# How can I calculate the empirical formula of copper sulfide?

Jan 10, 2015

In order to determine the empirical formula for copper sulfide (or for any compound, for that matter) you need to have some information about either the mass of one reactant and the mass of the product, or about the percent composition of the copper sulfide.

For the first case, let's assume you are doing a lab experiment in which you heat a mixture of copper and sulfur in order to produce a sample of copper sulfide.

You then weigh the empty crucible and find a weight of 2.077 g, the crucible + the copper and find a weight of 2.289 g, and the crucible + the copper sulfide and find a weight of 2.396 g.

Let's find the percent composition of the copper sulfide and its empirical formula. You know that the mass of copper is equal to

${m}_{\text{copper}} = 2.289 - 2.077 = 0.212$ $\text{g}$

The copper sulfide mass is equal to

${m}_{\text{copper sulfide}} = 2.396 - 2.077 = 0.319$ $\text{g}$

This means that the mass of sulfur is

${m}_{\text{sulfur") = m_("copper sulfide") - m_("copper}} = 0.319 - 0.212 = 0.107 g$

The percentages of copper and sulfur in the copper sulfide are

"%copper" = 0.212/(0.319)*100% = 66.5%

"%sulfur" = 0.107/(0.319)*100% = 33.5%

This means that, for every 100g of copper sulfide, you get 66.5g of copper and 33.5g of sulfur. You now divide each element's percentage by its molar mass to get the mole ratio of the two elements in the formula

$66.5$ "g" * ("1 mole")/("63.55 g") = 1.05 $\text{moles}$

$33.5$ "g" * ("1 mole")/("32.0 g") = 1.05 $\text{moles}$

Since the mole ratio is $1 : 1$, your empirical formula is

$C {u}_{1} {S}_{1}$, or $C u S$.