The solution would be like this for this specific problem:
Â
(0.3395 mg CO2) /
(44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) × (12.01078 g C/mol) =
0.0926538 mg CÂ
(0.1391 mg H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol
H2O) × (1.007947 g H/mol) = 0.0155651 mg HÂ
(0.1206 mg total) - (0.0926538 mg C) - (0.0155651 mg H) =
0.0123811 mg OÂ
(0.0926538 mg C) / (12.01078 g C/mol) = 7.71422 × 10^-3 mmol
CÂ
(0.0155651 mg H) / (1.007947 g H/mol) = 1.54424 × 10^-2 mmol
HÂ
(0.0123811 mg O) / (15.99943 g O/mol) = 7.73846 × 10^-4 mmol
OÂ
Now, we divide by the smallest number of moles:Â
(7.71422 × 10^-3 mmol C) / (7.73846 × 10^-4 mmol) = 9.969Â
(1.54424 × 10^-2 mmol H) / (7.73846 × 10^-4 mmol) = 19.955Â
(7.73846 × 10^-4 mmol O) / (7.73846 × 10^-4 mmol) = 1.000Â
After rounding to the nearest whole numbers, we can find the
empirical formula which is:
C10H20O