Chapter 7

- Chapter 7 -52. 1 mol Ag ¼ 107.9 g Ag1cm 3 (a) ð107:9gAgÞ¼ 10:3cm 3 ðvolume of cubeÞ10:5g(b) 10:3cm 3 ¼ volume of cube ¼ ðsideÞ 3side ¼3p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10:3cm 3 ¼ 2:18 cm53. (a) Determine the molar mass of each compound.CO 2 , 44.01 g; O 2 , 32.00 g; H 2 O, 18.02 g; CH 3 OH, 32.04 g. The 1.00 gram sample with the lowestmolar mass will contain the most molecules. Thus, H 2 O will contain the most molecules. 1 mol ðÞ6:022 3 10(b)atoms ð1:00 g H 2 OÞ¼ 1:00 10 23 atoms18:02 gmol 1 mol ðÞ6:022 6 10 23 atoms ð1:00 g CH 3 OHÞ¼ 1:13 10 23 atoms32:04 gmol 1 mol ðÞ6:022 3 10 23 atoms ð1:00 g CO 2 Þ¼ 4:10 10 22 atoms44:01 gmol 1 mol ðÞ6:022 2 10 23 atoms ð1:00 g O 2 Þ¼ 3:76 10 22 atoms32:00 gmolThe 1.00 g sample of CH 3 OH contains the most atoms54. 1 mol Fe 2 S 3 ¼ 207.9 g Fe 2 S 3 ¼ 6.022 10 23 formula units6:022 10 23 atoms 1 formula unit 207:9gFe 2 S 35 atoms 6:022 10 23 formula units55. The conversion is g P ! mol P ! mol Ca ! gCa 1 mol P 3 mol Ca 40:08 g Cað1:00 g PÞ¼ 1:94 g Ca30:97 g P 2 mol P 1 mol Ca1.94 g Ca combines with 1.00 g P.56. Grams of Fe per ton of ore that contains 5% FeSO 4 .¼ 41:58 g Fe 2 S 3The conversion is: ton ! lb ! g ! g FeSO 4 ! gFe 2000 lb 453:6g55:85 g Fe4ð1:0 tonÞð0:05 FeSO 4 Þ¼ 1.7 10 g Feton lb151:9 g FeSO 41.0 ton of iron ore contains 2 10 4 g Fe.57. From the formula, 2 Li (13.88 g) combine with 1 S (32.07 g). 13:88 g Lið20:0gSÞ ¼ 8:66 g Li32:07 g S-68-