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Quantities in Chemical Reactions (4.1/4.5) Proportions in Compounds and Percentage Composition.

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Presentation on theme: "Quantities in Chemical Reactions (4.1/4.5) Proportions in Compounds and Percentage Composition."— Presentation transcript:

1 Quantities in Chemical Reactions (4.1/4.5) Proportions in Compounds and Percentage Composition

2 Law of Definite Proportions a specific compound always contains the same elements in definite proportions by mass, regardless of how it is synthesized compounds with the same mass proportions must be the same compound the proportions are found by calculating the percent by mass.

3 Percent by Mass (percentage composition) based on the law of conservation of mass % by mass = MASS element x 100% MASS compd MASS compd = sum of MASSES elements

4 Percent by Mass example: H 2 O made up of 2 atoms of hydrogen and 1 atom of oxygen to find percent by mass of each element: H= (mass H / mass of water) x 100% O= (mass O / mass of water) x 100%

5 Percent by Mass of H 2 O Mass % of H = mass of H (X2) X 100 mass of H 2 O =1.01u X 2 X 100 1.01u X 2 + 16.00 =2.02u X 100 18.02u =11.2% Mass % of O= 100% - 11.2% = 88.8%

6 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%

7 Practice Problems Q: How many grams of oxygen can be produced from the decomposition of 100.0 g of KClO 3 ? A: % mass O = mass O x 100% mass KClO 3 = 3(16.00)u x 100% [39.10+35.45+3(16.00)]u = 39.17%

8 Practice Problems Q: How many grams of oxygen can be produced from the decomposition of 100.0 g of KClO 3 ? A (continued): mass O= %O x mass KClO 3 = 0.3917 X 100.0g = 39.17g

9 Practice Problems Q: Two 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 are equal, then they are the same

10 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

11 Homework  Read pg. 160 – 162 & pg. 178 - 184  Finish “Percent Composition Worksheet”  pg. 184 “Section 4.5 Questions” #3 - 5


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