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 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

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 7 of 22 14. A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula? Solution To determine the empirical formula, a relationship between percent composition and atom composition must be established. The percent composition of each element in a compound can be found either by dividing its mass by the total mass of compound or by dividing the molar mass of that element as it appears in the formula (atomic mass times the number of times the element appears in the formula) by the formula mass of the compound. From this latter perspective, the percent composition of an element can be converted into a mass by assuming that we start with a 100-g sample. Then, multiplying the percentage times 100 g gives the mass in grams of that component. Division of each mass by its respective atomic mass gives the relative ratio of atoms in the formula. From the numbers so obtained, the whole-number ratio of elements in the compound can be found by dividing each ratio by the number representing the smallest ratio. Generally, this process can be done in two simple steps (a third step is needed if the ratios are not whole numbers). Step 1: Divide each element’s percentage (converted to grams) by its atomic mass: This operation established the relative ration of carbon to hydrogen in the formula. Step 2: To establish a whole-number ratio of carbon to hydrogen, divide each factor by the smallest factor. In this case, both factors are essentially equal; thus the ration of atoms is 1 to 1: The empirical formula is CH. Since the molecular mass of the compound is 78.1 amu, some integer times the sum of the mass of 1C and 1H in atomic mass units (12.011 amu + 1.00794 amu = 13.019 amu) must be equal to 78.1 amu. To find this number, divide 78.1 amu by 13.019 amu: The molecular formula is (CH) 6 = C 6 H 6 . 15. Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula? Solution The molecular formula is a whole-number multiple of the empirical formula, and the molecular mass is a whole-number multiple of the empirical mass. The solution sequence is to determine the empirical formula, the formula mass, and then the molecular formula. Step 1: 1 1 92.3 g C: = 7.68 mol 12.011 g mol 7.7 g H: = 7.6 mol 1.00794 g mol - - 7.68 C: = 1 7.6 7.6 H: = 1 7.6 78.1 amu = 5.9989 6 13.019 amu ¾¾®

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 8 of 22 Step 2: The empirical formula is CH 2 Cl; the empirical formula mass is 49.5. Molecular mass = (empirical formula mass) (number of formula units) 99 = 49.5 no. of formula units Solve for the number of formula units: Molecular formula: 2(CH 2 Cl) = C 2 H 4 Cl 2 16. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol. Solution (2)(Mg 1.5 Si 1 H 1.5 O 4 ) = Mg 3 Si 2 H 3 O 8 (empirical formula), empirical mass of 260.1 g/unit so (2)(Mg 3 Si 2 H 3 O 8 ) = Mg 6 Si 4 H 6 O 16 17. Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: (a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (b) Saran; 24.8% C, 2.0% H, 73.1% Cl (c) polyethylene; 86% C, 14% H (d) polystyrene; 92.3% C, 7.7% H (e) Orlon; 67.9% C, 5.70% H, 26.4% N 24.3 C: = 2.02 mol 12.011 4.1 H : = 4.07 mol 1.0079 71.6 Cl: = 2.02 mol 35.453 2.02 C: = 1.0 2.02 4.07 H: = 2.0 2.02 2.02 Cl: = 1.0 2.02 ´ ´ 99 = 2 49.5 ( ) 1 mol Mg 1.153 28.03 g Mg = 1.153 mol Mg = 1.512 mol Mg 24.30 g 0.769 æ ö ç ÷ è ø ( ) 1 mol Si 0.769 21.60 g Si = 0.769 mol Si = 1.00 mol Si 28.09 g Si 0.769 æ ö ç ÷ è ø ( ) 1 mol H 1.149 1.16 g H = 1.149 mol H = 1.49 mol H 1.01 g H 0.769 æ ö ç ÷ è ø ( ) 1 mol O 3.076 49.21 g O = 3.076 mol O = 4.00 mol O 16.00 g O 0.769 æ ö ç ÷ è ø MM 520.8 = = 2.00, EM 260.1

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 10 of 22 The empirical formula is CHCl. (c) Step 1: Step 2: The empirical formula is CH 2 . (d) Step 1: Step 2: The empirical formula is CH. (e) 2.06 mol C: 2.06 mol = 1 2.0 mol H: 2.06 mol = 1 2.06 mol Cl: 2.06 mol = 1 86 g C: 12.011 g 1 = 7.2 mol mol 14 g H: - 1.00794 g 1 = 14 mol mol - 7.2 mol C: 7.2 mol = 1 14 mol H: = 1.9 2 7.2 mol » C: 92.3 g 1 mol 12.011 g ´ = 7.68 mol H: 7.7 g 1 mol 1.00794 g ´ = 7.64 mol 7.68 mol C: 7.64 mol = 1 7.64 mol H: = 1 7.64 mol

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 11 of 22 Step 1: Step 2: The empirical formula is C 3 H 3 N. 18. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye. Solution Assume 100.0 g; the percentages of the elements are then the same as their mass in grams. Divide each mass by the molar mass to find the number of moles. Step 1: Step 2: Divide each by the smallest number. The answers are 5C, 1N, and 5H. The empirical formula is C 5 H 5 N, which has a molar mass of 79.10 g/mol. To find the actual molecular formula, divide 240, the molar mass of the compound, by 79.10 to obtain 3. So the formula is three times the empirical formula, or C 15 H 15 N 3 . This resource file is copyright 2019, Rice University. All Rights Reserved. Chemistry: Atoms First 2e 6: Composition of Substances and Solutions 6.3: Molarity 67.9 g C: 12.011 g 1 = 5.65 mol mol 5.70 g H: - 1.00794 g 1 1 = 5.66 mol mol 26.4 g N: = 1.88 mol 14.0067 g mol - - 5.65 mol C: 1.88 mol = 3.01 5.66 mol H: 1.88 mol = 3.01 1.88 mol N: 1.88 mol = 1.00 1 1 1 75.95 g = 6.323 mol C 12.011 g mol 17.72 g = 1.265 mol N 14.0067 g mol 6.33 g = 6.28 mol H 1.00794 g mol - - -

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 13 of 22 ; (f) Molar mass (HCl) = 36.46 g/mol 23. Determine the molarity of each of the following solutions: (a) 1.457 mol KCl in 1.500 L of solution (b) 0.515 g of H 2 SO 4 in 1.00 L of solution (c) 20.54 g of Al(NO 3 ) 3 in 1575 mL of solution (d) 2.76 kg of CuSO 4 •5H 2 O in 1.45 L of solution (e) 0.005653 mol of Br 2 in 10.00 mL of solution (f) 0.000889 g of glycine, C 2 H 5 NO 2 , in 1.05 mL of solution Solution Make any conversions of units as necessary before setting up the problem and calculate the molar mass where necessary. (a) ; (b) ; (c) ; (d) ; (e) ; (f) 24. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C 6 H 12 O 6 , used for intravenous injection? (a) Outline the steps necessary to answer the question. (b) Answer the question. Solution (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 0.500 L contains 0.30 M × 0.500 L = 1.5 × 10 –1 mol. Molar mass (glucose): 6 × 12.0011 g + 12 × 1.00794 g + 6 × 15.9994 g = 180.158 g, 1.5 × 10 –1 mol × 180.158 g/mol = 27 g. 25. Consider this question: What is the mass of solute in 200.0 L of a 1.556- M solution of KBr? 2 7.00 mmol I = 0.070 100 mL M 1 1 mol mass (HCl) = 1.8 10 g HCl = 0.49 mol HCl 36.46 g ´ ´ 0.49 mol HCl = 6.6 0.075 L M 1.457 mol KCl = 0.9713 1.500 L M 0.515 g 98.079 g 1 mol = 0.00525 1.00 L M - 20.54 g 213.00 g 1 mol = 0.06123 1.575 L M - 3 2.76 10 g ´ 429.69 g 1 mol = 7.62 1.45 L M - 0.005653 mol = 0.5653 0.01000 L M 0.000889 g 75.0675 g 1 mol = 0.0113 0.00105 L M -

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 14 of 22 (a) Outline the steps necessary to answer the question. (b) Answer the question. Solution (a) Determine the number of moles of KBr in 200.0 L of a 1.556- M solution. Determine the formula mass of KBr. Then determine the mass of KBr from the number of moles and its formula mass. (b) mol KBr = 200.0 L × 1.556 M = 311.2 mol Mass (formula) = 39.0983 g/mol + 79.904 g/mol = 119.002 g/mol Mass (KBr) = 311.2 mol × 119.002 g/mol = 3.703 × 10 4 g 26. Calculate the number of moles and the mass of the solute in each of the following solutions: (a) 2.00 L of 18.5 M H 2 SO 4 , concentrated sulfuric acid (b) 100.0 mL of 3.8 × 10 –6 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum (c) 5.50 L of 13.3 M H 2 CO, the formaldehyde used to “fix” tissue samples (d) 325 mL of 1.8 × 10 –6 M FeSO 4 , the minimum concentration of iron sulfate detectable by taste in drinking water Solution The molarity must be converted to moles of solute, which is then converted to grams of solute: (a) ; (b) ; (c) ; (d) 27. Calculate the number of moles and the mass of the solute in each of the following solutions: (a) 325 mL of 8.23 × 10 –5 M KI, a source of iodine in the diet (b) 75.0 mL of 2.2 × 10 –5 M H 2 SO 4 , a sample of acid rain (c) 0.2500 L of 0.1135 M K 2 CrO 4 , an analytical reagent used in iron assays mol = or mol = liter liter M M ´ 2 4 mol H SO = 2.00 L 18.5 mol L ´ 2 4 2 4 = 37.0 mol H SO 37.0 mol H SO 2 4 2 4 98.08 g H SO 1 mol H SO ´ 3 2 4 = 3.63 10 g H SO ´ mol NaCN = 0.1000 L × 3.8 × 10 − 6 mol L = 3.8 × 10 − 7 mol NaCN 3.8 × 10 − 7 mol NaCN × 49.01 g 1 mol NaCN = 1.9 × 10 − 5 g NaCN 2 mol H CO = 5.50 L 13.3 mol L ´ 2 2 = 73.2 mol H CO 73.2 mol H CO 2 30.026 g 1 mol H CO ´ 2 2 = 2198 g H CO 2.20 kg H CO = 4 mol FeSO = 0.325 L 6 1.8 10 mol L - ´ ´ 7 4 7 4 = 5.9 10 mol FeSO 5.85 10 mol FeSO - - ´ ´ 4 151.9 g 1 mol FeSO ´ 5 4 = 8.9 10 g FeSO - ´

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 16 of 22 30. Calculate the molarity of each of the following solutions: (a) 0.195 g of cholesterol, C 27 H 46 O, in 0.100 L of serum, the average concentration of cholesterol in human serum (b) 4.25 g of NH 3 in 0.500 L of solution, the concentration of NH 3 in household ammonia (c) 1.49 kg of isopropyl alcohol, C 3 H 7 OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol (d) 0.029 g of I 2 in 0.100 L of solution, the solubility of I 2 in water at 20 ° C Solution (a) ; (b) ; (c) ; (d) 31. Calculate the molarity of each of the following solutions: (a) 293 g HCl in 666 mL of solution, a concentrated HCl solution (b) 2.026 g FeCl 3 in 0.1250 L of a solution used as an unknown in general chemistry laboratories (c) 0.001 mg Cd 2+ in 0.100 L, the maximum permissible concentration of cadmium in drinking water (d) 0.0079 g C 7 H 5 SNO 3 in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink. Solution (a) ; (b) ; (c) ; (d) 32. There is about 1.0 g of calcium, as Ca 2+ , in 1.0 L of milk. What is the molarity of Ca 2+ in milk? Solution 27 46 0.195 g mol C H O = = M V 27 46 C H O 386.660 g 1 3 27 46 mol C H O = 5.04 10 0.100 L M - - ´ 3 4.25 g mol NH = = M V 3 NH 17.0304 g 1 3 mol NH = 0.499 0.500 L M - 3 7 1.49 kg mol C H OH = = M V 3 7 1000 g C H OH ´ 1 kg 3 7 1 mol C H OH 60.096 g ´ = 9.92 2.50 L M 2 0.029 g mol I = = M V 2 I 253.8090 g 1 3 2 mol I = 1.1 10 0.100 L M - - ´ 293 g HCl mol HCl = M V = 1 mol × 36.4606 g HCl = 12.1 0.666 L M 3 3 2.026 g FeCl mol FeCl = = M V 3 1 mol × 162.205 g FeCl = 0.09992 0.1250 L M 2+ 0.001 g Cd mol Cd = M V = 2 0.001 g × 1 mg + 1 mol × 112.411 g Cd 2 8 = 9 × 10 0.100 L M + - 7 5 3 7 5 3 0.0079 g C H SNO mol C H SNO = = M V 7 5 3 1 mol × 183.188 g C H SNO 3 = 1.5 × 10 0.0296 L M -

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 17 of 22 33. What volume of a 1.00- M Fe(NO 3 ) 3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M ? Solution 34. If 0.1718 L of a 0.3556- M C 3 H 7 OH solution is diluted to a concentration of 0.1222 M , what is the volume of the resulting solution? Solution 35. If 4.12 L of a 0.850 M -H 3 PO 4 solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution? Solution 36. What volume of a 0.33- M C 12 H 22 O 11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M ? Solution 37. What is the concentration of the NaCl solution that results when 0.150 L of a 0.556- M solution is allowed to evaporate until the volume is reduced to 0.105 L? Solution 38. What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume? (a) 1.00 L of a 0.250- M solution of Fe(NO 3 ) 3 is diluted to a final volume of 2.00 L (b) 0.5000 L of a 0.1222- M solution of C 3 H 7 OH is diluted to a final volume of 1.250 L (c) 2.35 L of a 0.350- M solution of H 3 PO 4 is diluted to a final volume of 4.00 L 1.0 g mol = = M V 40.08 g 1 mol = 0.025 1.0 L M - 2 2 1 2 × 0.250 = 1.00 L × = 0.250 L 1.00 V M M V M M = 1 1 2 2 2 2 0.3556 mol 0.1718 L L 0.1222 mol L 0.5000 L CV V C V V = ´ = = 1 1 2 2 2 2 0.850 mol 4.12 L L 10.00 L 0.350 CV C V C M C = ´ = = 2 2 1 2 × 0.025 = 25 mL × = 1.9 mL 0.33 V M M V M M = 1 1 2 2 × 0.556 = 0.150 L × = 0.794 0.105 L V M M M M V =

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 19 of 22 Chemistry: Atoms First 2e 6: Composition of Substances and Solutions 6.4: Other Units for Solution Concentrations 44. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO 3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO 3 by mass? (a) Outline the steps necessary to answer the question. (b) Answer the question. Solution (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: %mass 1 × mass 1 = %mass 2 × mass 2 This equation can be rearranged to isolate mass 1 and the given quantities substituted into this equation. (b) 45. What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH? Solution 46. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL. Solution . The mass of pure NaOH required is . This mass of NaOH must come from the 97.0% solution: 47. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm –3 and contains 37.21% HCl by mass? Solution The solution contains 37.21% HCl by mass, and the remainder is water. Calculate the mass of the 45.0-mL sample, and then multiply by the percentage to obtain the mass of HCl. Since 1 cm 3 = 1 mL. Mass (sample) = 45.0 mL × 1.19 g/mL = 53.55 g; Mass (HCl) = 53.55 g × 0.3721 = 19.9 g 48. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO 3 , which is equivalent to milligrams of CaCO 3 per liter of water. What is the molar concentration of Ca 2+ ions in a water sample with a hardness count of 175 mg CaCO 3 /L? 2 2 1 1 %mass mass 10.0 % 400.0 g mass = = = 58.8 g %mass 68.0 % ´ ´ 4.00% mass (NaOH solution) = 15.0 g 100.0% 15.0 g mass (NaOH solution) = 375 g 0.0400 ´ ´ 1000 cm 3 1.109 g × cm 3 3 = 1.11 × 10 g ( ) 3 2 10.0% mass NaOH × 1.11 × 10 g = 1.11 × 10 g 100.0% = 2 2 97.0% mass (NaOH solution) × = 1.11 × 10 g 100.0% 1.11 × 10 g mass (NaOH solution) = = 114 g 0.970

OpenStax Chemistry: Atoms First 2e 6.1: Formula Mass and the Mole Concept Page 20 of 22 Solution Since CaCO 3 contains 1 mol Ca 2+ per mol of CaCO 3 , the molar concentration of Ca 2+ equals the molarity of CaCO 3 : 49. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream. Solution Convert 0.68 parts per billion to mass in grams: 50. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 m M is observed, what is the concentration of glucose (C 6 H 12 O 6 ) in mg/dL? Solution 1 mg/dL = 0.01 g/L and 1 L = 10 dL 5.3 mmol/L × 180.158 mg/mmol = 9.5 × 10 2 mg/L 9.5 × 10 2 mg/L × = 95 mg/dL 51. A throat spray is 1.40% by mass phenol, C 6 H 5 OH, in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution. Solution Assume 1.000 L of solution. Then 1000 mL × 0.9956 g/mL = 995.6 g 995.6 g × 1.40% = 995.6 g × 0.0140 = 13.9 g. The molar mass of phenol is: ; Since this value is the number of moles in 1 L, the molarity is 0.148 M . 52. Copper(I) iodide (CuI)is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass? 2+ 3 175 mg mol CaCO Ca = = L M 1 mol 100.0792 g ´ 1 g æ ö ´ ç ÷ è ø 1000 mg 3 = 1.75 10 1 L M - æ ö ç ÷ è ø ´ 10 9 0.68 g Hg g Hg mass Hg = 6.8 × 10 1 × 10 g solution g solution - = 10 12 1 g Hg 6.8 × 10 mol Hg g solution mol Hg = 3.4 × 10 200.59 g mol g solution - - - = 12 9 3.4 10 mol Hg 1.0 g solution 1000 mL molarity of Hg = = 3.4 10 g solution mL solution 1 L M - - ´ ´ ´ ´ 1 L 10 dL 1 6 × C = 6 × 12.011 = 72.066 6 × H = 6 × 1.00794 = 6.04764 1 × O = 1 × 15.9994 = 15.9994 94.113 g mol - 13.9 g mol phenol = 94.113 g 1 = 0.148 mol mol -