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 ch₂o

o c₃h₈o

o c₄h₆o₅

o c₂h₄o₂

o cho₃

Explanation:

Step1: Assume 100g of the compound

This allows us to convert mass - percentages to masses. So, we have 40.0 g of C, 6.71 g of H, and 53.29 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.
For C: $n_{C}=\frac{40.0\ g}{12.01\ g/mol}\approx3.33\ mol$
For H: $n_{H}=\frac{6.71\ g}{1.01\ g/mol}\approx6.64\ mol$
For O: $n_{O}=\frac{53.29\ g}{16.00\ g/mol}\approx3.33\ mol$

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (3.33 mol in this case).
For C: $\frac{3.33\ mol}{3.33\ mol}=1$
For H: $\frac{6.64\ mol}{3.33\ mol}\approx2$
For O: $\frac{3.33\ mol}{3.33\ mol}=1$
The empirical formula is $CH_{2}O$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $CH_{2}O$ is $12.01\ g/mol+(2\times1.01\ g/mol)+16.00\ g/mol = 30.03\ g/mol$

Step5: Find the ratio of molar mass to empirical - formula mass

$n=\frac{60.05\ g/mol}{30.03\ g/mol}\approx2$

Step6: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by $n$.
The molecular formula is $C_{2}H_{4}O_{2}$

Answer:

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