Question

One of the components that make up common table sugar is fructose, a compound that containsonly carbon, hydrogen, and oxygen. Complete combustion of 1.50 g of fructose produced 2.20 g of carbon dioxide and 0.900 g of water. What is the empirical formula of fructose?

Short answer

The empirical formula of fructose can be found as $C{H}_{2}O$.

Step-by-step solution

Chemical equation of combustion of fructose

The equation can be derived as,

$aCHO+b{O}_2\to cC{O}_2+d{H}_{2}O$

First, need to find the coefficient of reactants and products to get the fructose empirical formula. Coefficient is a=1. The mass on both sides should be same to balance the equation.

Finding moles of each compounds

The mass of oxygen can be $2.2g+0.9g-1.5g=1.6g$. The moles can be denoted as,

$\begin{array}{rcl}molof{O}_2:\frac{1.6g}{32{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}& =& 0.05\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ molofC{O}_2:\frac{2.2g}{43.01{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}& =& 0.05\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ molof{H}_{2}O:\frac{0.9g}{18{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}& =& 0.05\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\end{array}$

Here, see that three compounds are equal. Coefficients are 1 and fructose have 1C, 1O and 2H. The empirical formula of fructose can be found as $C{H}_{2}O$.