Question

A compound is 54.53% C, 9.15% H, and 36.32% O. What is its empirical formula?

The molar mass of the compound is 132 amu. What is its molecular formula?

Answer 1

i got it nevermind

Answer 2

This requires a couple steps and a little imagination: assume you've got 100g of the substance. This turns the percentages into masses:
54.53g C, 9.15g H, 36.32g O

Turn each mass into moles using the molar mass of each:
\[54.53g C * (\frac{1mol C}{12g C}) = 4.54mol C\]\[9.15g H * (\frac{1mol H}{1g H}) = 9.15mol H\]\[36.32g O * (\frac{1mol O}{16g O}) = 2.27mol O\]

In order to turn each of these fractional moles into whole #s, divide them all by the smallest among them. In this case, 2.27 is the smallest:
\[\frac{4.54mol}{2.27mol} = 2mol C\]\[\frac{9.15mol}{2.27mol} = 4mol H\]\[\frac{2.27mol}{2.27mol} = 1mol O\]
So your empirical formula is:\[C{_2}H{_4}O\]

Answer 3

Thanks!