From the centesimal composition of a chemical compound we can determine its empirical formula. The molecular formula is obtained from the empirical one using the molecular mass of the compound, which can be determined by mass spectrometry.
Determination of the empirical and molecular formula of cortisone
The percentage composition of cortisone is: 69.96% of C; 7.83% H and 22.21% O. Determine the empirical formula. Knowing that the molecular mass is 360.43, obtain the molecular formula.
1. Divide the mass percentages by the atomic mass of the element.
\(C:\frac{69.96}{12.01}=5.825;\;\;\; H:\frac{7.83}{1.008}=7.768;\;\;\; O:\frac {22,217}{16,00}=1,388\)
2. Divide the previous results by the smallest.
\(C:\frac{5.825}{1.388}=4.20;\;\;\; H:\frac{7.768}{1.388}=5.60;\;\;\; O:\frac{1.388 }{1,388}=1\)
3. If the previous values are not integers, they are multiplied by a factor that converts them into integers. In this case the factor is 5.
\(C:4.20\times 5 = 21;\;\;\; 5.60\times 5 = 28;\;\;\; 1\times 5 =5\)
4. We write the empirical formula \((C_{21}H_{28}O_5)_n\)
The parameter n is determined by dividing the molecular mass of the compound by the molecular mass of the empirical formula
\(n=\frac{360.43}{21\times 12.01+28\times 1.008+5\times 16.00}=1\)
5. We write the molecular formula C 21 H 28 O 5
There are thousands of molecules that meet the above formula (isomers). The spectra (IR, NMR) make it possible to determine which of these molecules corresponds to cortisone.
