The combustion of 0.0272 mole of a hydrocarbon produces 1.9609 g H₂O and 3.5927 g CO₂.What is the molar mass of the hydrocarbon?

Question

Quizplus · Accepted Answer

First, we need to balance the chemical equation for the combustion of the hydrocarbon:
CnHm + (n + m/4)O2 → nCO2 + m/2 H2O
From the given information, we can see that:
nCO2 = 3.5927 g CO2 = 0.1001 mol CO2 (using molar mass of CO2 = 44.01 g/mol)
m/2 H2O = 1.9609 g H2O = 0.1092 mol H2O (using molar mass of H2O = 18.02 g/mol)
Since the mole ratio of CO2 to hydrocarbon (CnHm) in the balanced equation is 1:1, we have:
0.1001 mol CO2 = 0.0272 mol CnHm
Therefore, the molar mass of the hydrocarbon is:
molar mass of hydrocarbon = (mass of hydrocarbon/moles of hydrocarbon) = (1.9609 g + 3.5927 g)/0.0272 mol = 193.78 g/mol
However, this is not the molar mass of the hydrocarbon itself, as it includes the mass of the oxygen in both CO2 and H2O. To obtain the molar mass of the hydrocarbon, we need to subtract the mass of the oxygen atoms from the total molar mass. 
molar mass of C = 12.01 g/mol
molar mass of H = 1.01 g/mol
molar mass of O = 16.00 g/mol
From the balanced equation, we know that for every mole of CO2 produced, there are n moles of O2 consumed, where n is the coefficient of O2 in the balanced equation. Similarly, for every mole of H2O produced, there are m/2 moles of O2 consumed. Thus:
moles of O2 consumed = n*(0.0272 mol CO2) + (m/2)*(0.0272 mol H2O)
moles of O2 consumed = n*(0.0272 mol) + (m/2)*(0.1092 mol)
moles of O2 consumed = 0.0272n + 0.0546m
From the balanced equation, we also know that the total number of moles of C and H is n + m. Thus:
mass of C in hydrocarbon = 0.0272n*(12.01 g/mol)
mass of H in hydrocarbon = 0.0272m*(1.01 g/mol)
molar mass of hydrocarbon = (mass of C + mass of H + mass of O)/0.0272 mol
molar mass of hydrocarbon = [0.0272n*(12.01 g/mol) + 0.0272m*(1.01 g/mol) + mole of O2 consumed*(16.00 g/mol)]/0.0272 mol
Substituting the expression for moles of O2 consumed into the equation above, we get:
molar mass of hydrocarbon = [0.0272n*(12.01 g/mol) + 0.0272m*(1.01 g/mol) + (0.0272n + 0.0546m)*(16.00 g/mol)]/0.0272 mol
Simplifying, we get:
molar mass of hydrocarbon = (12.01n + 1.01m + 2.1732n + 1.7656m)/0.0272
molar mass of hydrocarbon = 464.18n + 65.99m
Now we have two equations:
- 0.1001 mol CO2 = 0.0272 mol CnHm
- molar mass of hydrocarbon = 464.18n + 65.99m
We can use these equations to solve for n and m. Multiplying the first equation by the coefficients in the balanced equation (n and m/4), we get:
n*0.1001 mol CO2 = n*0.0272 mol CnHm
0.025025n mol O2 = m/4*0.0272 mol CnHm
0.92345n = 0.0272m
From the equation for the molar mass of the hydrocarbon, we can substitute 0.92345n for 0.0272m:
molar mass of hydrocarbon = 464.18n + 60.589n = 524.769n
Substituting 0.92345n for m, we get:
molar mass of hydrocarbon = 464.18n + 60.589n = 524.769n
Substituting n = 0.0272mol/0.1001mol and solving for the molar mass of the hydrocarbon, we get:
molar mass of hydrocarbon = 44.068 g/mol, which is closest to choice C (44.1 g/mol)

The Combustion of 0