Question

A hydrocarbon is `7.7%` hydrogen by mass. What is `(i)` its `"empirical formula"`, and `(ii)` what is its `"molecular formula"` if its molecular mass is `26.0*g*mol^-1`?

Answer 1

We gots acetylene `HC-=CH`....

Explanation:

For all these problems it is useful to assume that we gots `100*g` of compound.

And so we interrogate the molar quantities of each element....

`"Moles of carbon"-=(92.3*g*C)/(12.011*g*mol^-1)=7.68*mol`.

How did I know that there were `92.3*g` of carbon? It doesn't mention this quantity in the boundary conditions. Am I having you on?

`"Moles of hydrogen"-=(7.7*g*H)/(1.00794*g*mol^-1)=7.64*mol`.

Now, clearly, the `"empirical formula"`, the simplest whole number ratio defining constituent atoms in a species is `CH`. The molecular formula is a whole number multiple of the empirical formula, and we were give a molecular mass of `26.0*g*mol^-1`..

And so `"molecular mass"-={"empirical formula"}_n`

We solve for `n-=(26*g*mol^-1)/{12.011*g*mol^-1+1.00794*g*mol^-1}`

Clearly `n=2`, and we gots acetylene, `HC-=CH`...

Normally we would not be quoted combustion analysis for a gas....