Question

1. calculate the empirical formula of a compound that is 62.1% c, 13.8% h, and 24.1% n.

Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of the compound. Then, the mass of C is \(m_{C}=62.1g\), the mass of H is \(m_{H} = 13.8g\), and the mass of N is \(m_{N}=24.1g\).

Step2: Calculate the number of moles of each element

The molar mass of C is \(M_{C}=12.01g/mol\), the molar mass of H is \(M_{H}=1.008g/mol\), and the molar mass of N is \(M_{N}=14.01g/mol\).
The number of moles of C: \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{62.1g}{12.01g/mol}\approx5.17mol\)
The number of moles of H: \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{13.8g}{1.008g/mol}\approx13.7mol\)
The number of moles of N: \(n_{N}=\frac{m_{N}}{M_{N}}=\frac{24.1g}{14.01g/mol}\approx1.72mol\)

Step3: Divide each number of moles by the smallest number of moles

The smallest number of moles is \(n_{min}\approx1.72mol\)
For C: \(\frac{n_{C}}{n_{min}}=\frac{5.17mol}{1.72mol}\approx3\)
For H: \(\frac{n_{H}}{n_{min}}=\frac{13.7mol}{1.72mol}\approx8\)
For N: \(\frac{n_{N}}{n_{min}}=\frac{1.72mol}{1.72mol} = 1\)

Answer:

The empirical formula of the compound is \(C_{3}H_{8}N\)