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compound has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen.
Laboratory data shows that the compound's molar mass is 62.0 g/mol. What is the
molecular formula of the compound?


Sagot :

The molecular formula for a compound with a percent composition of 38.7% C, 9.76% H, 51.5% O, and a molar mass of 62.0 g/mol is C₂H₆O₂.

What is a molecular formula?

A molecular formula is a chemical formula indicating the kinds and number of atoms of each kind in a molecule of a compound.

To determine the molecular formula, we need to follow a series of steps.

  • Step 1. Divide the percentage of each element by its molar mass.

C: 38.7/12.01 = 3.22

H: 9.76/1.01 = 9.66

O: 51.5/16.00 = 3.22

  • Step 2. Divide all the numbers by the smallest one, i.e. 3.22.

C: 3.22/3.22 = 1

H: 9.76/3.22 = 3

O: 3.22/3.22 = 1

The empirical formula is CH₃O and its molar mass is 31.04 g/mol.

  • Step 3. Determine the molecular formula.

First, we will calculate how many times the empirical formula is contained in the molecular formula by dividing their molar masses.

n = (62.0 g/mol) / (31.04 g/mol) = 2

The molecular formula is 2 times the empirical formula, that is, C₂H₆O₂.

The molecular formula for a compound with a percent composition of 38.7% C, 9.76% H, 51.5% O, and a molar mass of 62.0 g/mol is C₂H₆O₂.

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