Question

a compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? insert subscripts as needed. empirical formula: cho the molecular mass of the compound is 132 amu. what is its molecular formula? insert subscripts as needed. molecular formula: cho

Explanation:

Step1: Assume 100 g of the compound

Since the percentages are by mass, if we assume 100 g of the compound, we have 54.53 g of C, 9.15 g of H, and 36.32 g of O.

Step2: Calculate the number of moles of each element

The molar mass of C is 12.01 g/mol, of H is 1.01 g/mol, and of O is 16.00 g/mol.
The number of moles of C: $n_{C}=\frac{54.53\ g}{12.01\ g/mol}=4.54\ mol$
The number of moles of H: $n_{H}=\frac{9.15\ g}{1.01\ g/mol}=9.06\ mol$
The number of moles of O: $n_{O}=\frac{36.32\ g}{16.00\ g/mol}=2.27\ mol$

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (2.27 mol in this case).
For C: $\frac{4.54\ mol}{2.27\ mol}=2$
For H: $\frac{9.06\ mol}{2.27\ mol}=4$
For O: $\frac{2.27\ mol}{2.27\ mol}=1$
The empirical formula is $C_{2}H_{4}O$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $C_{2}H_{4}O$ is $2\times12.01 + 4\times1.01+1\times16.00=44.06\ g/mol$

Step5: Determine the molecular formula

Let $n=\frac{\text{molecular mass}}{\text{empirical - formula mass}}$. Given the molecular mass is 132 amu (which is equivalent to 132 g/mol).
$n=\frac{132\ g/mol}{44.06\ g/mol}\approx3$
Multiply the sub - scripts in the empirical formula by $n$.
The molecular formula is $C_{6}H_{12}O_{3}$.

Answer:

empirical formula: $C_{2}H_{4}O$
molecular formula: $C_{6}H_{12}O_{3}$