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QUESTION

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

In order to determine the 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 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 of the copper sulfide and its empirical formula. You know that the mass of copper is equal to

m_("copper") = 2.289-2.077 = 0.212 "g"

The copper sulfide mass is equal to

m_("copper sulfide") = 2.396-2.077 = 0.319 "g"

This means that the mass of sulfur is

m_("sulfur") = m_("copper sulfide") - m_("copper") = 0.319-0.212 =0.107g

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 of the two in the formula

66.5 "g" * ("1 mole")/("63.55 g") = 1.05 "moles"

33.5 "g" * ("1 mole")/("32.0 g") = 1.05 "moles"

Since ratio is 1:1, your empirical formula is

Cu_(1)S_(1), or CuS.