Question

An unknown compound is composed of 47.08% C, 6.59% H, and 46.33% Cl. What mass (in g) of the compound will have 3.65 moles of carbon? The molar mass of the compound is 153 g/mol, what are the empirical and molecular formulas of the compound?

Please show every step. Thank you!

Answer 1

Let us say that the total mass of the compound = 100 g

Hence, the mass of C in the compound = 100 g x 47.08 % = 100 g x 47.08/100 = 47.08 g

Similarly, the mass of hydrogen and chlorine in the compound are 6.59 g and 46.33 g.

Molar mass of C, H and Cl are 12 g/mol, 1 g/mol and 35.5 g/mol respectively.

Hence the mole ratio of C : H : Cl = 47.08/12 : 6.59/1 : 46.33/35.5

                                                      = 3.92 : 6.59 : 1.31

                                                      = 3 : 5 : 1                  (dividing all with 1.31)

Hence, the empirical formula of the compound = C₃H₅Cl

Let us say that the molecular formula of the compound = (C₃H₅Cl)_n     (where, n is an integer)

Molar mass from the molecular formula = n x (3 x 12 g/mol + 5 x 1 g/mol + 1 x 35.5 g/mol)

                                                               = 76.5n g/mol

The molar mass given = 153 g/mol

Hence,

76.5n = 153

or, n = 2

Hence, the molecular formula of the compound = (C₃H₅Cl)₂ = C₆H₁₀Cl₂

Thus, 6 moles of carbon are there in 153 g compound

Therefore, 3.65 moles of carbon are there in = (153 g x 3.65 mol)/6 mol = 93.075 g compound

Hence, 93.075 g of the compound will have 3.65 mol of carbon.