1. Law of Definite Proportions Regardless of the amount, a compound is always composed of the same elements in the same proportion by mass. The proportions are found by calculating the percent by mass.

2. Percent by Mass Based on the law of conservation of mass % by mass = MASS element x 100% MASS compd MASS compd = sum of MASSES elements

3. Percent by Mass Example on page 75 Sucrose = Carbon, hydrogen, oxygen To find percent by mass of each element: C= (mass C / mass of sucrose) x 100% H= (mass H / mass of sucrose) x 100% O= (mass O / mass of sucrose) x 100%

4. Percent by Mass (20g of sucrose) ElementAnalysis by mass (g) Percent by mass (%) Carbon8.48.4 x 100 = 42% 20.0 Hydrogen1.31.3 x 100 = 6.5% 20.0 Oxygen10.310.3 x 100 = 51.5% 20.0 Total (Sucrose) 20.0100%

5. Percent by Mass (500g sucrose) ElementAnalysis by mass (g) Percent by mass (%) Carbon210.5210.5 x 100 = 42% 500.0 Hydrogen32.432.4 x 100 = 6.5% 500.0 Oxygen257.1257.1 x 100 = 51.5% 500.0 Total (Sucrose) 500.0100%

6. Law of Definite Proportions Therefore, mass percentages of elements in a compound do NOT depend on amount. Compounds with the same mass proportions must be the same compound

7. Practice Problems Q: A 78.0 g sample of an unknown compound contains 12.4g of hydrogen. What is the percent by mass of hydrogen in the compound? A: % Mass H = mass H x 100% mass comp = 12.4g x 100% 78.0g = 15.9%

8. Practice Problems Q: If 3.5 g of X reacts with 10.5g of Y to form the compound XY, what is the percent by mass of X in the compound? A: % Mass X = mass X x 100% mass XY = 3.5g x 100% (3.5 + 10.5)g = 25%

9. Practice Problems Q: If 3.5 g of X reacts with 10.5g of Y to form the compound XY, how many grams of Y would react to form XY 2 ? A: Mass Y XY = 2 ( Mass Y XY 2 ) = 2 ( 10.5g) = 21.0 g

10. Practice Problems Q: 2 unknown compounds are tested. Compound 1 contains 15.0g of hydrogen and 120.0g oxygen. Compound 2 contains 2.0g of hydrogen and 32.0g oxygen. Are the compounds the same? HINT!! If % Masses =, then they are the same

11. Practice Problems A: Compd 1- %H = [15.0 / (15.0+120.0)] x 100% = 11.1% %O = [120.0 / (15.0+120.0)] x 100% = 88.9% Compd 2- %H = [2.0 / (2.0+32.0)] x 100% = 5.9% %O = [32.0 / (2.0+32.0)] x 100% = 94.1% NOT THE SAME COMPOUNDS

12. More Practice Complete #s 10 – 13 on page 28-29 of Solving Problems: A Chemistry Handbook.

13. More Practice (ANSWERS) 10. a) % Bromine = 72.9% b) Total = 100% 11. % Hydrogen = 4.48% % Carbon = 60.00% % Oxygen = 35.52% 12. % Sulfur = 33.6% % Copper = 66.4% 13. Mass of oxygen = 14.6g

14. Handout Key 1. Since the mass percentages in the unknown compound are the same as in sucrose, the compound must be sucrose. 2. Since amount does NOT affect mass percentages, the mass percentage of carbon is 42.40% in 5, 50, and 500 g of sucrose. 3. 51.30% = [Mass O / 50.00g sucrose] x 100% Mass O = [51.30 / 100] x 50.00 = 25.65g 4. 42.40% = [Mass C / 100.0g sucrose] x 100% Mass C = [42.40 / 100] x 100.0 = 42.40g 5. 6.50% = [Mass H / 6.0 g sucrose] x 100% Mass H = [6.50 / 100] x 6.0 = 0.39g 7. % Cl= [12.13g Cl / 20.00g salt] x 100% = 60.65% % Na= [7.87g Na / 20.00g salt] x 100% = 39.4%