Question

in the given example, what is the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
a. c₂h₂o₂
b. ch₂o
c. c₂ho
d. ch₃o

Explanation:

Step1: Assume 100g of the compound

If we assume we have 100g of the compound, then the mass of carbon ($C$) is $m_{C}=40g$, the mass of hydrogen ($H$) is $m_{H} = 6.7g$, and the mass of oxygen ($O$) is $m_{O}=53.3g$.

Step2: Calculate the number of moles of each element

The molar mass of carbon $M_{C}=12g/mol$, the molar mass of hydrogen $M_{H}=1g/mol$, and the molar mass of oxygen $M_{O}=16g/mol$.
The number of moles of carbon $n_{C}=\frac{m_{C}}{M_{C}}=\frac{40g}{12g/mol}=\frac{10}{3}mol$.
The number of moles of hydrogen $n_{H}=\frac{m_{H}}{M_{H}}=\frac{6.7g}{1g/mol}=6.7mol$.
The number of moles of oxygen $n_{O}=\frac{m_{O}}{M_{O}}=\frac{53.3g}{16g/mol}=\frac{53.3}{16}mol\approx3.33mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (in this case, the number of moles of oxygen $n_{O}\approx3.33mol$).
For carbon: $\frac{n_{C}}{n_{O}}=\frac{\frac{10}{3}}{3.33}\approx1$.
For hydrogen: $\frac{n_{H}}{n_{O}}=\frac{6.7}{3.33}\approx2$.
For oxygen: $\frac{n_{O}}{n_{O}} = 1$.

Answer:

B. $CH_{2}O$