Question

When 1.64 g sample of an unknown hydrocarbon was completely burned in oxygen, it produced 2.36 g of H2O. What is the percentage by mass of carbon in this hydrocarbon?

Answer 1

Approx. `84%`....

Explanation:

We got a molar quantity of `(2.36*g)/(18.01*g*mol^-1)=0.131*mol` with respect to water...agreed.

Now oxygen WAS ADDED when we performed the combustion, but hydrogen was NOT added, and thus all the hydrogen in that water is presumed to derive from the starting hydrocarbon, i.e. a mass of `2xx0.131*molxx1.00794*g*mol^-1=0.264*g`.

The balance of the mass MUST be due to carbon (because a starting hydrocarbon was SPECIFIED). And so `%H=(0.264*g)/(1.64*g)xx100%=16.1%`...and `%C=100%-16.1%=83.9%`...

We could go farther with this problem and specify the empirical formula, by ASSUMING an `100*g` mass of compound and dividing the constituent masses thru by their atomic mass....

`"Moles of carbon"=(83.9*g)/(12.01*g*mol^-1)=7*mol`

`"Moles of hydrogen"=(16.1*g)/(1.01*g*mol^-1)=16*mol`

And thus an empirical formula of `C_7H_16`.