M2: Exam Chem 103

M2: Exam Chem 103 - Requires Respondus LockDown Browser 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. N=28.02 H=8.064 C=12.01 O=48 add =96.09 MW 2. C= 96.08 H=6.048 N=14.01 O=64 Br=79.90 add =260.04 MW 1. 2N + 8H + C + 3O = 96.09 2. 8C + 6H + N + 4O + Br = 260.04 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(SO4)3 2. 16.0 grams of C7H5NOBr Your Answer: 1. Al2(SO4)3 Al =53.96 S= 96.21 O=192 =342.17MW moles = 13.0g/342.17 =.0380 moles 2. C7H5NOBr C=84.07 H=5.04 N=14.01 O=16 Br=79.90 =199.02 MW moles = 16g/199.02 MW =.0804 moles 1. Moles = grams / molecular weight = 13.0 / 342.17 = 0.0380 mole 2. Moles = grams / molecular weight = 16.0 / 199.02 = 0.0804 mole Question 3 Not yet graded / 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. 1.20 moles of Al2(CO3)3 2. 1.04 moles of C8H6NO4Cl Your Answer: 1. Al2(CO3)3 Al= 53.96 C= 36.03 O= 144 = 233.99 MW grams = Moles X MW g= 1.20moles X 233.99 grams 280.8 2. C8H6NO4Cl C=96.08 H=6.048 N= 14.01 O=64 Cl=35.48 MW= 215.588 215.588x 1.04 moles =224.2g 1. Grams = Moles x molecular weight = 1.20 x 233.99 = 280.8 grams 2. Grams = Moles x molecular weight = 1.04 x 215.59 = 224.2 grams 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: 1. Al2(SO4)3 Al =53.96 S= 96.21 O=192 =342.17MW Al = 53.96 / 342.17MW X100 = 15.75% S =96.21 /342.17MW X100 = 28.12% O =192/342.17MW X100 = 56.11% 2. C7H5NOBr C=84.07 H=5.04 N=14.01 O=16 Br=79.90 =199.02 MW C = 84.07/199.02 X100 =42.24% H= 5.04/199.02 X100 =2.53% N= 14.01/199.02 X100 =7.04% O=16/199.02 X100 =8.04% Br= 79.90 /199.02 X100 =40.15% 1. %Al = 2 x 26.98/342.17 x 100 = 15.77% %S = 3 x 32.07/342.17 x 100 = 28.12% %O = 12 x 16/342.17 = 56.11% 2. %C = 7 x 12.01/ 199.02 x 100 = 42.24% %H = 5 x 1.008/ 199.02 x 100 = 2.53% %N = 1 x 14.01/199.02 = 7.04% %O = 1 x 16.00/199.02 x 100 = 8.03% %Br = 79.90/199.02 x 100 = 40.15% Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 27.60% Mn, 24.17% S, 48.23% O Your Answer: Mn (0.50) 1.00 S (.75) 1.50 O (3.01) 6.00 Multiply ratio by 2 to have full numbers Mn2S3O12 27.60% Mn / 54.94 = 0.5024 / 0.5024 = 1 x 2 = 2 24.17% S / 32.07 = 0.7536 / 0.5024 = 1.5 x 2 = 3 48.23% O / 16.00 = 3. / 0.5024 = 6 x 2 = 12 → Mn2S3O12 Click this link to access the Periodic Table. This may be helpful throughout the exam. Balance each of the following equations by placing coefficients in front of each substance. 1. C6H6 + O2 → CO2 + H2O 2. As + O2 → As2O5 3. Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4 Your Answer: 1. C6H6 + O2 → CO2 + H2O C 6-12 1-12 H 6- 12 2-12 O 2-30 3-30 2C6H6 + 15 O2 → 12 CO2 + 6H2O 2. As + O2 → As2O5 As 1 → 4 2→4 O 2 →10 5→10 4 As + 5 O2 → 2 As2O5