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?

1 Answer
Feb 7, 2016

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#.