Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen

Question

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations

Answer 1

Hi Selene,

For the first part of the question, we need to divide the mass composition of each element by its molecular mass, and then normalize the result by the lowest value.

C H O

Mass (g) 47.37 10.59 42.04

Molar mass (g/mol) 12.011 1.01 16.0

Mass/molar mass 3.944 10.485 2.628

Divide by smallest 1.5 3.99 1

Normalize to nearest ~3 ~8 2

whole number

Empirical formula: C3H8O2

For the second part, a constant n is introduced, which is required to scale up the molecular mass of the empirical representation, i.e.,

(C3H8O2) * n = 228.276 g/mol

(3*12.011 + 8*1.01 + 2*16) * n = 228.276

76.113 * n = 228.276 g/mol

n = 3

Thus, the molecular formula is (C3H8O2) * 3

= C9H24O6

47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen