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Mass Relationships in
Chemical Reactions
Chapter 3
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Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
3.1
Average atomic mass (6.941)
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
3.2
How many atoms of He are in 5.36 moles of He?
23 atoms He
6.022
x
10
5.36 mol He x
=
1 mol He
3.23 x 1024 atoms He
How many atoms of O are in 5.36 moles of O2?
5.36 mol O2 6.022 x 1023 molecules O2
2 atoms O
=
x
x
molecules
1 O2 molecule
1 mol O2 molecules
6.46 x 1024 atoms O
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2
One Mole of:
S
C
Hg
Cu
Fe
3.2
1 12C atom
12.00 g
1.66 x 10-24 g
x
=
23
12
12.00 amu
6.022 x 10
C atoms
1 amu
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
M = molar mass in g/mol
NA = Avogadro’s number
3.2
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
3.2
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
3.3
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
HNO3
1H
1N
1.008 amu
14.0067 amu
3O + 3 x 16.00 amu
HNO3
63.01 amu
1 molecule HNO3 = 63.01 amu
1 mole HNO3 = 63.01 g HNO3
3.3
Ba(NO3)2
1Ba
2N
137.327 amu
2 x 14.0067 amu
6 O + 6 x 16.00 amu
Ba(NO3)2 261.33 amu
Ba
137.327 amu
- OR - 2 NO3 + 2 x 62.00 amu
Ba(NO3)2
261.33 amu
1 Ba(NO3)2 formula unit = 261.33 amu
1 mole Ba(NO3)2 = 261.33 g Ba(NO3)2
How many atoms are in 1.00 mol of Ba(NO3)2 ?
1.00 mol
Ba(NO3)2
23 Ba(NO )2
6.022
x
10
3
x
1 mol Ba(NO3)2
9 atoms
=
x
1 Ba(NO3)2
5.42 x 1024 atoms
3.3
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
3.3
+
Can look at atoms,
molecules or fragments
of molecules.
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
Can see isotopes of
F=qxvxB
atoms.
Heavy
Different
masses
deflect
differently
Heavy
Light
Very small
amount of
sample
Light
Similar to JJ Thomson’s Experiment
3.4
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
3.5
Empirical Formulas
Weight % of Elements
Ratio of Elements
1 - Start with weight percent of elements (make sure they
total to 100%).
2 - Assume 100 gram sample and convert weight percent
of each element to grams of each element.
3 - Convert to moles of each element using atomic
weights.
4 - Calculate relative ratio of elements by dividing by the
smallest number of moles.
5 - Multiply to get rid of any fractions, if necessary.
When in doubt, convert to moles !!!
A compound is found to contain:
20.2 wt. % Al and 79.8 wt. % Cl.
What is its empirical formula?
1 - Make sure total is 100%.
20.2 % + 79.8 % = 100.0%
2 - Convert weight percent of each element to grams of each
element.
20.2 g Al and 79.8 g Cl
3 - Convert to moles of each element using atomic weights.
20.2 g Al
= 0.749 mol Al
79.8 g Cl
= 2.25 mol Cl
26.98 g/mol Al
35.345 g/mol Cl
4 - Calculate relative ratio of elements by dividing by the
smallest number of moles. => 0.749 mol Al is smaller
0.749 mol Al
0.749 mol Al
= 1.00
2.25 mol Cl
0.749 mol Al
=>
AlCl3
= 3.01mol Cl / mol Al
Round to 3
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
3.7
reactants
products
Starting materials
Substances formed
aA+bB
plus
cC +dD
to yield
plus
Conservation of Mass
Must have the same number of atoms on
each side of the reaction !!
3.7
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
3.7
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
3.7
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
remove fraction
multiply both sides by 2
4. Remove all fractions (generally by multiplying
everything by 2) and reduce all stoichiometric
coefficients to try to get one equal to 1 (generally by
dividing everything by 2).
C2H6 + 7 O2
2
2C2H6 + 7O2
2CO2 + 3H2O
4CO2 + 6H2O
3.7
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
14 O (7 x 2)
12 H (2 x 6)
4 C (2 x 2)
14 O (4 x 2 + 6)
12 H (6 x 2)
4C
Always show a
check of your
solution on a quiz or
exam !!
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
3.7
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left
side and the correct formula(s) for the product(s) on the
right side of the equation.
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of each
element the same on both sides of the equation. Do not
change the subscripts.
3. Start by balancing those elements that appear in only one
reactant and one product.
4. Balance those elements that appear in two or more
reactants or products.
4. Remove all fractions (generally by multiplying everything by 2)
and reduce all stoichiometric coefficients to try to get one
equal to 1 (generally by dividing everything by 2).
5. Check to make sure that you have the same number of
each type of atom on both sides of the equation.
Stoichiometry
Quantitative study of chemical reactions.
How much in ? => How much out ?
You may have used these principles
baking cookies, doing carpentry, ………..
Chemicals react in ratios of moles,
NOT in weight ratios !!
When in doubt, convert to moles !!!
CH4 + 2 O2
CO2 + 2 H2O
Get a set of relationships between all reactants and products:
1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
These are exact conversions !!!
Mass Changes in Chemical Reactions
When in doubt,
convert to
moles !!!
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8
How many mole of CO2 can be generated from
10.31 moles of O2?
CH4 + 2 O2
CO2 + 2 H2O
1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
10.31 mol O2 x 1 mol CO2
2 mol O2
=
5.155 mol CO2
How many grams of CH4 are needed to produce
3.85 mol of CO2?
3.85 mol CO2 x 1 mol CH4 x 16.04 g CH4
1 mol CO2
1 mol CH4
= 61.8 g CH4
MW CH4 = 12.01 + 4(1.008) = 16.04 g / mol CH4
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
3.8
Limiting Reagents
Who gets used up first?
6 green used up
6 red left over
When one reactant runs out the reaction stops.
Excess reagent is the reagent left over when the
3.9
reaction is complete.
How many mole of NH3 can be generated by
mixing 2.35 mol N2 and 6.50 mol of H2?
N2 + 3 H2
2 NH3
1 mol N2 = 3 mol H2 = 2 mol NH3
2.35 mol N2 x 2 mol NH3
1 mol N2
=
4.70 mol NH3
6.50 mol H2 x 2 mol NH3
3 mol H2
=
4.33 mol NH3
Lower number
H2 is limiting reagent and the amount of NH3 that
can be generated is 4.33 mol
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
g Fe2O3
124 g Al x
mol Fe2O3 needed
OR
mol Al needed
mol Fe2O3
1 mol Al
27.0 g Al
x
g Fe2O3 needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
3.9
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
x 100 %
Theoretical Yield
Did you get as much as you thought
you were going to get ???
Can calculate using weight or moles?
NOTE: Law of conservation of mass dictates
that % Yield < 100 %
3.10
If the theoreticall yield is 4.33 mol NH3 and 3.62
mol NH3 was recovered, what is the % yield ?
N2 + 3 H2
% Yield =
2 NH3
Actual Yield
x 100 %
Theoretical Yield
3.62 mol NH3
4.33 mol NH3
X 100 % = 83.6 %
If 26.5 g Mg was reacted with an excess of O2
and a 75.2 % yield was achieved, how much
MgO was recovered ?
2 Mg + O2
26.5 g Mg x 1 mol Mg
24.30 g Mg
2 MgO
= 1.09 mol Mg
2 mol Mg = 2 mol MgO
1.09 mol Mg x 2 mol MgO = 1.09 mol MgO theoretical yield
2 mol Mg
1.09 mol MgO x 0.752 = 0.820 mol MgO
MW MgO = (24.30 + 16.00) g/mol = 40.30 g/mol MgO
0.820 mol MgO x 40.30 g MgO
1 mol MgO
=
33.0 g MgO