MCAT Chemistry Question — Molecular and Empirical Formulas | Next Step Test Prep MCAT Chemistry Question — Molecular and Empirical Formulas | Next Step Test Prep

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One mole of an unknown compound is determined to have a molecular weight of 86 grams per mole. The mass percent of the compound is as follows: 55.81% carbon, 6.98% hydrogen, and 37.21% oxygen. What is the empirical formula and molecular formula, respectively, of the unknown compound.

 

  1. C2H3O and C4H6O2
  2. C2H3O and C6H9O3
  3. C5H8O and C5H8O
  4. C5H8O and C10H16O2

 

Explanation

 

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.

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