% composition of a mixture

Question

A mixture of propane and methane occupied 122 mL at 25⁰C and 1.0 atm. This mixture was totally burned in excess O2 to produce CO2(g) and H2O(g). The mass of the CO2produced was 0.506 g. What is the mass % composition of the original mixture? Show all work.

Answer 1

CH₄ + 2O₂ = CO₂ + 2H₂O ... combustion of methane

C₃H₈ + 5O₂ = 3CO₂ + 4H₂O ... combustion of propane

Ideal gas law: PV = nRT and solve for n (moles) of CO₂

n = PV/RT = (1 atm)(0.122 L)/(0.0821 Latm/Kmol)(298K) = 0.0050 moles CH₄ + moles C₃H₈

Let x = moles of CH₄ and y = moles of C₃H₈

x+y = 0.0050 moles of gas in the mixture

0.506 g CO₂ x 1 mol CO₂/44 g = 0.0115 moles CO₂ produced from the mixture

x + 3y = 0.0115

y = 0.005 - x

x + 0.015 - 3x = 0.0115

-2x = -0.0035

x = 0.00175 moles CH₄

y = 0.00325 moles C₃H₈

mass CH₄ = 0.00175 mol x 16 g /mol = 0.028 g CH₄

mass C₃H₈ = 0.00325 mol x 44 g/mol = 0.143 g C₃H₈

% CH₄ = 0.028 / 0.171 (x100%) = 16.4% CH₄

% C₃H₈ = 0.143 g / 0.171 g (x1005) = 83.6% C3H₈

Answer 2

The complete reaction of CH4 with 2 O2 gives 1 CO2 + 2 H2O for every CH4.

The complete reaction of C3H8 with 5 CO2 gives 3 CO2 + 4 H2O for every C3H8.

The sum of the reactions is CH4 + C3H8 + 7 O2 ---> 4 CO2 + 6 H2O

As 0.506 g of CO2 is produced, this corresponds to 0.506 g/ 44.01 g/mole = 0.0115 moles CO2.

Of this, 1/4 x 0.0115 = 0.00288 moles CO2 is produced from CH4; therefore, this comes from burning 0.00288 moles of CH4 as the ratio of the moles CO2 to moles CH4 is 1:1.

3/4 x 0.0115 = 0.00863 moles of CO2 is produced by burning 1/3 x 0.00863 = 0.00288 moles of C3H8 as the ratio of CO2 to C3H8 is 3 moles : 1 mole.

The mass of CH4 is 16 g/mole x 0.00288 moles = 0.046 g CH4 (approximately).

The mass of C3H8 is 44 g/mole x 0.00288 moles = 0.13 g C3H8 (approximately).

From these results you can calculate the mass% of each.