1. Mass Relationships in Chemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Atomic mass is the mass of an atom in atomic mass units (amu)
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 = amu
16O = amu
3.1

3. Average atomic mass of lithium:
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x x 7.016
100
= amu
3.1

4. Average atomic mass (6.941)

5. Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)
3.2

6. How many atoms of He are in 5.36 moles of He?
6.022 x 1023 atoms He
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 molecules
6.022 x 1023 molecules O2
2 atoms O = x x
1 O2 molecule
1 mol O2 molecules
6.46 x 1024 atoms O

7. Molar mass is the mass of 1 mole of in grams marbles atoms
eggs
shoes
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2

8. One Mole of: S C Hg Cu Fe
3.2

9. 1 12C atom 12.00 amu 12.00 g 6.022 x 1023 12C atoms = 1.66 x 10-24 g
1 amu = 1.66 x g or 1 g = x 1023 amu M
= molar mass in g/mol
NA = Avogadro’s number
3.2

10. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
1 mol K
39.10 g K x x
6.022 x 1023 atoms K
1 mol K =
0.551 g K
8.49 x 1021 atoms K
3.2

11. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2
3.3

12. Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
HNO3 1H
1.008 amu 1N
amu 3O
+ 3 x amu
HNO3
63.01 amu
1 molecule HNO3 = amu
1 mole HNO3 = g HNO3
3.3

13. 1 Ba(NO3)2 formula unit = 261.33 amu
2 x amu
2 NO3
+ 2 x amu
- OR -
6 O
+ 6 x amu
Ba(NO3)2
amu
Ba(NO3)2
amu
1 Ba(NO3)2 formula unit = amu
1 mole Ba(NO3)2 = g Ba(NO3)2
How many atoms are in 1.00 mol of Ba(NO3)2 ?
1.00 mol Ba(NO3)2
6.022 x 1023 Ba(NO3)2
9 atoms = x x
1 Ba(NO3)2
1 mol Ba(NO3)2
5.42 x 1024 atoms
3.3

14. 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 = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 atoms H
3.3

15. Similar to JJ Thomson’s Experiment
Heavy
Similar to JJ Thomson’s Experiment
Light
Different masses deflect differently
Very small amount of sample +
Can look at atoms, molecules or fragments of molecules.
Can see isotopes of atoms.
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F = q x v x B
3.4

16. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%
3.5

17. When in doubt, convert to moles !!!
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 !!!

18. 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% % % = 100.0%
2 - Convert weight percent of each element to grams of each element g Al and g Cl
3 - Convert to moles of each element using atomic weights.
20.2 g Al = mol Al g Cl = 2.25 mol Cl
26.98 g/mol Al g/mol Cl
4 - Calculate relative ratio of elements by dividing by the smallest number of moles. => mol Al is smaller
0.749 mol Al = mol Cl = 3.01mol Cl / mol Al
0.749 mol Al mol Al
=> AlCl3
Round to 3

19. g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H
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

20. 3 ways of representing the reaction of H2 with O2 to form H2O
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

21. Must have the same number of atoms on each side of the reaction !!
reactants
products
Starting materials Substances formed
a A + b B c C + d D
plus
plus
to yield
Conservation of Mass
Must have the same number of atoms on each side of the reaction !!
3.7

22. How to “Read” Chemical Equations
2 Mg + O 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 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7

23. Balancing Chemical Equations
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
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

24. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O
3.7

25. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
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).
2C2H6 + 7O2
4CO2 + 6H2O
3.7

26. Always show a check of your solution on a quiz or exam !!
Balancing Chemical Equations
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
Always show a check of your solution on a quiz or exam !!
14 O (7 x 2)
14 O (4 x 2 + 6)
12 H (2 x 6)
12 H (6 x 2)
4 C (2 x 2)
4 C
Reactants
Products
4 C
12 H
14 O
3.7

27. Balancing Chemical Equations
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.
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.
Start by balancing those elements that appear in only one reactant and one product.
Balance those elements that appear in two or more reactants or products.
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).
Check to make sure that you have the same number of each type of atom on both sides of the equation.

28. 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 !!!
CH O CO 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 !!!

29. When in doubt, convert to moles !!!
Mass Changes in Chemical Reactions
When in doubt, convert to moles !!!
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units
3.8

30. 1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
How many mole of CO2 can be generated from moles of O2?
CH O CO H2O
1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
10.31 mol O2 x 1 mol CO = mol CO2
2 mol O2
How many grams of CH4 are needed to produce 3.85 mol of CO2?
3.85 mol CO2 x 1 mol CH4 x g CH4 =
61.8 g CH4
1 mol CO2
1 mol CH4
MW CH4 = (1.008) = g / mol CH4

31. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
3.8

32. 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 reaction is complete.
3.9

33. How many mole of NH3 can be generated by mixing 2. 35 mol N2 and 6
How many mole of NH3 can be generated by mixing 2.35 mol N2 and 6.50 mol of H2?
N H NH3
1 mol N2 = 3 mol H2 = 2 mol NH3
2.35 mol N2 x 2 mol NH = mol NH3
1 mol N2
6.50 mol H2 x 2 mol NH = mol NH3
3 mol H2
Lower number
H2 is limiting reagent and the amount of NH3 that can be generated is 4.33 mol

34. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9

35. Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
3.9

36. 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
Theoretical Yield
x 100 %
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

37. If the theoreticall yield is 4. 33 mol NH3 and 3
If the theoreticall yield is 4.33 mol NH3 and 3.62 mol NH3 was recovered, what is the % yield ?
N H NH3
% Yield =
Actual Yield
Theoretical Yield
x 100 %
3.62 mol NH3
X 100 % = %
4.33 mol NH3

38. If 26. 5 g Mg was reacted with an excess of O2 and a 75
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 + O MgO
26.5 g Mg x 1 mol Mg = mol Mg
24.30 g 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 = mol MgO
MW MgO = ( ) g/mol = g/mol MgO
0.820 mol MgO x g MgO = g MgO
1 mol MgO