CHEM 103
MODULE 2 EXAM 2022
Question 1
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Show the calculation of
... [Show More] the molecular weight for the following
compounds, reporting your answer to 2 places after the decimal.
1. Al2(CO3)3
2. C8H6NO4Cl
1. 2Al + 3C + 9O = 233.99
2. 8C + 6H + N + 4O + Cl = 215.59
Question 2
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 (NH4)2CO3
2. 16.0 grams of C8H6NO4Br
1. Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole
2. Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 mole
Question 3
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 (NH4)2CO3
2. 1.04 moles of C8H6NO4Br
1. Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams
2. Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 grams
Question 4
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. (NH4)2CrO4
2. C8H8NOI
1. %N = 2 x 14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152.08
x 100 = 5.30%
%Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x
16.00/152.08 x 100 = 42.08%
2. %C = 8 x 12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261.05
x 100 = 3.09%
%N = 1 x 14.01/261.05 x 100 = 5.37% %O = 1 x
16.00/261.05 x 100 = 6.13%
%I = 1 x 126.9/261.05 x 100 = 48.61%
Question 5
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.
38.76% Ca, 19.87% P, 41.27% O
38.76% Ca / 40.08 = 0.9671 / 0.6416 = 1.5 x 2 = 3
19.87% P / 30.97 = 0.6416 / 0.6416 = 1 x 2 = 2
41.27% O / 16.00 = 2.579 / 0.6416 = 4 x 2 = 8 → Ca3P2O8
Question 6
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
1. 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O
2. 4 As + 5 O2 → 2 As2O5
3. Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4
Question 7
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Classify each of the following reactions as either:
Combination
Decomposition
Combustion
Double Replacement
Single Replacement
1. H2SO4 → SO3 + H2O
2. S + 3 F2 → SF6
3. H2 + NiO → Ni + H2O
1. H2SO4 → SO3 + H2O = Decomposition, One reactant → Two Products
2. S + 3 F2 → SF6 = Combination. Two reactants→ One product [Show Less]