Question

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)

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

b) `2.20/12*44*100/1.5`

c)`12/44*1.50*100/2.2`

d)`12/60*1.50*100/2.2`

Answer 1

Can be solved partially

Explanation:

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`