What is the % of carbon in the hydrocarbon, if $2.20 g$ of $"CO"_2$ is formed by complete combustion of $1.50 g$ of a hydrocarbon? ( C= 12, O= 16, H= 1) ( $M_r$ of hydrocarbon is=60)

Question

a) $12/44* 2.20/1.5 100$

b) $2.20/1244100/1.5$

c) $12/441.50100/2.2$

d) $12/601.50*100/2.2$

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Answer 1 · 2018-02-07T14:19:40

Can be solved partially

First we need to find the number of moles of hydrocarbon and carbon dioxide.

$n("Hydrocarbon")=(m("Hydrocarbon"))/(M_r("Hydrocarbon"))=1.5/60=0.025mol$

$n("CO"_2)=(m("CO"_2))/(M_r("CO"_2))=2.2/44=0.05mol$

The mole ratio of the hydrocarbon to the carbon dioxide is $1:2$ , so this gives us a reaction of:
$"C"_a"H"_b+x"O"_2->2"CO"_2+y"H"_2"O"$

Since we have two carbons on the right hand side, we need 2 carbons on the left hand side.

$a=2$

Our hydrocarbon has a formula of $"C"_2"H"_b$ , however, the highest value for $b$ is $6$ , which gives an $M_r$ of $30$ .

This could mean an error in the question where the $M_r$ is meant to be 30, or the mass used being $0.75g$