# Serotonin (molar mass 176 g/mol) is a compound that conducts nerve impulses in brain and muscle. It contains 68.2% C, 6.86% H, 15.9% N and 9.08% O. What is the molecular formula?

Feb 16, 2017

$\text{Serotonin} \equiv {C}_{10} {H}_{12} {N}_{2} O$

#### Explanation:

As with all these problems, we ASSUME $100 \cdot g$ of compound, and we work out the percentages of each element on a molar basis:

$\text{Moles of carbon}$ $=$ $\frac{68.2 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.68 \cdot m o l$.

$\text{Moles of hydrogen}$ $=$ $\frac{6.86 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 6.81 \cdot m o l$.

$\text{Moles of nitrogen}$ $=$ $\frac{15.9 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 1.14 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\frac{9.08 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.568 \cdot m o l$.

Note that normally the analyst would not give you %O; you would be expected to work it out by difference.

And now we divide thru by the SMALLEST molar quantity, that of oxygen, to give an empirical formula of:

${C}_{10} {H}_{12} {N}_{2} O$.

And this is the empirical formula, the simplest whole number ratio defining constituent atoms in a species.

Now the molecular formula is ALWAYS a multiple of the empirical formula. That is ${\text{Molecular formula"-="(empirical formula)}}_{n}$.

Thus the molecular mass must show this proportionality:

${\left({C}_{10} {H}_{12} {N}_{2} O\right)}_{n} = 176 \cdot g \cdot m o {l}^{-} 1$

So $n \times \left(10 \times 12.011 + 12 \times 1.00794 + 2 \times 14.01 + 15.999\right) \cdot g \cdot m o {l}^{-} 1 = 176 \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = 1$, and the $\text{empirical formula}$ is the same as the $\text{molecular formula}$.

This approach is standard, however, most of the time you would not be quoted a percentage oxygen composition. You would be expected to calculate this yourself by difference.