30.4 mg of an organic compound known to contain C, H, and O was combusted and produced 70.4 mg of `CO_2` and 14.4 mg of `H_2O` .
70.4 mg of `CO_2` has `(70.4/44)*12` = 19.2 mg of carbon. 30.4 mg of the compound has 19.2 mg of carbon.
14.4 mg of `H_2O` has `(14.4/18)*2` = 1.6 mg of hydrogen. 30.4 mg of the compound has 1.6 mg of hydrogen.
The rest of the mass is made up by oxygen. 30.4 mg of the compound has 9.6 mg of oxygen.
19.2 mg of carbon is `1.6/1000` mole of carbon, 1.6 m of hydrogen is `1.6/1000` mole of hydrogen and 9.6 mg of oxygen is `0.6/1000` mole of oxygen.
The empirical formula of the compound is `C_((n*1.6)/1000)H_((n*1.6)/1000)O_((n*0.6)/1000)` . The smallest value of n that makes all the subscripts integers is n = 5
The empirical formula of the compound is `C_8H_8O_3` .
The molecular weight of the compound is 152 a.m.u.
This gives the molecular formula of the compound as `C_8H_8O_3`