# If a compound has empirical formula of C2H3O, but has a molecular mass of 86.1 what is the molecular formula?

Jul 16, 2014

1) Write the empirical formula:

${C}_{2}$ ${H}_{3}$ O

2) Compute the "empirical formula mass:" by taking the mass of each mole of the given element. One mole of Carbon has mass 12 g, Hydrogen has mass 1 g, and O has mass of 16 g.

$$2x 12 + 3x1 + 1x16 = 43g


3) Divide the molecular mass by the "Empirical formula mass:"

$$86.1 / 43 = 2 (approximately)


4) The molecular mass is twice of empirical formula mass , it means that the ratio of elements in molecular formula is twice the ratio of elements in empirical formula.

${C}_{4}$ ${H}_{6}$ ${O}_{2}$