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Question

Jazmine analyzes a compound having the following percentage composition by mass: carbon, 53.31%; hydrogen, 11.19%; oxygen, 35.51%. In a separate experiment, she finds that the molar mass of the compound is approximately 90 g.

a) Assuming 100.0 g of this compound, how many grams of C, H, and O are there?

b) Determine the empirical formula of this compound. Show your calculations. 

c) Determine the molecular formula of this compound.

Answer 1

a) Assuming 100.0 g of the compound,

mass of C = 53.31% x 100.0 g = 53.51 g C

mass of H = 11.19% x 100.0 g = 11.19 g H

mass of O = 35.51% x 100.0 g = 35.51 g O

b) Convert the g of C, H, and O in the compound from above to moles using their MW.

mol C = 53.31 g C x (1 mol C / 12.01 g C) = 4.439 mol C

mol H = 11.19 g H x (1 mol H / 1.01 g H) = 11.08 mol H

mol O = 35.51 g O x (1 mol O / 16.00 g O) = 2.219 mol O

Divide the moles of C, H, and O by the lowest # of moles (2.219) to obtain the mole ratio of each element in the compound.

mol C = 4.439 / 2.219 = 2.000 mol C ~ 2 mol C

mol C = 11.08 / 2.219 = 4.993 mol H ~ 5 mol H

mol O = 2.219 / 2.219 = 1.000 mol O ~ 1 mol O

Empirical formula = C₂H₅O

c) To determine the molecular formula, you first have to calculate the empirical formula mass (EFM) of C₂H₅O.

EFM = (2 x 12.01) + (5 x 1.01) + (1 x 16.00) = 45.07 g/mol

Divide molar mass (MW) of compound by EFM to get a multiple, which is then used to convert the empirical formula to the molecular formula using the equation,

Molecular formula = multiple x Empirical formula

multiple = MW / EFM = 90.00 / 45.07 = 1.997 ~ 2 (multiple MUST be a whole #).

Molecular formula = 2 x C₂H₅O = C₄H₁₀O₂