1. Empirical Formulas from Analysis When iron reacts with oxygen in the air a compound (rust) is formed that is 69.9% Fe and 30.1% O. Determine the empirical formula of rust and write a balanced chemical equation for the process. The mole ratios should be rounded to the nearest whole number OR to the nearest common fraction (1/4, 1/3, ½) and multiplied by the least common denominator.

2. Combustion analysis Experimental method for determining empirical formulas When a compound containing carbon and hydrogen is completely combusted, all the carbon in the compound is converted to CO 2 and all the hydrogen is converted to H 2 O. The CO 2 and H 2 O that is produced is weighed to determine the empirical formula of the compound. “Empirical” means “based on observation and experiment”.

3. Combustion analysis Experimental method for determining empirical formulas Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO 2 and 0.306 g H 2 O. Determine the empirical formula.

4. Combustion analysis Experimental method for determining empirical formulas Caproic acid (stinky socks) is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO 2 and 0.209 g H 2 O. What is empirical formula of caproic acid? Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

5. Quantitative Information from Balanced Equations The coefficients in a balanced chemical equation tell us the relative number of molecules (or relative number of moles) involved in the reaction. 2H 2 (g) + O 2 (g)  2H 2 O(l) 2 molecules 2(6.022x10 23 molecules) 2 mol 1 molecule 1(6.022x10 23 molecules) 1 mol 2 molecules 2(6.022x10 23 molecules) 2 mol

6. Quantitative Information from Balanced Equations 2H 2 (g) + O 2 (g)  2H 2 O(l) 2 mol H 2, 1 mol O 2, and 2 mol H 2 O are stoichiometrically equivalent quantities. 2 mol H 2  1 mol O 2  2 mol H 2 O These stoichiometric relations can be used to convert between quantities of reactants and products. 2 molecules 2(6.022x10 23 molecules) 2 mol 1 molecule 1(6.022x10 23 molecules) 1 mol 2 molecules 2(6.022x10 23 molecules) 2 mol

7. Stoichiometric Calculations 2H 2 (g) + O 2 (g)  2H 2 O(l) How many moles of H 2 O are produced from 1.57 mol of O 2 ? From stoichiometry 1 mol O 2  2 mol H 2 O

8. Stoichiometric Calculations Calculate the mass of CO 2 produced when 1.50 g of butane (C 4 H 10 ) is burned. 1. Obtain balanced chemical equation for reaction. 2C 4 H 10 (l) + 13O 2 (g)  8CO 2 (g) + 10H 2 O(g)

9. Stoichiometric Calculations Calculate the mass of CO 2 produced when 1.50 g of butane (C 4 H 10 ) is burned. 2. Use molar mass to determine the moles of butane consumed.

10. Stoichiometric Calculations Calculate the mass of CO 2 produced when 1.50 g of butane (C 4 H 10 ) is burned. 3. Use stoichiometry from the balanced chemical equation to find the moles of CO 2. 2C 4 H 10 (l) + 13O 2 (g)  8CO 2 (g) + 10H 2 O(g) 2 mol C 4 H 10  8 mol CO 2

11. Stoichiometric Calculations Calculate the mass of CO 2 produced when 1.50 g of butane (C 4 H 10 ) is burned. 4. Use molar mass to convert moles of CO 2 to grams.

12. Stoichiometric Calculations Summary of procedure A  B

13. Stoichiometric Calculations The decomposition of KClO 3 is commonly used to prepare small amounts of O 2 in the laboratory. How many grams of O 2 can be prepared from 4.50 g of KClO 3 ? 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

14. Stoichiometric Calculations Propane, C 3 H 8, is commonly used for cooking and home heating. What mass of O 2 is consumed when 5.00 g of propane is burned? What mass of CO 2 is produced when 5.00 g of propane is burned?

15. Stoichiometric Calculations The complete combustion of octane (C 8 H 18 ), a component of gasoline, proceeds as follows: 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18 H 2 O(g) How many grams of CO 2 are produced when 1.00 gallon of octane (density = 0.692 g/mL) is burned? 1 gallon =3.7854 L How many gallons of octane must be burned to produce 1.00 gallons of water (density = 1.00 g/mL)?