1. 10.3 Percent Composition and Chemical Formulas 1 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Chapter 10 Chemical Quantities 10.1 The Mole: A Measurement of Matter 10.2 Mole-Mass and Mole-Volume Relationships 10.3 Percent Composition and Chemical Formulas

2. 2 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Do Now: 1.There are 35 days left in the school year. If there are 180 school days, what percentage of the school year is over? 2.The molecular formula for table sugar is C 12 H 22 O 11. What is the % Hydrogen in sugar?

3. 10.3 Percent Composition and Chemical Formulas 3 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. percent composition: percent by mass of each element in a compound Percent Composition of a Compound

4. 10.3 Percent Composition and Chemical Formulas 4 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Percent Composition from the Chemical Formula Percent Composition of a Compound % by mass of element mass of element in 1 mol compound molar mass of compound × 100% =

5. 10.3 Percent Composition and Chemical Formulas 5 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Calculate the percent composition of a) Al(OH) 3 b) K 2 S Percent Composition from the Chemical Formula Percent Composition of a Compound

6. 10.3 Percent Composition and Chemical Formulas 6 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Calculate the percent composition of a) Al(OH) 3 Al: 34.6%, O: 61.5%, H: 3.9% b) K 2 S K: 70.9%, S: 29.1% Percent Composition from the Chemical Formula Percent Composition of a Compound

7. 10.3 Percent Composition and Chemical Formulas 7 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Percent Composition from Mass Data Percent Composition of a Compound % by mass of element = mass of element mass of compound × 100%

8. 10.3 Percent Composition and Chemical Formulas 8 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. When a 14.2-g sample of HgO is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percent composition of the compound?

9. 10.3 Percent Composition and Chemical Formulas 9 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Sample Problem 10.9 % Hg = mass of Hg mass of compound × 100% = 93.0% Hg = × 100% 13.20 g 14.20 g

10. 10.3 Percent Composition and Chemical Formulas 10 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Sample Problem 10.9 % O = mass of O mass of compound × 100% 1.0 g = 7.0% O = × 100% 14.2 g

11. 10.3 Percent Composition and Chemical Formulas 11 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Check that percents add up to 100%. 93.0% + 7.0% = 100% Sample Problem 10.9 Does the result make sense?

12. 10.3 Percent Composition and Chemical Formulas 12 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Calculate the mass of nitrogen and the mass of hydrogen in 125 g of NH 3 fertilizer. Sample Problem 10.11 Calculating the Mass of an Element in a Compound Using Percent Composition

13. 10.3 Percent Composition and Chemical Formulas 13 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. 1. Calculate the % composition for each component. 14.0 g N 17.0 g NH 3 Sample Problem 10.11 2. Multiply each % composition by the mass of NH 3. 125 g NH 3 × 0.8235 = 103 g N % N = = 82.35% H = =17.65% 3.0 g H 17.0 g NH 3 125 g NH 3 × 0.1765= 22 g H

14. 10.3 Percent Composition and Chemical Formulas 14 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. The sum of the two masses equals 125 g, (103 g N + 22 g H = 125 g NH 3 ). Sample Problem 10.11 Does the result make sense?

15. 10.3 Percent Composition and Chemical Formulas 15 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Calculate the masses of hydrogen, sulfur, and oxygen in 90.0 g of H 2 SO 4.

16. 10.3 Percent Composition and Chemical Formulas 16 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Empirical Formulas CompoundMolecular Formula Empirical Formula Hydrogen PeroxideH2O2H2O2 HO WaterH2OH2OH2OH2O SucroseC 6 H 12 O 6 CH 2 O AcetyleneC2H2C2H2 CH

17. 10.3 Percent Composition and Chemical Formulas 17 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. empirical formula: lowest whole-number ratio of atoms or moles of elements in a compound. Empirical Formulas An empirical formula may or may not be the same as a molecular formula. CompoundMolecular Formula Empirical Formula Hydrogen PeroxideH2O2H2O2 HO WaterH2OH2OH2OH2O SucroseC 6 H 12 O 6 CH 2 O AcetyleneC2H2C2H2 CH

18. 10.3 Percent Composition and Chemical Formulas 18 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Ethyne (C 2 H 2 ), also called acetylene, is a gas used in welders’ torches. Styrene (C 8 H 8 ) is used in making polystyrene. Empirical Formulas

19. 10.3 Percent Composition and Chemical Formulas 19 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Empirical Formula Determination Start with percent composition 1. Assume you have 100 g of compound. The % composition will equal the mass of each element. 2. Convert each mass to moles. 3. Determine lowest whole number mole ratio Empirical Formulas

20. 10.3 Percent Composition and Chemical Formulas 20 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Sample Problem 10.12 Determining the Empirical Formula of a Compound

21. 10.3 Percent Composition and Chemical Formulas 21 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. KNOWNS percent by mass of N = 25.9% N percent by mass of O = 74.1% O Assuming there is 100 g of compound, then you would have 25.9 g N and 74.1 g O UNKNOWN empirical formula = N ? O ? Sample Problem 10.12 Analyze List the knowns and the unknown. 1

22. 10.3 Percent Composition and Chemical Formulas 22 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Convert mass to moles. 25.9 g N × 1 mol N 14.0 g N = 1.85 mol N74.1 g O × 1 mol O 16.0 g O = 4.63 mol O The mole ratio of N to O is N 1.85 O 4.63. Sample Problem 10.12 Calculate Solve for the unknown. 2

23. 10.3 Percent Composition and Chemical Formulas 23 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles. The mole ratio of N to O is N 1 O 2.5. 4.63 mol O 1.85 = 2.50 mol O 1.85 mol N 1.85 = 1 mol N Sample Problem 10.12 Calculate Solve for the unknown. 2

24. 10.3 Percent Composition and Chemical Formulas 24 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. 1 mol N × 2 = 2 mol N 2.5 mol O × 2 = 5 mol O The empirical formula is N 2 O 5. Sample Problem 10.12 Calculate Solve for the unknown. 2

25. 10.3 Percent Composition and Chemical Formulas 25 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 36.1% Ca 63.9% Cl What is the empirical formula for the compound?

26. 10.3 Percent Composition and Chemical Formulas 26 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 36.1% Ca 63.9% Cl What is the empirical formula for the compound? CaCl 2

27. 10.3 Percent Composition and Chemical Formulas 27 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 40% C6.7% H53.3% O What is the empirical formula for the compound?

28. 10.3 Percent Composition and Chemical Formulas 28 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 40% C6.7% H53.3% O What is the empirical formula for the compound? CH 2 O

29. 10.3 Percent Composition and Chemical Formulas 29 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 3.7% H44.4% C51.9% N What is the empirical formula for the compound?

30. 10.3 Percent Composition and Chemical Formulas 30 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is analyzed and found to contain: 3.7% H44.4% C51.9% N What is the empirical formula for the compound? HCN

31. 10.3 Percent Composition and Chemical Formulas 31 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. 1. Calculate the empirical formula mass. 2.Divide the molar mass by the empirical formula mass. 3.Multiply the empirical formula subscripts by the answer from step 2. Sample Problem 10.13 Finding the Molecular Formula of a Compound

32. 10.3 Percent Composition and Chemical Formulas 32 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. Sample Problem 10.13 Finding the Molecular Formula of a Compound

33. 10.3 Percent Composition and Chemical Formulas 33 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. 1. Calculate the empirical formula mass (efm). efm of CH 4 N = 12.0 g/mol + 4(1.0 g/mol) + 14.0 g/mol = 30.0 g/mol Sample Problem 10.13

34. 10.3 Percent Composition and Chemical Formulas 34 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. 2. Divide the molar mass by the empirical formula mass. Sample Problem 10.13 molar mass efm = 60.0 g/mol 30.0 g/mol = 2 3. Multiply the formula subscripts by this value. (CH 4 N) × 2 = C 2 H 8 N 2

35. 10.3 Percent Composition and Chemical Formulas 35 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. The molecular formula has the molar mass of the compound. Evaluate Does the result make sense? 3 Sample Problem 10.13

36. 10.3 Percent Composition and Chemical Formulas 36 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is composed of carbon and hydrogen and has a molar mass of 86.0 g/mol. It’s empirical formula is C 3 H 7. What is the molecular formula?

37. 10.3 Percent Composition and Chemical Formulas 37 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. A compound is composed of carbon and hydrogen and has a molar mass of 86.0 g/mol. It’s empirical formula is C 3 H 7. What is the molecular formula? C 6 H 14

38. 10.3 Percent Composition and Chemical Formulas 38 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. END OF 10.3

39. 10.3 Percent Composition and Chemical Formulas 39 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Do Now: Dry air is about 20.95% oxygen by volume. Assuming STP, how many oxygen molecules are in a 75.0 g sample of air? The density of air is 1.19 g/L.

40. 10.3 Percent Composition and Chemical Formulas 40 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Do Now: How many oxygen molecules are in a 13.2L sample of O 2 ?

41. 10.3 Percent Composition and Chemical Formulas 41 > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Do Now: 13.2 L O 2 x 1 mol O 2 x 6.02 x 10 23 molecules 22.4 L O 2 1 mol O 2 = 3.5 x 10 23 molecules O 2.