Question

A `10.5*g` mass of a hydrocarbon contains `1.5*g` of hydrogen. If the molecular mass of the hydrocarbon in `210*g*mol^-1`, what are the empirical and molecular formulae of the hydrocarbon?

Answer 1

`1.5g` of H would be `1.5div1=1.5mol`
`9g` of C would be `9div12=0.75mol`

Explanation:

So the mol-ratio of `CdivH=0.75div1.5=1div2`
And the empirical formula is `(CH_2)_n`

One of those units has a mass of `12+2xx1=14u`

So there are `210div14=15` of those units, and the molecular formula will be:

`C_15H_30`

Answer 2

We find `"(i) the empirical formula"`, and then `"(ii) the molecular formula"`.

Explanation:

`"Moles of hydrogen"` `=` `(1.5*g)/(1.00794*g*mol^-1)=1.49*mol`

`"Moles of carbon"` `=` `(9.0*g)/(12.011*g*mol^-1)=0.75*mol`.

We divide thru by the smallest molar quantity, and (clearly) we get an empirical formula of `CH_2`.

But we know that the `"molecular formula"` is whole number multiple of the `"empirical formula"`;

i.e. `"molecular formula"=nxx"empirical formula"`

And thus, `210*g*mol=nxx(12.011+1.00794)*g*mol^-1`, and we solve for `n`, to get `n=16.15`. This value should be closer to an integer, however, this gives a molecular formula of `C_16H_32` based on the figures you quoted.