When burning a substance with the molecular formula of the type C x H y O z , CO 2 and H 2 O are obtained. All the carbon in the sample will go to CO 2 , while the hydrogen will be transformed into H2O, according to the chemical equation:
C x H y O z + O 2 -------> x CO 2 + y/2 H 2 O
The carbon dioxide generated by the combustion is absorbed on sodium hydroxide, determining its mass by difference in weight. While the water vapor is adsorbed on magnesium perchlorate.
Once the masses of CO 2 and H 2 O have been determined, the empirical formula is calculated. Let's see an example:
The combustion of a sample of 0.2000 g of vitamin C produced 0.2998 g of CO 2 and 0.0819 g of H 2 O. Obtain the empirical formula of vitamin C.
- Calculate the amounts in moles of carbon and hydrogen
$$0.2998\;g\;CO_2\cdot\frac{1\;mol\;CO_2}{44.010\;g\;CO_2}\cdot\frac{1\;mol\;C}{1\;mol \;CO_2}=0.006812\;mol\;C$$
$$0.0819\;g\;H_2O\cdot\frac{1\;mol\;H_2O}{18.02\;g\;H_2O}\cdot\frac{2\;mol\;H}{1\ ;mol\;H_2O}=0.00909\;mol\;H$$
- Calculate the masses of carbon and hydrogen
$$0.006812\;mol\;C\cdot\frac{12.011\;g\;C}{1\;mol\;C}=0.08182\;g\;C$$
$$0.00909\;mol\;H\cdot\frac{1.008\;g\;H}{1\;mol\;H}=0.00916\;g\;H$$
- Obtaining the mass of oxygen by difference
$$0.2000\;g\;sample - 0.08182\;g\;C - 0.00916\;g\;H=0.1090\;g\;O$$
- Calculation of moles of oxygen
$$0.1090\;g\;O\cdot\frac{1\;mol\;O}{15.999\;g\;O}=0.000813\;mol\;O$$
- Divide the moles of C, H, and O by the smallest value
$$C:\frac{0.006812}{0.006182}=1\;\;\; H:\frac{0.00909}{0.006812}=1.33\;\;\;\; Or:\frac{0.006813}{0.006812}=1$$
- Multiply by a small number that converts the above values to integers (x3)
C: 1 x 3 =3; H: 1.33 x 3 = 4; Or: 1 x 3 = 3
- Empirical formula: C 3 H 4 O 3