1. The Mathematics of Chemistry Stoichiometry

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

3. Calculations MOLE P A R T I C L E S GRAMSGRAMS Avogadro’s Number Molar Mass

4. Molar Mass The mass in grams of 1 mole of the compound C 10 H 6 O 3 10 C = 10 X 12.01g = 120.10 6 H = 6 X 1.00g = 6.00 3 O = 3 X 16.00g = 48.00 TOTAL = 174.1 grams

5. Molar Mass The mass in grams of 1 mole of the compound CaCO 3 Ca 1 X 40.0 grams = 40.0 grams C 1 X 12.0 grams = 12.0 grams O 3 X 16.0 grams = 48.0 grams TOTAL = 100.0 grams

6. Chemical Equations Recipe for a Chemical Reaction Relative number of Reactants and Products –Coefficients – relative numbers CH 4 + O 2  CO 2 + H 2 O

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 C 2 H 5 OH (l) + O 2 (g)  CO 2 (g) + H 2 O (g)

8. Balancing Chemical Equations C 2 H 5 OH (l) + O 2 (g)  CO 2 (g) + H 2 O (g) Count Atoms: Carbon 2 Carbon 1 Hydrogen 6Hydrogen 2 Oxygen 3

9. Balancing Chemical Equations C 2 H 5 OH (l) + O 2 (g)  CO 2 (g) + H 2 O (g) C 2 H 5 OH (l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O (g)

10. Stoichiometric Calculations 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

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 LiOH + CO 2 (g)  Li 2 CO 3 + H 2 O (g) 2. Balance the Chemical Equation. 1. Write the Chemical Equation. 2 LiOH + CO 2 (g)  Li 2 CO 3 + H 2 O (g)

13. Stoichiometric Calculations The Plan 3. Grams known  Moles known 48 grams X 1 mole = 2.00 moles of LiOH 24 grams 2.00 mole LiOH X 1 mole CO 2 = 1.00 moles of CO 2 2 mole LiOH 2 LiOH + CO 2 (g)  Li 2 CO 3 (g) + H 2 O (g) 4. Moles known  Moles unknown

14. Stoichiometric Calculations The Plan 1.00 mole CO 2 X 44.0 grams = 44.0 grams of CO 2 1 mole 5. Moles unknown  Grams unknown

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.