Question

A 42.0 mg sample of carbon reacts with sulfur to form 117 mg of the compound.​ What is the empirical formula of the carbon sulfide?​

Express your answer as a chemical formula.

Answer 1

mass of C = 42mg = 0.042g

mass of S   = mass of compound - mass of C

                    = 0.117-0.042   = 0.075 g

no of moles of C   = W/G.A.Wt =0.042/12   = 0.0035 moles

no of moles of S   = W/G.A.Wt   = 0.075/32   = 0.00234375moles

C :S     = 0.0035 : 0.00234375

            = 1.5      : 1

             = 3 : 2

C3S2

The empirical formula = C3S2