Question

A compound with molar mass 180.1 g/mol has the following composition by mass:

C	40.0%
H	6.70%
O	53.3%

Determine the empirical and molecular formulas of the compound.

Short answer

The empirical and molecular formulasof the compound are CH₂O and C₆H₁₂O₆.

Step-by-step solution

Definition

A chemical formula that shows the simplest ratio of elements in a compound rather than the total number of atoms in the molecule is known as an empirical formula.

Determination of moles of each component

From the given,

The molar mass of the compound is 180.1g/mol.

Whether we have 100g of the compound, later we have 40g of c, 6.7g of H, and 53.3g of O

Let’s calculate moles of each component,

$\begin{array}{rcl}n\left(C\right)& =& \frac{m\left(C\right)}{Mr\left(C\right)}\\ & =& \frac{40g}{12{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}=3.33mol\\ n\left(H\right)& =& \frac{m\left(H\right)}{Mr\left(H\right)}\\ & =& \frac{6.7g}{1{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}=6.7mol\\ n\left(O\right)& =& \frac{m\left(O\right)}{Mr\left(O\right)}\\ & =& \frac{53.3g}{16{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}=3.33mol\end{array}$

Determination of ratio between elements

$\begin{array}{rcl}n\left(C\right):n\left(H\right):n\left(O\right)& =& 3.33:6.7:3.33\\ & =& 1:2:1\end{array}$

Therefore the empirical formula is CH₂O

$\begin{array}{rcl}Mr\left(C{H}_{2}O\right)& =& Ar\left(C\right)+2Ar\left(H\right)+Ar\left(O\right)\\ & =& 12\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+2\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+16\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ & =& 30\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\end{array}$

Molecular formula: $\begin{array}{rcl}\frac{Mr\left(Compound\right)}{Mr\left(C{H}_{2}O\right)}& =& \frac{180.1{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}{30{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}\\ & =& 6\end{array}$

Therefore the molecular formula is $6\times \left(C{H}_{2}O\right)=C_6{H}_{12}{O}_6$.