Question

1. a compound contains c, h, and o. the percentages of h and c are 6.73% h, and 39.99% c by mass. the molar mass of the compound is 60.06 g mole⁻¹. what is the molecular formula of the compound?

a) c₂h₄o₂
b) c₂h₃o₂
c) ch₂o
d) c₃h₆o₃
e) c₄h₈o₄

Explanation:

Step1: Calculate the percentage of O

Since the compound only contains C, H, and O, the percentage of O is $100-(6.73 + 39.99)=53.28\%$.

Step2: Calculate the mass of each element in 1 - mole of the compound

For H: $m_H=60.06\times6.73\% \approx 4.04$ g. Since the molar - mass of H is 1 g/mol, the number of moles of H, $n_H=\frac{4.04}{1}=4$ mol.
For C: $m_C=60.06\times39.99\% \approx 24$ g. Since the molar - mass of C is 12 g/mol, the number of moles of C, $n_C=\frac{24}{12}=2$ mol.
For O: $m_O=60.06\times53.28\% \approx 32$ g. Since the molar - mass of O is 16 g/mol, the number of moles of O, $n_O=\frac{32}{16}=2$ mol.

Answer:

A. $C_2H_4O_2$