Chem1000B
Fall 2003
Assignment 5 - Solutions
Two or three questions will be marked from each assignment.
DUE ON Oct. 12, 2003 (Friday) 1:00 PM
To be dropped off at my office (C886)
1. A sample of 1.053 g benzoic acid (CxHyOz) is burned in oxygen and produces 0.466 g water and
2.656 g of carbon dioxide. What is the empirical formula of benzoic acid? The molar mass of benzoic
acid was determined to be 122.1 g/mol. What is the molecular formula for benzoic acid?
M(CO2) = 44.0098 g/ mol; M(H2O) = 18.01534 g/mol
n(H2O) = 0.466 g/(18.01534 g/mol) =0.0259 mol
n(CO2) = 2.656 g/(44.0098 g/ mol) = 0.06035 mol
n(H) = 0.0259 mol H2O × 2 mol H/1 mol H2O = 0.0518 mol H
n(C) = 0.06035 mol CO2× 1 mol C/1 mol CO2 =0.06035 mol C
m(H) = 0.0518 mol ×1.0079 g/mol =0.0522 g
m(C) = 0.06035 mol × 12.011 g/mol =0.7249 g
m(O) = m(benzoic acid = CxHyOz) – m(H) – m(C) = 1.053 g -0.0522 g – 0.7249 g = 0.276 g
n(O) = 0.276 g/(15.999 g/mol) = 0.0172 mol
n(C) : n(H) : n(O) = 0.0604 mol: 0.0518 mol : 0.0172 mol
= 3.51 : 3.01 : 1.00
≈7:6:2
empirical formula: C7H6O2
molar mass of C7H6O2: M(C7H6O2) = 122.122 g mol-1
experimental molar mass of benzoic acid: M(benzoic acid) = 122.1 g mol-1
molecular formula of benzoic acid: C7H6O2
2. A sample of 0.523 g naphthalene, a hydrocarbon, is burned in oxygen and produces 1.795 g of
carbon dioxide and 0.294 g of water. What is the empirical formula? The molar mass of naphthalene
is 128.2 g/mol. What is the molecular formula?
M(CO2) = 44.0098 g/ mol; M(H2O) = 18.01534 g/mol
Since naphthalene is a hydrocarbon, it contains only carbon and hydrogen.
n(H2O) = 0.294 g/(18.01534 g/mol) =0.0163 mol
n(CO2) =1.795 g/(44.0098 g/ mol) = 0.0408 mol
n(H) = 0.0163 mol H2O × 2 mol H/1 mol H2O = 0.0326 mol H
n(C) = 0.0408 mol CO2× 1 mol C/1 mol CO2 =0.0408 mol C
n(C) : n(H)= 0.0408 mol: 0.0326 mol = 1.25 : 1.00 ≈ 5 : 4
empirical formula: C5H4
molar mass of C5H4: M(C5H4) = 64.087 g mol-1
experimental molar mass of naphthalene: M(naphthalene) = 128.2 g mol-1
M(naphthalene)/ M(C3H4) = 128.2 g mol-1/64.087 g mol-1 = 2.000
molecular formula of naphthalene: C10H8
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3. Balance the complete combustion reactions of the following organic compounds with excess
oxygen.
(a) 1-propanol, C3H7OH
2C3H7OH + 9O2 → 6CO2 + 8H2O
(b) acetylene, HC≡CH
2HC≡CH + 5O2 → 4CO2 + 2H2O
(c) phenol C6H5OH
C6H5OH + 7O2 → 6CO2 + 3H2O
4. Balance the following reaction equation. The product OsO3FSb3F16 contains OsO3F+ cations and
Sb3F16- anions which have close contacts in the solid state. You want to prepare exactly 100 mg of
OsO3FSb3F16. How much (in mg) of the reactants do you need?
OsO3F2(s) + 3SbF5(l)
→ OsO3FSb3F16(s)
M(OsO3F2) = 276.23 g/ mol; M(SbF5) = 216.752 g/mol; M(OsO3FSb3F16) = 926.48 g/mol
n(OsO3FSb3F16) = 0.1 g /926.48 gmol-1 = 1.0794 × 10-4 mol
n(OsO3F2) = 1.0794 × 10-4 mol OsO3FSb3F16 × 1 mol OsO3F2/1 mol OsO3FSb3F16
= 1.0794 × 10-4 mol OsO3F2
n(SbF5) = 1.0794 × 10-4 mol OsO3FSb3F16 × 3 mol SbF5/1 mol OsO3FSb3F16
= 3.2381 × 10-4 mol SbF5
m(OsO3F2) = 1.0794 × 10-4 mol ×276.23 g/ mol =0.029815 g = 29.815 mg
m(SbF5) = 3.2381 × 10-4 mol ×216.752 g/mol =0.070186 g =70.186 mg
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