Answer
Empirical formula: $$C_6H_{10}S_2O$$
Real molecular formula:$$C_6H_{10}S_2O$$
Work Step by Step
1. Divide all mass percentages by the molar mass of each compound:
$$C_{\frac{44.4}{12.01}}H_{\frac{6.21}{1.008}}S_{\frac{39.5}{32.07}}O_{\frac{9.86}{16.00}}$$
$$C_{3.70}H_{6.16}S_{1.23}O_{0.616}$$
2. Divide all numbers by the smaller one (0.616):
$$C_{6.01}H_{10.0}S_{2.00}O_{1}$$
Rounding the number of carbons, we get the empirical formula:
$$C_6H_{10}S_2O$$
3. Calculate the molar mass for the empirical formula:
$ C_6H_{10}S_2O $ : ( 1.008 $\times$ 10 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 1 )+ ( 32.07 $\times$ 2 )= 162.28 g/mol
It is approximately equal to Allicin's molar mass. Therefore, this is the formula for allicin:
$$C_6H_{10}S_2O$$
