1. The Mathematics of Chemistry
Stoichiometry

2. 1 mole of an element is equal to its atomic mass in grams
The Mole
1 mole of an element is equal to its atomic mass in grams

3. Calculations M O L E C U S
Molar
Mass
Avogadro’s
Number G R A M S
MOLE

4. The mass in grams of 1 mole of the compound
Molar Mass
The mass in grams of 1 mole of the compound
C10 H6 O3
10 C = 10 X 12.01g =
6 H = 6 X g =
3 O = 3 X 16.00g =
TOTAL = grams

5. The mass in grams of 1 mole of the compound
Molar Mass
The mass in grams of 1 mole of the compound
CaCO3
Ca 1 X 40.0 grams = grams
C X 12.0 grams = grams
O X 16.0 grams = grams
TOTAL = grams

6. Chemical Equations CH4 + O2  CO2 + H2O Recipe for a Chemical Reaction
Relative number of Reactant and Product
Coefficients – relative numbers
CH4 + O2  CO2 + H2O

7. Balancing Chemical Equations
Atoms are Conserved in a Chemical Reaction
When an Equation is Balanced
Never change the IDENTITIES of the Reactants or Products
C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g)

8. Balancing Chemical Equations
C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g)
Count Atoms:
Carbon 2
Carbon 1
Hydrogen 6
Hydrogen 2
Oxygen 3
Oxygen 3

9. Balancing Chemical Equations
C2H5OH (l) + O2 (g)  CO2 (g) + H2O (g)
C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (g)

10. Stoichiometric Calculations The Plan
Write the Chemical Equation
Balance the Chemical Equation
Grams known  Moles known
Moles known  Moles unknown
5. Moles unknown  Grams unknown

11. Stoichiometric Calculations The Problem
Lithium hydroxide is used in an outer space environment to remove excess exhaled carbon dioxide from the living environment. The products of the reaction are lithium carbonate and water. If 48.0 grams of lithium hydroxide are used in a small scale experimental device, how much carbon dioxide will the device process?
The Plan
1. Write the Chemical Equation
2. Balance the Chemical Equation
3. Grams known  Moles known
4. Moles known  Moles unknown
5. Moles unknown  Grams unknown

12. Stoichiometric Calculations The Plan 2. Balance the Chemical Equation
1. Write the Chemical Equation
LiOH + CO2 (g)  Li2CO H2O (g)
2. Balance the Chemical Equation
2 LiOH + CO2 (g)  Li2CO H2O (g)

13. Stoichiometric Calculations The Plan
3. Grams known  Moles known
48 grams X 1 mole = moles of LiOH
24 grams
4. Moles known  Moles unknown
2 LiOH + CO2 (g)  Li2CO3 (g) + H2O (g)
2.00 mole LiOH X 1 mole CO2 = moles of CO2
2 mole LiOH

14. Stoichiometric Calculations The Plan 5. Moles unknown  Grams unknown
1.00 mole CO2 X grams = grams of CO2
1 mole

15. Stoichiometric Calculations The Problem
Lithium hydroxide is used in an outer space environment to remove excess exhaled carbon dioxide from the living environment. The products of the reaction are lithium carbonate and water. If 48.0 grams of lithium hydroxide are used in a small scale experimental device, how much carbon dioxide will the device process?
When 48.0 grams of lithium hydroxide are available for use in a reaction, 44 grams of carbon dioxide can be processed by the reaction.