CHEM 103 MODULE 2 EXAM

Question 1 10 / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. (NH4)2CO3 2. C8H6NO4Br Your Answer: 1. (NH4)2CO3 =96.09grams N = 14.01 x 2 2(14.01) + 8(1.008)+3(16.0) = 96.09grams H = 1.008 x 8 C = 12.01 x 1 O = 16 x 3 = 96.09grams 2. C8H6NO4Br = 260.04grams C= 12.01 x 8 8(12.01) + 6(1.008)+(14.01) + 4(16) = 260.04grams H=1.008 x 6 N = 14.01 O = 16 x 4 Br = 79.90 Added = 260.04 1. 2N + 8H + C + 3O = 96.09 2. 8C + 6H + N + 4O + Br = 260.04 Question 2 10 / 10 pts This study source was downloaded by from CourseH on :24:57 GMT -05:00 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 1. 13.0 grams of Al2(CO3)3 2. 16.0 grams of C8H6NO4Cl Your Answer: 1. 13.0 grams of Al2(CO3)3 Mole = Grams/ MW 13.0 / 233.99 = .0555 moles 2. 16.0 grams of C8H6NO4Cl Mole = Grams/ MW 16/215.59 = .0742moles 1. Moles = grams / molecular weight = 13.0 / 233.99 = 0.0556 mole 2. Moles = grams / molecular weight = 16.0 / 215.59 = 0.0742 mole Question 3 10 / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1.05 moles of (NH4)2SO4 2. 1.18 moles of C9H9NO3 Your Answer: This study source was downloaded by from CourseH on :24:57 GMT -05:00 1. 1.05 moles of (NH4)2SO4 Grams = MW x Mole 132.154 x 1.05 = 138.8 grams 2. 1.18 moles of C9H9NO3 179.172 x 1.18 = 211.4 grams 1. Grams = Moles x molecular weight = 1.05 x 132.15 = 138.8 grams 2. Grams = Moles x molecular weight = 1.18 x 179.17 = 211.4 grams Question 4 10 / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. Al2(SO4)3 2. C7H5NOBr Your Answer: MW of Al2(SO4)3= 2(26.98) + 3(32.07) + 12(16.0) = 342.17 %Al = 53.96/342.17 x 100 = 15.77% %S = 96.21 / 342.17 x 100 = 28.12% %O= 192/342.17 x 100 = 56.11% 2. C7H5NOBr =7(12.01)+5(1.008)+(14.01)+(16)+(79.90) = 199.02 %C=84.07/199.02 x 100 = 42.24% %H= 5.04/199.02 x 100 = 2.53% %N= 14.01/199.02 x 100 = 7.04% %O= 16/199.02 x 100 = 8.04% %Br= 79.90/199.02 x 100 = 40.15%