Question

02 question (2 points)
1st attempt
part 1 (1 point)
the percent composition by mass of an unknown compound with a molecular mass of 120.104 u is 40.002% c, 6.7135% h, and 53.284% o. determine the compounds empirical and molecular formulas.
empirical formula:
part 2 (1 point)
molecular formula:

Explanation:

Step1: Assume 100 g of the compound

If we assume 100 g of the compound, then the masses of C, H, and O are:
Mass of C = 40.002 g, Mass of H = 6.7135 g, Mass of O = 53.284 g

Step2: Calculate the moles of each element

Moles of C: $n_{C}=\frac{40.002\ g}{12.01\ g/mol}\approx 3.331\ mol$
Moles of H: $n_{H}=\frac{6.7135\ g}{1.008\ g/mol}\approx 6.660\ mol$
Moles of O: $n_{O}=\frac{53.284\ g}{16.00\ g/mol}\approx 3.330\ mol$

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (3.330 mol in this case).
For C: $\frac{3.331\ mol}{3.330\ mol}\approx 1$
For H: $\frac{6.660\ mol}{3.330\ mol}= 2$
For O: $\frac{3.330\ mol}{3.330\ mol}= 1$
The empirical formula is $CH_{2}O$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $CH_{2}O$ is $M_{empirical}=12.01\ g/mol+(2\times1.008\ g/mol)+16.00\ g/mol = 30.03\ g/mol$

Step5: Determine the multiple (n)

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{120.104\ g/mol}{30.03\ g/mol}\approx 4$

Step6: Find the molecular formula

Multiply the sub - scripts in the empirical formula by n.
The molecular formula is $(CH_{2}O)_4 = C_{4}H_{8}O_{4}$

Answer:

Empirical formula: $CH_{2}O$
Molecular formula: $C_{4}H_{8}O_{4}$