Question

what is the empirical formula of an unknown compound containing 63.13 % c, 8.83 % h, and 28.04 % o by mass?
o c6h10o2
o c5h9o2
o c3h5o
o c5h8o2
o c3h6o

Explanation:

Step1: Assume 100g of the compound

So, we have 63.13g C, 8.83g H, and 28.04g O.

Step2: Calculate the moles of each element

Moles of C: $n_{C}=\frac{63.13g}{12.01g/mol}\approx5.26mol$
Moles of H: $n_{H}=\frac{8.83g}{1.01g/mol}\approx8.74mol$
Moles of O: $n_{O}=\frac{28.04g}{16.00g/mol}\approx1.75mol$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (1.75mol).
Ratio of C: $\frac{5.26mol}{1.75mol}\approx3$
Ratio of H: $\frac{8.74mol}{1.75mol}\approx5$
Ratio of O: $\frac{1.75mol}{1.75mol}=1$

Answer:

C. $C_{3}H_{5}O$