Chapter 6 Solutions (1)

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 1 of 22 Chemistry: Atoms First 2e 6: Composition of Substances and Solutions 6.1: Formula Mass and the Mole Concept 1. What is the total mass (amu) of carbon in each of the following molecules? (a) CH 4 (b) CHCl 3 (c) C 12 H 10 O 6 (d) CH 3 CH 2 CH 2 CH 2 CH 3 Solution (a) 1 × 12.01 amu = 12.01 amu; (b) 1 × 12.01 amu = 12.01 amu; (c) 12 × 12.01 amu = 144.12 amu; (d) 5 × 12.01 amu = 60.05 amu 2. What is the total mass of hydrogen in each of the molecules? (a) CH 4 (b) CHCl 3 (c) C 12 H 10 O 6 (d) CH 3 CH 2 CH 2 CH 2 CH 3 Solution (a) 4 × 1.008 amu = 4.032 amu; (b) 1 × 1.008 amu = 1.008 amu; (c) 10 × 1.008 amu = 10.080 (significant figures) amu; (d) 12 × 1.008 amu = 12.096 amu 3. Calculate the molecular or formula mass of each of the following: (a) P 4 (b) H 2 O (c) Ca(NO 3 ) 2 (d) CH 3 CO 2 H (acetic acid) (e) C 12 H 22 O 11 (sucrose, cane sugar) Solution (a) 4 × 30.974 amu = 123.896 amu; (b) 2 × 1.008 amu + 15.999 amu = 18.015 amu; (c) 40.078 amu + 2 × 14.007 amu + 6 × 15.999 amu = 164.086 amu; (d) 2 × 12.011 amu + 4 × 1.008 amu + 2 × 15.999 amu = 60.052 amu; (e) 12 × 12.011 amu + 22 × 1.008 amu × 11 × 15.999 amu = 342.297 amu 4. Determine the molecular mass of the following compounds: (a) (b) (c) (d) Solution

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 2 of 22 (a) Cl 2 CO ; (b) C 2 H 2 ; (c) C 2 H 2 Br 2 ; (d) H 2 SO 4 5. Determine the molecular mass of the following compounds: (a) (b) (c) (d) Solution (a) C 4 H 8 1 1 1 1 1C 12.011 = 12.011 g mol 1O 15.9994 = 15.9994 g mol 2Cl 35.4527 = 70.9054 g mol = 98.916 g mol - - - - ´ ´ ´ 1 1 1 2C × 12.011 = 24.022 g mol 2H × 1.0079 = 2.0158 g mol = 26.038 g mol - - - 1 1 1 1 2C × 12.011 = 24.022 g mol 2H × 1.0079 = 2.0158 g mol 4O × 79.904 = 159.808 g mol = 185.846 g mol - - - - 1 1 1 1 2H × 1.0079 = 2.0158 g mol 1S × 32.066 = 32.066 g mol 2Br × 15.9994 = 63.9976 g mol = 98.079 g mol - - - -

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 3 of 22 ; (b) C 4 H 6 ; (c) H 2 Si 2 Cl 4 ; (d) H 3 PO 4 6. Which molecule has a molecular mass of 28.05 amu? (a) (b) (c) Solution (b) 2 × 12.011 amu + 4 × 1.008 amu = 28.05 amu C 2 H 4 This resource file is copyright 2019, Rice University. All Rights Reserved. Chemistry: Atoms First 2e 6: Composition of Substances and Solutions 6.2: Determining Empirical and Molecular Formulas 7. What information is needed to determine the molecular formula of a compound from the empirical formula? Solution After determining the empirical formula, additional information such as the molar mass, or the moles of an element per mole of the compound, must be given. 8. Calculate the following to four significant figures: (a) the percent composition of ammonia, NH 3 (b) the percent composition of photographic fixer solution (“hypo”), Na 2 S 2 O 3 4C 12.011 48.044 amu 8H 1.0079 8.06352 amu 56.107 amu ´ = ´ = = 4C × 12.011 = 48.044 amu 6H × 1.0079 = 6.0474 amu = 54.091 amu 2H × 1.0079 = 2.01558 amu 2Si × 28.0855 = 56.1710 amu 4Cl × 35.4527 = 141.8108 amu = 199.9976 amu 3H 1.0079 = 3.0237 amu 1P 30.973762 = 30.973762 amu 4O 15.9994 = 63.9976 amu = 97.9950 amu ´ ´ ´

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 4 of 22 (c) the percent of calcium ion in Ca 3 (PO 4 ) 2 Solution In each of these exercises asking for the percent composition, divide the molecular weight of the desired element or group of elements (the number of times it/they occur in the formula times the molecular weight of the desired element or elements) by the molecular weight of the compound. (a) ; (b) ; (c) 9. Determine the following to four significant figures: (a) the percent composition of hydrazoic acid, HN 3 (b) the percent composition of TNT, C 6 H 2 (CH 3 )(NO 2 ) 3 (c) the percent of in Al 2 (SO 4 ) 3 Solution (a) ; (b) ; (c) 10. Determine the percent ammonia, NH 3 , in Co(NH 3 ) 6 Cl 3 , to three significant figures. 1 1 1 1 1 1 14.0067 g mol × 100% 14.0067 g mol % N = = = 82.24% [3(1.007940 + 14.0067)] g mol 17.0305 g mol 3 × 1.00794 g mol % H = × 100% = 17.76% 17.0305 g mol - - - - - - 2 × 22.989768 45.9795 % Na = × 100% = × 100 = 29.08% 2 × 22.989768 + 2 × 32.066 + 3 × 15.9994 158.1097 64.132 % S = × 100% = 40.56% 158.1097 47.9982 % O = × 100% = 30.36% 158.1097 2 3 × 40.078 120.234 % Ca = × 100% = × 100% = 38.76% 3 × 40.078 + 2 × 30.973762 + 8 × 15.9994 310.1816 + 2 4 SO - 1.008 % H = × 100 = 2.34% 43.029 42.021 % N = × 100 = 97.66% 43.029 84.077 % C = × 100 = 37.01% C 227.132 5.040 % H = × 100 = 2.219% H 227.132 95.994 % O = × 100 = 42.26% O 227.132 42.021 % N = × 100 = 18.50% N 227.132 æ ö ç ÷ è ø æ ö ç ÷ è ø æ ö ç ÷ è ø æ ö ç ÷ è ø 2 4 3(32.066 + 4 15.999) 100% 288.186 100% % SO = = 84.23% 2 26.982 + 3(32.066 + 4 15.999) 342.15 - ´ ´ ´ = ´ ´

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 5 of 22 Solution 11. Determine the percent water in CuSO 4 •5H 2 O to three significant figures. Solution 12. Determine the empirical formulas for compounds with the following percent compositions: (a) 15.8% carbon and 84.2% sulfur (b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen Solution (a) The percent of an element in a compound indicates the percent by mass. The mass of an element in a 100.0-g sample of a compound is equal in grams to the percent of that element in the sample; hence, 100.0 g of the sample contains 15.8 g of C and 84.2 g of S. The relative number of moles of C and S atoms in the compound can be obtained by converting grams to moles as shown. Step 1: Step 2: The empirical formula is CS 2 . (b) Step 1: Step 2: 3 6(14.007 3 × 1.008) 102.186 % NH = × 100% = × 100% = 38.2% 58.933 + 6(14.007 + 3 × 1.008) + 3(35.453) 267.478 + 2 5(2 1.008 + 15.999) % H O = 63.546 + 32.066 + 4(15.999) + 5(2 1.008 + 15.999) 90.075 90.075 = = 100% = 36.1% 159.608 + 90.075 249.683 ´ ´ ´ 1 mol C: 15.8 g × = 1.315 mol 12.011 g 1 mol S: 84.2 g × = 2.626 mol 32.066 g 1.315 mol C: = 1.000 1.315 mol 2.626 mol S: = 1.997 1.315 mol 1 mol C: 40.0 g × = 3.330 mol 12.011 g 1 mol H: 6.7 g × = 6.647 mol 1.00794 g 1 mol O: 53.3 g × = 3.331 mol 15.9994 g