Question

1. 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. c₂ho
c. ch₃o
d. ch₂o

Explanation:

Step1: Assume 100g of the compound

If we assume 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 $C$ is $M_{C} = 12g/mol$, the molar mass of $H$ is $M_{H}=1g/mol$, and the molar mass of $O$ is $M_{O}=16g/mol$.
The number of moles of $C$: $n_{C}=\frac{m_{C}}{M_{C}}=\frac{40g}{12g/mol}\approx3.33mol$
The number of moles of $H$: $n_{H}=\frac{m_{H}}{M_{H}}=\frac{6.7g}{1g/mol}=6.7mol$
The number of moles of $O$: $n_{O}=\frac{m_{O}}{M_{O}}=\frac{53.3g}{16g/mol}\approx3.33mol$

Step3: Find the mole - ratio

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

Answer:

d. $CH_{2}O$