MCAT Chemistry Question — Molecular and Empirical Formulas

To answer this question, you must determine the empirical formula of the unknown compound by treating the mass percent as the number of grams of each element out of a 100g sample. The number of moles of each element is determined by the following calculations:

Moles of C = (55.81g C) x (1mol C/12g C) = 4.65 mol C

Moles of H = (6.98g H) x (1 mol H/1g H) = 6.98 mol H

Moles of O = (37.21g O) x (1 mol O/16g O) = 2.33 mol O

Dividing each by the largest common denominator of 2.33 gives the empirical formula of the unknown compound: C₂H₃O

(4.65/2.33) mol C = 2 mol C

(6.98/2.33) mol H = 3 mol H

(2.33/2.33) mol O = 1 mol O

Since the molecular weight of the compound is 86 g/mol, the correct answer must be A. Note that if you were in a hurry on Test Day, or if you forgot how to solve this problem, you could instead use the periodic table to calculate the molar mass of the molecular formula given in each answer choice. If only one answer gave a molar mass of 86 g/mol (as is the case here, with choice A), then you could be certain that this option was correct. This is true of many MCAT questions: there may be multiple ways to solve the same problem!