Question

a compounds percent composition is 40.0% carbon, 6.71% hydrogen, and 53.29% oxygen by mass. if its molar mass is approximately 60.05 g/mol, what is its molecular formula? o c₂h₄o₂ o cho₃ o ch₂o o c₃h₈o o c₄h₈o₅

Explanation:

Step1: Calculate moles of each element

Assume 100 g of the compound. So, mass of C = 40.0 g, mass of H = 6.71 g, mass of O = 53.29 g.
Moles of C: $n_{C}=\frac{40.0\ g}{12.01\ g/mol}\approx3.33\ mol$
Moles of H: $n_{H}=\frac{6.71\ g}{1.008\ g/mol}\approx6.66\ mol$
Moles of O: $n_{O}=\frac{53.29\ g}{16.00\ g/mol}\approx3.33\ mol$

Step2: Find the empirical - formula ratio

Divide each number of moles by the smallest number of moles (3.33 mol in this case).
Ratio of C : H : O = $\frac{3.33}{3.33}:\frac{6.66}{3.33}:\frac{3.33}{3.33}=1:2:1$
So, the empirical formula is $CH_{2}O$.

Step3: Calculate the empirical - formula mass

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

Step4: Determine the molecular formula

Given molar mass = 60.05 g/mol.
$n=\frac{\text{Molar mass}}{\text{Empirical - formula mass}}=\frac{60.05\ g/mol}{30.03\ g/mol}\approx2$
Molecular formula = $n\times$(Empirical formula) = $C_{2}H_{4}O_{2}$

Answer:

E. $C_{2}H_{4}O_{2}$