The chemical compounds

At this point we will determine the formula of a chemical compound from its percentage composition obtained experimentally. Let's see an example:

The percentage composition of methyl succinate is 62.58% C; 9.63% H and 27.79% O. Its molecular mass is 230 amu. Determine the empirical and molecular formula.

• In 100 g of methyl succinate we have: 62.58 g of C; 9.63 g of H and 27.7 g of O
• Conversion of the masses of the elements to moles

$$62.58\;g\;C\cdot\frac{1\;mol\;C}{12.011\;g\;C}=5.210\;mol\;C$$

$$9.63\;g\;H\cdot\frac{1\;mol\;H}{1.008\;g\;H}=9.55\;mol\;H$$

$$27.97\;g\;O\cdot\frac{1\;mol\;O}{15.999\;g\;O}=1.757\;mol\;O$$

• Divide the moles obtained by the lowest value

C: 5,210/1,757 = 2.99 ;

H: 9.55/1.757 = 5.49;

Or: 1,757/1,757 = 1;

• Multiply all the values by a small number that converts them to integers, in this case x2.

2.99 x 2 = 6 C atoms;

5.49 x 2 = 11 H atoms;

1 x 2 = 2 O atoms;

• Write the empirical formula: C ₆ H ₁₁ O ₂
• Calculation of "n" to obtain the molecular formula (C ₆ H ₁₁ O ₂ ) _n

$$n=\frac{12,011\cdot 6+ 1,008\cdot 11 + 15,999\cdot 2}{230}=2$$

The calculation of "n" is made by quotient between the mass of the empirical formula and the molecular

• Molecular formula: C ₁₂ H ₂₂ O ₄