Question

A compound is 31.1 % S, the rest is chlorine. The compound has a molecular mass of 205.94 amu. What is its molecular formula?

Answer 1

The molecular formula of the compound would be  $S_2Cl_4$

The compound has 31.1% S, thus, the chlorine would be:

    100 - 31.1 = 68.9%

Calculate the mole of each element in the compound:

• mole = mass/molar mass

• S = 31.1/32 = 0.97

• Cl = 68.9/35.4 = 1.95

Divide the moles by the smallest:

• S = 0.97/0.97 = 1
• Cl = 1.95/0.97 = 2

Thus, the empirical formula of the compound is SCl2

Molecular formula = n(SCl2)

n = molar mass of compound/molar mass of its empirical formula

molar mass of SCl2 = 32 + (35.4x2)

                                 = 102.8

molar mass of compound = 205.94

n = 205.94/102.8

        = 2

Thus, the molecular formula becomes 2(SCl2), which is equivalent to $S_2Cl_4$

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