Question

What is molecular formula of compound that weighs 46 g/mol and analyzes to 52.2% C, 34.8% O, and 13% H?

Answer 1

`"Molecular formula"` `-=` `C_2H_6O`

Explanation:

As with all these problems, we assume that there are `100*g` unknown material:

First we calculate an empirical formula...........

`"Moles of carbon:"` `=` `(52.2*g)/(12.011*g*mol^-1)=4.35*mol`.

`"Moles of hydrogen:"` `=` `(13.0*g)/(1.079*g*mol^-1)=12.04*mol`.

`"Moles of oxygen:"` `=` `(34.8*g)/(15.999*g*mol^-1)=2.18*mol`.

Note (i) that here we have simply divided the atomic masses thru by the `"atomic mass"`, and (ii), normally `"% oxygen content"` would not be measured. You would be given `"% carbon content"`, and `"% hydrogen content"`, and `"% nitrogen content"`, and `"% oxygen content"` would be assessed by the balance.

And now we normalize the formula by dividing thru by the lowest molar ratio, that of oxygen to get the empirical formula:

`O=(2.18*mol)/(2.18*mol)` `=` `1`

`C=(4.35*mol)/(2.18*mol)` `=` `2`

`H=(12.04*mol)/(2.18*mol)` `=` `6`

And thus our empirical formula is `C_2H_6O`.

But the molecular formula is always a mulitple of the empirical formula:

i.e. `"(empirical formula)"xxn="(molecular formula)"`.

We KNOW the molecular mass because it has been measured in a separate experiment, and provided for us:

So `nxx(2xx12.011+6xx1.00794+15.999)*g*mol^-1=46.0*g*mol^-1`,

and we solve for `n`.

Clearly `n=1`, and here

`"molecular formula "-=" empirical formula"=C_2H_6O`.