Question

A sample of a compound is 80% carbon and 20% hydrogen by mass. Its formula mass is 30 amu. What is the molecular formula?

Answer 1

We have `"ethane"`.

Explanation:

As with all these problems, it is usually assumed that we have a `100*g` mass of unknown compound, and we work out the molar quantities:

And thus `"moles of carbon" -=(80*g)/(12.011*g*mol^-1)=6.66*mol`.

And thus `"moles of hydrogen" -=(20*g)/(1.008*g*mol^-1)=19.8*mol`.

We divide the molar quantities thru by the SMALLER molar quantity:

`C:(6.66*mol)/(6.66*mol)=1`

`H:(19.8*mol)/(6.66*mol)=2.97`

And thus the `"empirical formula"` `=` `CH_3`

And we know that the molecular formula is always a multiple of the empirical formula, and thus is terms of mass:

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

But we have a molecular mass of `30*"amu"`

So `30*"amu"` `=` `nxx(12.01+3xx1.01)*"amu"`

Clearly, `n=2`, and the MOLECULAR FORMULA is `C_2H_6`.

This is not a realistic problem, as few analysts would perform combustion on a liquid, and no analyst could perform combustion on a gas.