# A sample of an unknown metal chlorate, weighing 5.837 g, is heated until all of the oxygen is driven off. The residue remaining in the container weighs 1.459 g. What is the percentage of oxygen in this metal chlorate?

${M}_{n} {\left(C l {O}_{3}\right)}_{m} \left(s\right) + \Delta \rightarrow {M}_{n} C {l}_{m} \left(s\right) + \frac{3 m}{2} {O}_{2} \left(g\right)$
While I have given the stoichiometric equation for the decomposition of the metal chlorate, this reaction is simpler than it seems. We know that a mass of $5.837 - 1.459 = 4.378$ $g$ was lost during the reaction. This must represent a mass of oxygen.
Given that we started with a mass of $5.837$ $g$, this represents a percentage of (4.378*g)/(5,837*g) xx 100% $=$ ?? in the initial sample.