Question

A yellow substance was found to contain 54.5% Carbon, 9.10% Hydrogen and 34.6% Oxygen by mass. What is the molecular and empirical formula for the compound if it has a molar mass of approximately 135g/mole?

Answer 1

Here the addition of all the percentage cmposition is not 100 %

54.5 % + 9.10 % + 34.6 % = 98.2 %

Hence there is something that is wrong in the question.

However if we add the rest of the 1.8 % to Oxygen and make the molar mass 135 g/mol we obtain the following result.

Let's consider 100g of the substance.

Mass of C = 54.5 g

molecular mass of C = 12.0 g/mol

Hence moles of C in the compound = mass / molecular mass = 54.5 g / 12.0 g/mol = 4.54 mol

Mass of H = 9.10 g

molecular mass of C = 1.00 g/mol

Hence moles of H in the compound = mass / molecular mass = 9.10 g / 1.00 g/mol = 9.10 mol

Mass of O = 36.4 g

molecular mass of O = 16.0 g/mol

Hence moles of O in the compound = mass / molecular mass = 36.4 g / 16.0 g/mol = 2.275 mol

moles of C : moles of H : moles of O

= 4.54 mol : 9.10 mol : 2.275 mol

Dividing the above ratio by the smallest number 2.275 mol we get

= (4.54 mol/2.275 mol) : (9.10 mol/2.275 mol) : (2.275 mol/2.275 mol)

= 2.00 mol C : 4.00 mol H : 1 mol O

Hence empirical formulae = C₂H₄O

Empirical formulae mass = 24 + 4 + 16 = 44 g

Hence n = 132 g/mol / 44 g = 3

Hence molecular formulae = 3 x( C₂H₄O) = C₆H₁₂O₃