Question

A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula?

Answer 1

The empirical formula is `CH_2O`, and the molecular formula is some multiple of this.

Explanation:

In 100 g of the unknown, there are `(40.0*g)/(12.011*g*mol^-1)` `C`; `(6.7*g)/(1.00794*g*mol^-1)` `H`; and `(53.5*g)/(16.00*g*mol^-1)` `O`.

We divide thru to get, `C:H:O` `=` `3.33:6.65:3.34`. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of `CH_2O`, i.e. near enough to WHOLE numbers.

Now the molecular formula is always a multiple of the empirical formula; i.e. `("EF")_n = "MF"`.

So `60.0*g*mol^-1 = nxx(12.011+2xx1.00794+16.00)g*mol^-1`.

Clearly `n=2`, and the molecular formula is `2xx(CH_2O)` `=` `C_xH_yO_z`.