Question

A hydrocarbon is `83.33%` by mass with respect to carbon. What is (i) its empirical formula, and (ii) its molecular formula if the hydrocarbon has a mass of `72*g*mol^-1`?

Answer 1

I gets `"pentane"`...................

Explanation:

As always with these problems it is useful to ASSUME a mass of `100*g` of sample, and then interrogate its ATOMIC composition.

`"Moles of carbon"=(83.33*g)/(12.011*g*mol^-1)=6.94*mol*C`

The balance `100*g-83.33*g` MUST represent the percentage of hydrogen, because we were told that we analyzed an HYDROCARBON.

`"Moles of hydrogen"=(100.0*g-83.33*g)/(1.00794*g*mol^-1)=16.54*mol*H`

We divide thru by the lowest molar quantity (`C`) to get a trial formula of:

`CH_(2.38)`; but we require the simplest WHOLE number ratio, and let us try `5xxCH_(2.38)~=C_5H_12`.

Now the MOLECULAR FORMULA is always a WHOLE number multiple of the EMPIRICAL FORMULA, and since we know the molecular mass:

`nxx(5xx12.011+12xx1.00794)*g*mol^-1=72.11*g*mol^-1`.

Clearly `n=1`, and we can quote the molecular formula as isomeric `C_5H_12`.