A compound with molar mass 110.1 g/mol has the following composition:

Question

C 65.5%

H 5.45%

O 29.1%

How would I do the math to determine the empirical formula?

Answer 1

Hey Jessica,

So to start off this question, what are we given? We know some percentage compositions and a total molar mass of a compound, but this information does not at first glance seem useful. However, what we can do is convert these percentage compositions to mass values for each of the three elements in the 110.1g/mol compound. To do so, we must multiply the percentage composition by the molar mass:

0.655C (110.1g/mol)= 72.12 grams C

0.0545H (110.1g/mol)= 6.00 grams H

0.291O (110.1g/mol)= 32.0 grams O

Now, in order to get the molecular formula, we must divide all of these mass values by the respective molar mass of each element.

72.12 grams C/12.01 grams/molC= 6 moles C

6.00 grams H/ 1.00 grams/molH= 6 moles of H

32.0 grams O/ 16 grams/molO= 2 moles of O

The molecular formula of the compound in question is C₆H₆O₂. The empirical formula however, is the simplest ratio of elements in the compound, so we get that the empirical formula is C₃H₃O.

Hope this helps!