1. Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Black pearls are composed of calcium carbonate, CaCO 3. The pearls can be measured by either weighing or counting. 7 Quantitative Composition of Compounds Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

2. 7.1 The Mole 7.2 Molar Mass of Compounds 7.3 Percent Composition of Compounds 7.4 Calculating Empirical Formulas 7.5 Calculating the Molecular Formula from the Empirical Formula Chapter Outline © 2014 John Wiley & Sons, Inc. All rights reserved.

3. Individual atoms are tiny and have such a small mass, more convenient units for atoms are needed to be useful on the macroscale. Analogy Fruit in a supermarket is “counted” by weighing the mass of fruit. If the average mass for a piece of fruit is known, the number of pieces of fruit can be calculated. Example 186 g 1 orange 75 oranges = 13,950 g = 13.95 kg × If one orange has a mass of 186 g, then 75 oranges have what mass? Chemists count atoms in a similar way, by weighing. The Mole © 2014 John Wiley & Sons, Inc. All rights reserved.

4. The standard unit of measurement for chemistry. The number represented by 1 mole, 6.022 x 10 23, is also called Avogadro’s number. Such a large number is useful because even the smallest amount of matter contains extremely large numbers of atoms. 1 mole = 6.022 x 10 23 objects The mole is similar to other common units of counting. 1 dozen = 12 objects Example The Mole © 2014 John Wiley & Sons, Inc. All rights reserved.

5. Moles can be used to describe elements, particles or compounds. 1 mol of atoms = 6.022 x 10 23 atoms 1 mol of molecules = 6.022 x 10 23 molecules 1 mol of electrons = 6.022 x 10 23 electrons Mole is often abbreviated as mol. Avogadro’s number can be used as a conversion factor. 1 mol 6.022 x 10 23 objects 1 mol 6.022 x 10 23 objects The Mole © 2014 John Wiley & Sons, Inc. All rights reserved.

6. How does the mol relate to masses of elements? The atomic mass of 1 mol of any element is defined as the amount of that substance that contains the same number of particles as exactly 12 g of 12 C. 1 mol of any element contains the same number of atoms, but can vary greatly in the overall mass. (Atoms of different elements have different masses) The Mole © 2014 John Wiley & Sons, Inc. All rights reserved.

7. Molar Mass: the atomic mass of an element or compound (in grams) which contains Avogadro’s number of particles. Determining Molar Mass Convert atomic mass units on the periodic table to grams and sum the masses of the total atoms present. Example CaF 2 Molar masses are expressed to 4 significant figures in the text. Molar Mass CaF 2 = 40.08 g + 2(19.00) g = 78.08 g Ca 2 F Molar Mass © 2014 John Wiley & Sons, Inc. All rights reserved.

8. 207.2 g Pb 1 mol Pb We can use both the mol and molar mass as conversion factors. How many moles of lead does 15.0 g of Pb represent? Calculate = 7.24 x 10 -2 mol Pb 15.0 g Pb × Solution Map g Pb mol Pb The conversion factor relates g of Pb to moles of Pb. 1 mol Pb 207.2 g Pb or 207.2 g Pb 1 mol Pb (Obtain molar mass from the periodic table.) Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

9. a. 4.62 x 10 3 mol Hg b. 1.15 x 10 -1 mol Hg c. 1.15 x 10 1 mol Hg d. 4.62 x 10 -3 mol Hg 23.0 g Hg 1 mol Hg 200.6 g Hg How many moles of mercury does 23.0 g of Hg represent? Calculate = 1.15 x 10 -1 mol Hg × Solution Map g Hg mol Hg The conversion factor needed: 1 mol Hg 200.6 g Hg or 200.6 g Hg 1 mol Hg Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

10. 1 mol Au 197.0 g Au 16.0 g Au How many Au atoms are contained in 16.0 g of Au? Calculate = 4.89 x 10 22 atoms Au Solution Map g Au mol Au atoms Au Two conversion factors are needed: 1 mol Au 197.0 g Au 1 mol Au 6.022 x 10 23 atoms Au and 6.022 x 10 23 atoms Au 1 mol Au × × Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

11. a. 2.71 x 10 -25 atoms Ti b. 2.25 x 10 26 atoms Ti c. 9.81 x 10 22 atoms Ti d. 6.20 x 10 -22 atoms Ti 1 mol Ti 47.87 g Ti How many Ti atoms are contained in 7.80 g of Ti? Calculate = 9.81 x 10 22 atoms Ti 7.80 g Ti Solution Map g Ti mol Ti atoms Ti 1 mol Ti 47.87 g Ti 1 mol Ti 6.022 x 10 23 atoms Ti and 6.022 x 10 23 atoms Ti 1 mol Ti × × Two conversion factors are needed: Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

12. What is the mass of 2.13 x 10 18 atoms of Li? Calculate = 2.46 x 10 -5 g Li 2.13 x 10 18 atoms Li Solution Map atoms Li mol Li grams Li Two conversion factors are needed: 1 mol Li 6.941 g Li 1 mol Li 6.022 x 10 23 atoms Li and 6.022 x 10 23 atoms Li × 1 mol Li × 6.941 g Li Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

13. a. 4.29 x 10 -15 g Ne b. 5.35 x 10 32 g Ne c. 3.06 x 10 -17 g Ne d. 1.11 x 10 31 g Ne What is the mass of 1.28 x 10 8 atoms of Ne? Calculate = 4.29 x 10 -15 g Ne 1.28 x 10 8 atoms Ne Solution Map atoms Ne mol Ne grams Ne Two conversion factors are needed: 1 mol Ne 20.18 g Ne 1 mol Ne 6.022 x 10 23 atoms Ne and 6.022 x 10 23 atoms Ne × 1 mol Ne × 20.18 g Ne Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

14. What is the mass of 1.05 mol of Ag? Calculate = 113. g Ag 1.05 mol Ag Solution Map mol Ag grams Ag One conversion factor is needed: 1 mol Ag 107.9 g Ag 1 mol Ag × 107.9 g Ag Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

15. a. 321. g K b. 2.10 x 10 -2 g K c. 113. g K d. 1.11 x 10 12 g K What is the mass of 8.21 mol of K? Calculate = 321. g K 8.21 mol K Solution Map mol K grams K One conversion factor is needed: 1 mol K 39.10 g K 1 mol K × 39.10 g K Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

16. How many hydrogen atoms are in 1.00 moles of H 2 molecules? Calculate = 1.20 x 10 24 atoms H 2 1.00 mol H 2 Solution Map mol H 2 molecules H 2 atoms H 2 Two conversion factors are needed: 1 mol H 2 6.022 x 10 23 molecules H 2 and 6.022 x 10 23 molecules H 2 × × 1 molecule H 2 2 atoms H 1 mol H 2 1 molecule H 2 2 atoms H Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

17. a. 1.33 x 10 -23 atoms S 8 b. 7.53 x 10 22 atoms S 8 c. 4.82 x 10 24 atoms S 8 d. 2.08 x 10 -25 atoms S 8 How many sulfur atoms are in 2.27 mol of S 8 molecules? Calculate = 4.82 x 10 24 atoms S 8 1.00 mol S 8 Solution Map mol S 8 molecules S 8 atoms S 8 Two conversion factors are needed: 1 mol S 8 6.022 x 10 23 molecules S 8 and 6.022 x 10 23 molecules S 8 × × 1 molecule S 8 8 atoms S 1 mol S 8 1 molecule S 8 8 atoms S Using the Mole and Molar Mass Concepts © 2014 John Wiley & Sons, Inc. All rights reserved.

18. Molar Mass (MM): mass of one mole of the formula unit of a compound. Much like an element, molar mass can be defined for a compound. The molar mass of a compound is equal to the sum of the molar masses of all the atoms in the molecule. Example H 2 O Molar Mass = MM O + 2MM H = 16.00 g + 2(1.008 g) = 18.02 g Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

19. a. 43.99 g b. 78.00 g c. 75.99 g d. 46.00 g What is the molar mass of aluminum hydroxide, Al(OH) 3 ? Using the atomic masses of each element: 1 Al = 1(26.98 g) = 26.98 g 3 O = 3(16.00 g) = 48.00 g 3 H = 3(1.008 g) = 3.024 g Al(OH) 3 78.00 g Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

20. The molar mass of a compound contains Avogadro’s number of formula units/molecules. H2H2 OH2OH2O 2 x (6.022 x 10 23 ) H atoms 6.022 x 10 23 O atoms 6.022 x 10 23 H 2 O molecules 2 mol H atoms1 mol O atoms1 mol H 2 O molecules 2 x 1.008 g = 2.016 g H16.00 g O18.016 g H 2 O Reminder: Pay close attention to whether the desired unit involves atoms or formula units/molecules. Example Cl 2 Contains 2 mol of Cl atoms but only 1 mol of Cl 2 molecules. Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

21. As for elements, we can use both the mol and molar mass of formula units/molecules as conversion factors. How many moles of NaCl are there in 253 g of NaCl? Calculate = 4.33 mol NaCl 253. g NaCl × Solution Map g NaCl mol NaCl 1 mol NaCl 58.44 g NaCl To convert between g of NaCl and moles, we must first calculate the molar mass of NaCl. MM = 22.99 g + 35.45 g = 58.44 g NaCl Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

22. a. 0.0659 mol TiCl 4 b. 0.0321 mol TiCl 4 c. 2.37 x 10 3 mol TiCl 4 d. 1.01 mol TiCl 4 How many moles of TiCl 4 are there in 12.5 g of titanium(IV) chloride? Calculate = 0.0659 mol TiCl 4 12.5 g TiCl 4 × Solution Map g TiCl 4 mol TiCl 4 1 mol TiCl 4 189.7 g TiCl 4 MM TiCl 4 = 47.87 g + 4(35.45) g = 189.7 g TiCl 4 Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

23. What is the mass of 3.45 mol of Li 2 O? Calculate = 103. g Li 2 O 3.45 mol Li 2 O × Solution Map mol Li 2 O g Li 2 O 1 mol Li 2 O To convert between mol of Li 2 O and g, we must first calculate the molar mass of Li 2 O. MM Li 2 O = 2(6.941) g + 16.00 g = 29.88 g Li 2 O 29.88 g Li 2 O Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

24. What is the mass of 1.23 mol of PH 3 ? Calculate = 41.8 g PH 3 1.23 mol PH 3 × Solution Map mol PH 3 g PH 3 1 mol PH 3 33.99 g PH 3 MM PH 3 = 30.97 g + 3(1.008) g = 33.99 g PH 3 a. 3.62 x 10 -2 g PH 3 b. 1.22 x 10 -6 g PH 3 c. 39.33 g PH 3 d. 41.8 g PH 3 Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

25. How many molecules of H 2 S are present in 7.53 g of H 2 S? How many atoms of H are present in the sample? Calculate = 1.33 x 10 23 molecules H 2 S 7.53 g H 2 S × Solution Map g H 2 S mol H 2 S molecules H 2 S atoms H 1 mol H 2 S 34.09 g H 2 S MM H 2 S = 2(1.008) g + 32.07 g = 34.09 g H 2 S 1 mol H 2 S 6.022 x 10 23 molecules H 2 S × 1.33 x 10 23 molecules H 2 S 1 molecule H 2 S 2 atoms H × = 2.66 x 10 23 atoms H Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

26. a. 4.29 x 10 -23 molecules H 2 O 2 b. 1.55 x 10 25 molecules H 2 O 2 c. 3.95 x 10 24 molecules H 2 O 2 d. 1.34 x 10 22 molecules H 2 O 2 How many molecules of H 2 O 2 are there in 0.759 g of the compound? Solution Map g H 2 O 2 mol H 2 O 2 molecules H 2 O 2 MM H 2 O 2 = 2(1.008) g + 2(16.00) g = 34.02 g H 2 O 2 = 1.34 x 10 22 molecules H 2 O 2 0.759 g H 2 O 2 × 1 mol H 2 O 2 34.02 g H 2 O 2 1 mol H 2 O 2 6.022 x 10 23 molecules H 2 O 2 × Molar Mass of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

27. Percent = parts per 100 parts Percent composition: mass percent of each element in a compound Molar mass: total mass (100%) of a compound % composition is independent of sample size % composition can be determined by: 1. Knowing the compound’s formula or 2. Using experimental data Percent Composition of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

28. Two Step Strategy 1. Calculate the molar mass of the compound. 2. Divide the total mass of each element by the compound’s molar mass and multiply by 100. Total element mass Compound molar mass % of the element = 100 × Percent Composition from the Compound’s Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

29. Calculate the percent composition of K 2 S. Step 1 Calculate compound molar mass MM K 2 S = 2(39.10) g + 32.07 g = 110.3 g Step 2 Calculate % composition of each element. 2(39.10) g K 110.3 g % K = 100 Notice the sum of the percentages must equal 100%. This provides another way of determining the % composition of a specific element, if the other %s are known. × 32.07 g S 110.3 g % S = 100 × = 70.90 % K = 29.10 % S Percent Composition of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

30. a. 94.07 % b. 5.93 % c. 88.9 % d. 11.1% Calculate the percent composition of O in H 2 O 2. Step 1 Calculate molar mass MM H 2 O 2 = 2(1.008) g + 2(16.00)g = 34.02 g Step 2 Calculate % composition O 2(16.00) g O 34.02 g % O = 100 × = 94.07 % O Percent Composition of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

31. Calculate the % composition of K 2 CrO 4. Step 1 Calculate compound molar mass MM K 2 CrO 4 = 2(39.10) g + 52.00 g + 4(16.00) g = 194.2 g Step 2 Calculate % composition 2(39.10) g K 194.2 g % K = 100 × = 40.27 % K 52.00 g Cr 194.2 g % Cr = 100 × = 26.78 % Cr 4(16.00) g O 194.2 g % O = 100 × = 32.95 % O Percent Composition of Compounds © 2014 John Wiley & Sons, Inc. All rights reserved.

32. Two Step Strategy 1. Calculate the mass of the compound formed. 2. Divide the mass of each element by the total mass and multiply by 100. Total element mass Total compound mass % of the element = 100 × Percent Composition from Experimental Data © 2014 John Wiley & Sons, Inc. All rights reserved.

33. When heated in air, 1.63 g of Zn reacts with 0.40 g of oxygen to give ZnO. Calculate the percent composition of the compound formed. Step 1 Calculate the mass of the compound formed. Mass compound = Mass Zn + Mass O = 1.63 g + 0.40 g Step 2 Calculate % composition Total should be +/0.5% of 100 100.3% = 2.03 g compound 1.63 g Zn 2.03 g % Zn = 100 × = 80.3 % Zn 0.40 g O 2.03 g % O = 100 × = 20. % O Percent Composition from Experimental Data © 2014 John Wiley & Sons, Inc. All rights reserved.

34. a. 53.2 % b. 79.8 % c. 20.2 % d. 46.8 % Aluminum chloride forms by reaction of 13.43 g of Al with 53.18 g of chlorine. What is the percent composition of Cl in the compound? Step 1 Calculate the mass of the compound formed. Step 2 Calculate % composition Mass compound = Mass Al + Mass Cl = 13.43 g + 53.18 g = 66.61 g compound 53.18 g Cl 66.61 g % Cl = 100 × = 79.8 % Percent Composition from Experimental Data © 2014 John Wiley & Sons, Inc. All rights reserved.

35. Empirical Formula: smallest whole number ratio of atoms in a compound Molecular Formula: actual formula of a compound. Represents the total number of atoms in one formula unit of the compound. Whole number multiple of the empirical formula Example Acetylene (C 2 H 2 ) and Benzene (C 6 H 6 ) Both have the same empirical formula CH. Each compound is a multiple of CH. Acetylene C 2 H 2 = (CH) 2 Benzene C 6 H 6 = (CH) 6 Empirical and Molecular Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

36. Each compound has very different chemical and physical properties even though they share the same empirical formula. Compounds with the same empirical formula have the same percent composition. FormulaComposition % C %H Molar Mass (g/mol) CH (empirical formula)92.37.713.02 C 2 H 2 (acetylene)92.37.726.04 (2 x 13.02) C 6 H 6 (benzene)92.37.778.16 (6 x 13.02) Molar mass = molar mass of the empirical unit multiple of the unit × Empirical and Molecular Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

37. To calculate an empirical formula, you need to know: 1. The elements present in the compound 2. The atomic masses of each element (from the Periodic Table) 3. The ratio (by mass or %) of the combined elements Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

38. Strategy to Calculate an Empirical Formula: 1. Assume a starting mass of the compound (usually 100.0 g) and express the mass of each element in grams. 2. Convert g of each element to mol using molar mass. (These numbers may or may not be whole numbers.) 3. Divide each of the mole amounts from Step 2 by the smallest mole amount. The new numbers are the subscripts in the empirical formula. Special Case: If fractions are encountered, multiply by a common factor to provide whole numbers for each subscript. Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

39. Calculate the empirical formula for a compound that contains 11.19% H and 88.79% O. Step 1 Find amounts of each element In a 100.0 g sample, there are 11.19 g H and 88.79 g O Step 2 Convert g to moles using element molar masses = 11.10 mol H 11.19 g H 1 mol H × 1.008 g H = 5.549 mol O 88.79 g O 1 mol O × 16.00 g O Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

40. Step 3 Convert to whole numbers by dividing by the smallest mole amount. = 2.000 11.10 mol H 5.549 mol O Empirical formula is H 2 O = 1.000 5.549 mol O Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

41. Calculate the empirical formula for a compound that contains 56.68% K, 8.68% C and 34.73% O. Step 1 Find amounts of each element In a 100.0 g sample, there are 56.68 g K, 8.68 g C and 34.73 g O Step 2 Convert g to moles a. K 3 C 2 O 3 b. K 4 C 2 O 6 c. K 2 CO 3 d. KCO 2 = 1.447 mol K 56.69 g K 1 mol K × 39.10 g K = 0.723 mol C 8.68 g C 1 mol C × 12.01 g C = 2.171 mol O 34.73 g O 1 mol O × 16.00 g O Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

42. Calculate the empirical formula for a compound that contains 56.68% K, 8.68% C and 34.73% O. Step 3 Convert to whole numbers by dividing by the smallest mole amount. = 2.000 1.447 mol K 0.723 mol Empirical formula is: K 2 CO 3 a. K 3 C 2 O 3 b. K 4 C 2 O 6 c. K 2 CO 3 d. KCO 2 = 1.000 0.723 mol C 0.723 mol = 3.000 2.171 mol O 0.723 mol Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

43. Calculate the empirical formula for a compound that contains 2.233 g Fe and 1.926 g S? Step 1 Find amounts of each element Already provided in problem 2.233 g Fe and 1.926 S Step 2 Convert g to moles = 0.03998 mol Fe 2.233 g Fe 1 mol Fe × 55.85 g Fe = 0.06006 mol S 1.926 g S 1 mol S × 32.07 g S a. FeS 2 b.Fe 3 S 2 c. FeS d. Fe 2 S 3 Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

44. Step 3 Convert to whole numbers by dividing by the smallest mole amount. = 1.000 0.03998 mol Fe 0.03998 mol = 1.502 0.06006 mol S 0.03998 mol To get a whole number, multiply the decimal by the corresponding number in the denominator of the fraction. Example After dividing, you get 0.75 (=3/4) Multiply by the denominator 4(0.75) = 4(3/4) = 3 Common Fractions DecimalFraction 0.251/4 0.33…1/3 0.51/2 0.66….2/3 0.753/4 2 = 2.000 2 = 3.000 Empirical formula is Fe 2 S 3 × × Calculating Empirical Formulas © 2014 John Wiley & Sons, Inc. All rights reserved.

45. If molar mass is known, the molecular formula can be calculated from the empirical formula. Molecular formula is a multiple of the empirical formula. Need to determine the value of n. Solving for n Molar mass Mass of empirical formula n = = number of empirical units in the molecular formula Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

46. The molecular formula can be calculated from the empirical formula if the compound’s molar mass is known. Molecular formula = multiple of the empirical formula (EF) n = MF Determining the multiple n gives the molecular formula Molar mass Mass of empirical formula n = = number of empirical units in the molecular formula Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

47. A compound with the empirical formula NH 2 was found to have a molar mass of 32.05 g. What is the molecular formula? Molecular formula = (NH 2 ) 2 = N 2 H 4 Molar mass Mass of empirical formula n = = number of empirical units in the molecular formula 32.05 14.01 + 2(1.008) n = = 2 Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

48. a) NO 2 b)N 2 O 4 c) N 3 O6 d) N 4 O 8 A compound with the empirical formula NO 2 was found to have a molar mass of 92.00 g. What is the molecular formula? Molecular formula = (NO 2 ) 2 = N 2 O 4 92.00 g 14.01 + 2(16.00) g n = = 2 Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

49. Propylene contains 14.3 % H and 85.7 % C and has a molar mass of 42.08 g. What is its molecular formula? Plan Calculate empirical formula and then determine the molecular formula Step 1 Find compound masses Step 2 Convert g to moles = 14.2 mol H 14.3 g H 1 mol H × 1.008 g H In 100.0 g of compound, 14.3 g H and 85.7 g C = 7.14 mol C 85.7 g C 1 mol C × 12.01 g C Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

50. Step 3 Convert to whole numbers by dividing by the smallest mole amount. = 1.99 14.2 mol H 7.14 mol = 1.00 7.14 mol C 7.14 mol Empirical formula = CH 2 With EF, calculate the molecular formula Molecular formula = (CH 2 ) 3 = C 3 H 6 42.08 g 12.01 + 2(1.008) g n = = 3 Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

51. Calculate the molecular formula for a compound that contains 80.0% C and 20.0% H with a molar mass of 30.00 g. Plan Calculate empirical and then molecular formula Step 1 Find compound masses Step 2 Convert g to moles = 19.8 mol H 20.0 g H 1 mol H × 1.008 g H = 6.66 mol C 80.0 g C 1 mol C × 12.01 g C In 100.0 g of compound, 20.0 g H and 80.0 g C a. CH 3 b. CH 2 c. C 2 H 6 d. C 2 H 4 Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

52. Step 3 Convert to whole numbers by dividing by the smallest mole amount. = 2.97 19.8 mol H 6.66 mol = 1.00 6.66 mol C 6.66 mol Empirical formula = CH 3 From empirical formula, calculate the molecular formula Molecular formula = (CH 3 ) 2 = C 2 H 6 30.00 g 12.01 + 3(1.008) g n = = 2 Calculating the Molecular Formula from the Empirical Formula © 2014 John Wiley & Sons, Inc. All rights reserved.

53. Apply the concept of the mole, molar mass, and Avogadro’s number to solve chemistry problems. 7.1 The Mole Calculate the molar mass of a compound. 7.2 Molar Mass of Compounds Calculate the percent composition of a compound from its chemical composition and from experimental data. 7.3 Percent Composition of Compounds Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.

54. Determine the empirical formula for a compound from its percent composition. 7.4 Calculating Empirical Formulas Compare an empirical formula to a molecular formula and calculate a molecular formula from an empirical formula, using the molar mass. 7.5 Calculating the Molecular Formula from the Empirical Formula Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.