Question

Combustion data of an organic compound yielded `74.0%, C;` `7.4%, H;` `8.6%, N;` and the balance oxygen. What is the empirical formula?

Answer 1

`C_10H_12NO`

Explanation:

As with all problems of this type, we assume a `100*g` mass of compound, and we determine the molar composition with respect to each element:

And thus:

`"Moles of carbon"` `=` `(74.0*g)/(12.011*g*mol^-1)` `=` `6.16*mol`

`"Moles of hydrogen"` `=` `(7.4*g)/(1.00794*g*mol^-1)` `=` `7.34*mol`

`"Moles of nitrogen"` `=` `(8.6*g)/(14.01*g*mol^-1)` `=` `0.614*mol`

`"Moles of oxygen"` `=` `(10.0*g)/(16.00*g*mol^-1)` `=` `0.625*mol`

So we have the molar composition. We simply divide thru by the smallest molar quantity to gets our empirical formula:

Dividing thru by `0.614*mol` we get `C_10H_12NO` (clearly, I have done some rounding off!), as the empirical formula.