
1 mole Cu = 6.022X1023 atoms Cu
)(
3.8X1021 atoms Cu
1 mol Cu
6.022X1023 atoms Cu
)
Click to see Solution
= 0.0063 mol Cu
Particles to Mass and Mass to Particles
How many atoms of Zn are contained in 14.3 g of Zn?
1 mole Zn = 6.022X1023 atoms Zn
Given: 14.3 g Zn
(
)(
14.3 g Zn
1 mol Zn
65.38 g Zn
)(
6.022X1023 atoms Zn
1 mol Zn
)
1 mol Zn = 65.38 g Zn
= 1.32X1023 atoms Zn
Click to see Solution
How many grams of copper are present if you have a sample
containing 7.430X1022 atoms of copper?
Given: 7.430X1022 atoms Cu
1 mole Cu = 6.022X1023 atoms Cu
1 mol Cu = 63.55 g Cu
(
)(
7.430X1022 atoms Cu
1 mol Cu
6.022X1023 atoms Cu
)(
)
63.55
g Cu
Click
to see =
Solution
7.841 g Cu
1 mol Cu
Example with an extension
A chemist has a sample of silicon (Si) that has a mass of 84.39 mg.
How many atoms of silicon are in the sample?
Given: 84.39 mg Si
1 mole Si = 6.022X1023 atoms Si
1 g Si = 1000 mg Si
1 mol Si = 28.09 g Si
Clickwe
to see
The extension is that we need to convert from mg to g before
startSolution
the rest of the problem.
(
)(
84.39 mg Si
1 g Si
1000 mg Si
)(
1 mol Si
28.09 g Si
)(
= 1.809X1021 atoms Si
6.022X1023 atoms Si
1 mol Si
)