Question

A compound has a percent composition of 54.5% carbon, 9.3% hydrogen and 36.2 % oxygen.If its molar mass is 88 g/mol, what is its molecular formula?

Complete the following:

Answer 1

C₄H₈O₂.

Step-by-step explanation:

• Firstly, we can calculate the no. of moles (n) of each component using the relation:

n = mass/atomic mass,

mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.

mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.

mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.

• To get the empirical formula, we divide by the lowest no. of moles (2.26 mol) of O:

∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.

∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) = 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol) = 44.0 g/mol.

∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.

∴ The molecular formula is, 2(C₂H₄O), that is; (C₄H₈O₂)