Answer :
294.2% is the percentage of the initial sample of pure hydrocarbon from the given impure sample of a hydrocarbon having a total mass of 39.1 g.
The given data is as follows:
Total mass = 39.1 g
Carbon-di-oxide weight = 74.4 g
Hydrogen = 15.2g
The equation should be balanced for the combustion process and for equal reactants. The equation is:
CnHm + (n + m/4) O2 → n CO2 + m/2 H2O
Number of moles in CO2 = 74.4 g / 44.01 g/mol
Number of moles in CO2 = 1.688 mol
Number of moles in H2O = 15.2 g / 18.02 g/mol
Number of moles in H2O = 0.844 mol
Number of moles in hydrocarbon = 1.688 mol
The molar mass of Hydrocarbons is calculated by:
Molar mass of hydrocarbon mixture= (39.1 g - 1.688 mol x 44.01 g/mol - 0.844 mol x 18.02 g/mol) / 1.688 mol
The molar mass of the hydrocarbon mixture= 68.06 g/mol
The weight of Pure hydrocarbon is calculated by:
Weight of pure hydrocarbon = 1.688 mol x 68.06 g/mol
weight of pure hydrocarbon = 115.02 g
The percentage of the initial sample in pure hydrocarbon is calculated by the formula:
Percentage of pure hydrocarbon = (115.02 g / 39.1 g) x 100%
Percentage of pure hydrocarbon = 294.2%
Therefore we can conclude that 294.2% is the percentage of the initial sample is a pure hydrocarbon.
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