1. Mathematics of Chemistry
Topic 3
Mathematics of Chemistry
aka Stoichiometry

2. Topic 3-Outline
Formula mass/Gram-formula mass
Molecular Mass/Gram-molecular mass
Percent Composition
-hydrates
Empirical and Molecular Formulas
Mole
grams
moles
liters
Math of Equations
a. mole-mole problems
b. gram-gram problems
c. volume-volume problems
d. mixed problems

3. A term used to refer to the mass of a mole of any substance
Definitions to Know
Molar mass-
A term used to refer to the mass of a mole
of any substance
aka: formula mass/gram-formula mass
molecular mass/gram-molecular mass
Mole-
The amount of a substance that contains
6.02 x 1023 (Avogadro’s #)
representative particles of that substance
Just like 1 dozen = 12 representative particles
1 baker’s dozen = 13 representative particles
1 gross = 144 representative particles
1 ream = 500 representative particles
1 mole = 6.02 x 1023 representative particles

4. How do we calculate a molar mass?
C6H12O6
1. Make sure the formula is written correctly
2. Determine the number of atoms of each element.
Carbon (C)-6 atoms total
Hydrogen (H)-12 atoms total
Oxygen (O)-6 atoms total
3. Multiply the # of atoms by the mass #
C = 6 atoms x 12 = 72
H = 12 atoms x 1 = 12
O = 6 atoms x 16 = 96
4. Add these numbers and this is the molar mass
= 180 g/mole

5. Let’s try: (NH4)2SO4
CH3COOH
N = 2 atom x 14 = 28
C = 2 atoms x 12 = 24
H = 4 atoms x 1 = 4
H = 8 atoms x 1 = 8
O = 2 atoms x 16 = 32
S = 1 atom x 32 =32
O = 4 atoms x 16 = 64
Total = 60 g/mol
Total = 132 g/mol
CuSO4.5H2O
Ca(NO3)2
Cu = 1 atom x 64 = 64
Ca = 1 atom x 40 = 40
S = 1 atom x 32 = 32
N = 2 atoms x 14 = 28
O = 9 atoms x 16 = 144
O = 6 atoms x 16 = 96
H = 10 atoms x 1 = 10
Total = 250 g/mol
Total = 164 g/mol

6. Percent Composition-see Table T
% element = mass of element in compound x 100
total mass of compound
Ex:
H2SO4
% H = x = 2.04 % H 98
% S = x 100 = % S 98
% O = x 100 = % O 98

8. Let’s try: (NH4)2SO4
% N = x 100 = % N
132
% H = x 100 = 6.06 % H
132
% S = x 100 = % S
132
% O = x 100 = % O
132

9. Let’s try it with information presented a different way!
How much gold can be obtained from 500 g of AuCl?
1. Find the percent of Au in the compound AuCl.
% Au = 197 x 100 = % Au
232
Nothing has changed so far but now here is the
additional step:
Multiply the mass of the sample by the % of element
and divide by 100
(500 g) (84.91)/100 = g Au

10. You must be able to do this with the information presented
Percent of Water in a Hydrate
You must be able to do this with the information presented
one of two ways:
Formula
2. Lab Data
Formula: CoCl2 . 6H2O
%H2O = x100 = % H2O
237

11. Lab Data: Mass of Hydrate = 5.00 g
Mass of Anhydrous = 3.61 g
% H2O = ?
%H2O = x 100 = 27.8 % H2O 5
Mass of water = 5.00 g – 3.61 g = 1.39 g H2O
Anhydrous-with all water removed
Na2SO4.10H2O-hydrated salt
Na2SO4 - anhydrous

12. A formula with the lowest whole number ratio of elements in a compound
Empirical Formula
A formula with the lowest whole number ratio of elements
in a compound
You must be able to calculate an empirical formula
A substance is made up of 94.1 % oxygen and 5.9 %
hydrogen. What is its empirical formula?
1. Divide by the mass number of each element
H = = 5.9 1
O = = 5.88 16

13. 2. Divide by the smaller number obtained in step #1.
5.88
H = = 1
5.88
These whole numbers are the subscripts: therefore: empirical formula = OH
1,6 diaminohexane is used to make nylon. What is
the empirical formula of this compound if it is
62.1% carbon, 13.8% hydrogen & 24.1% nitrogen?
C= = 5.18 12
H = = 13.8 1
N = = 1.72 14
5.18 = 3.01
1.72
13.8 = 8.02
1.72
1.72 = 1
1.72
C3H8N
ANSWER:

14. What if the numbers obtained in Step 2 are not whole numbers?
A compound is analyzed and found to contain 25.9 % nitrogen and 74.1 % oxygen. What is the empirical formula of the compound?
O = 74.1 = 4.63 16
1. N = 25.9 = 1.85 14
N = = 1
1.85
O = 4.63 = 2.50
1.85
NO2.5 but…definition of empirical formula states
WHOLE NUMBERS…this cannot be
the answer
So…what do we do?

15. Multiply the subscripts by the LOWEST possible number until we get a whole number….
These are our new
subscripts
N x 2 = 2
2.5 O x 2 = 5
ANSWER: N2O5

16. Remember: a molecular formula is the true ratio of atoms in a compound
How to determine a molecular formula
Remember: a molecular formula is the true ratio of atoms in a compound
Find the compound’s empirical formula
Divide the molar mass by the mass of the EF
Multiply the EF by the number obtained in Step 2.
This is your molecular formula

17. A compound is made up of 94.1 % O and 5.9 % H.
Its molar mass is 34 g. What is its empirical and molecular formula?
1. Find the empirical formula
O = 94.1/16 = 5.88
H = 5.9/1 = 5.9
Divide by the smaller number from step 1
O = 5.88/5.88 = 1
H = 5.9/5.88 = 1
EF = OH

18. Find mass of EF
OH = 17 g
Divide the molar mass (from problem) from the empirical formula mass (calculated)
34/17 = 2
This must be a whole number
Multiply the EF by the whole number obtained
OH x 2 = O2H2

19. The compound methyl butanoate smells like apples. Its
percent composition is 58.8 % C, 9.8 % H and 31.4 % O and
its molar mass is 102 g/mol. What is its empirical formula?
What is its molecular formula?
Has to be a whole number
so lets multiply
C = 2.5 x 2 = 5
H = 5 x 2 = 10
O = 1 x 2 = 2
EF = C5H10O2
C = 58.8/12 =4.9
H = 9.8/1 = 9.8
O = 31.4/16 = 1.96
Divide by smallest
C = 4.9/1.96 = 2.5
H = 9.8/1.96 = 5
O = 1.96/1.96 = 1
Find mass of EF
C = 5 atoms x 12 = 60
H = 10 atoms x 1 = 10
O = 2 atoms x 16 = 32
3. EF = C2.5H5O
6. Mass of EF = = 102 g

20. Here’s the new step:
Divide the molar mass (given in the problem) by
The mass of the EF (which you just calculated)
102/102 = 1
Multiply the EF by the number just obtained
C5H10O2 x 1 = C5H1OO2
C5H10O2 is both the empirical and molecular formula

21. Another way to do it comes from the fact that they give
you the empirical formula and the molar mass and you
have to find the molecular formula. These are real easy!
What is the molecular formula of the compound if its empirical formula is CH2O and its molar mass is 90 g/mol?
You have the empirical formula…you don’t have to
figure it out so just determine its mass.
C = 1 atom x 12 = 12
H = 2 atoms x 1 = 2
O = 1 atom x 16 = 16
Total = 30 g
2. Divide molar mass/EF mass
90/30 = 3
3. Multiply EF formula x number from Step 2
4. (CH2O) x 3 = C3H6O3 …this is molecular formula

22. Mole Island
/GMM
/22.4
Liters
Moles
Grams
X 22.4
X GMM
Multiply to get off Mole Island
Divide to go onto Mole Island

23. How many moles are in 100 g of C6H12O6?
START WITH WHAT YOU KNOW!!!!!!!!!!!!!
100 g C6H12O6 x 1mole C6H12O6 = .56 mole
180 g
How many liters are in 45 g of H2O
45 g H2O x 1 mol H2O x 22.4 L H2O = 56 L H2O
18 g mol

24. Stoichiometry-Math of Chemistry
There are 4 types of problems:
1. mole-mole
2. mass-mass
3. volume-volume
4. mixed
The key to any math problem in chemistry is to always get to the mole. This is the only means of comparison!

25. Mole-Mole Problems
How many moles of NaCl can be produced when 3 moles of Na reacts with chlorine?
1. Always write a balanced equation:
2Na + Cl2  2NaCl
2. Start with what you know
3 mol Na x 2 mol NaCl =
2 mol Na
3 mol NaCl

26. How many moles of oxygen can be made from the decomposition of 15 moles of KClO3?
1. Write a balanced equation
2KClO3  2KCl + 3O2
2. Start with what you know…..
15 mol KClO3 x 3 mol O2
2 mol KClO3
= 22.5 mol O2

27. How many grams of iron (III) chloride can be made from
Mass-Mass Problems
Now you are given a mass (grams) and asked to find an answer in mass (grams)
How many grams of iron (III) chloride can be made from
the reaction of 100 g of iron with chlorine?
1. Write a balanced equation
2Fe + 3Cl2 2 FeCl3
Start with what you know….but now you have to get into moles before you can do the math!
100 g Fe x 1 mol Fe x 2 mol FeCl3 x 161 g =
56 g Fe mol Fe mol FeCl3
287.5 g
of FeCl3

28. How many grams of acetylene are produced by adding water to 5
How many grams of acetylene are produced by adding water to 5.0 g of CaC2?
1. Write a balanced equation:
CaC2 + 2H2O  C2H2 + Ca(OH)2
2. Start with what you know…..
5 g CaC2 x 1 mol CaC2 x 1 mol C2H2 x 26 g =
64 g mol CaC2 1 mol C2H2
2.03 g Of
C2H2

29. What volume of CO reacting with oxygen, would
Volume-Volume Problems
Now the information is given in a volume unit and the
answer is wanted in a volume unit. Remember…you still
must enter the mole to solve any math question!!!!!!!!!!
What volume of CO reacting with oxygen, would
produce 200 liters of CO2?
1. Start with a balanced equation
2CO + O2  2CO2
2. Start with what you know!!!
200 L CO2 x 1 mol CO2 x 2 mol CO x 22.4 L O2 =
22.4 L CO2 2 mol CO2 1 mol O2
200 L of CO

30. Lets try another
How many liters of SO2 are needed to react with 216 liters of O2 to form SO3?
1. Write a balanced equation:
2SO2 + O2  2SO3
2. Start with what you know…………….
216 L O2 x 1 mol O2 x 2 mol SO2 x 22.4 L SO2 =
22.4 L O2 1 mol O mol SO2
432 L of
SO2

31. Now…they give you information in moles, mass or
Mixed Problems
Now…they give you information in moles, mass or
volume and want an answer in either moles, mass or volume
Calcium carbonate reacts with hydrochloric acid
to produce carbon dioxide, calcium chloride and water.
What volume of carbon dioxide will be produced when
80 g of calcium carbonate reacts?
1. Write a balanced equation
CaCO3 + 2HCl  CO2 + CaCl2 + H2O
2. Start with what you know
80 g CaCO3 x 1 mol CaCO3 x 1 mol CO2 x 22.4 L CO2 =
100 g CaCO mol CaCO3 1 mol CO2
17.92 L of CO2

32. Lets try another
For the reaction: 2Na + 2H2O  2NaOH + H2, suppose
23 grams of sodium reacts. How many moles of hydrogen will be produced?
1. Already balanced….HURRAH!
2. Start with what you know….
23 g Na x 1 mol Na x 1 mol H2 =
23 g Na mol Na
.5 mole H2