Answer to Question #311943 in General Chemistry for Zellie

Question

A 4.626 g sample of a hydrocarbon, upon combustion in a combustion analysis apparatus, yielded 6.484 g of carbon dioxide. Calculate the mass percent of carbon in the hydrocarbon.

Answer 1

$n(CO_2)=\dfrac{mass}{molar\ mass}=\dfrac{6.484\ g}{12 +2*16\ g/mol}=0.1474\ mol$

All of the carbon from the hydrocarbon is converted to carbon dioxide:

$n(C)=n(CO_2)=0.1474\ mol$

$m(C) =n(C)*Ar(C)=0.1474\ mol*12\ g/mol=1.768\ g$

Mass percent of carbon in the hydrocarbon:

$\omega(C)=\dfrac{m(C)}{m(hydrocarbon)}*100\% =\dfrac{1.768\ g}{4.626\ g}*100\%=38.2\%$