Question

A 40.8 mg sample of carbon reacts with sulfur to form 113 mg of the compound

What is the empirical formula of the carbon sulfide?

Answer 1

Given the mass of C - atom in the sulphide = 40.8 mg = 40.8x10^-3 g

Hence moles of C - atom in the sulphide = mass / atomic mass of C = 40.8x10^-3 g / 12.0 g/mol = 0.0034 mol

Given the mass of the sulphide = 113 mg

Hence mass of S - atom in the sulphide = 113 mg - 40.8 mg = 72.2 mg =  72.2x10^-3 g

Hence moles of S - atom in the sulphide = mass / atomic mass of S = 72.2x10^-3 g / 32.0 g/mol = 0.002256 mol

Now the ratio of moles of C to the moles of S

= 0.0034 mol C : 0.002256 mol S

dividing the above ratio by 0.002256

(0.0034 mol C / 0.002256) : (0.002256 mol S / 0.002256)

= 1.5 mol C : 1 mol S

multiplying the above ratio by 2.

= 3 mol C : 2 mol S

Hence empirical formulae of the carbon sulfide is C3S2 (answer)