Question

1. what is the empirical formula for a compound containing 67.9% c, 5.7% h, and 26.4% n?

Explanation:

Step1: Assume 100g of the compound

If we assume we have 100g of the compound, then the masses of \(C\), \(H\), and \(N\) are \(m_{C}=67.9g\), \(m_{H} = 5.7g\), \(m_{N}=26.4g\)

Step2: Calculate the number of moles

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 \(C\): \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{67.9g}{12.01g/mol}\approx5.65mol\)

The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{5.7g}{1.008g/mol}\approx5.66mol\)

The number of moles of \(N\): \(n_{N}=\frac{m_{N}}{M_{N}}=\frac{26.4g}{14.01g/mol}\approx1.88mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n_{N} = 1.88mol\))

For \(C\): \(\frac{n_{C}}{n_{N}}=\frac{5.65mol}{1.88mol}\approx3\)

For \(H\): \(\frac{n_{H}}{n_{N}}=\frac{5.66mol}{1.88mol}\approx3\)

For \(N\): \(\frac{n_{N}}{n_{N}} = 1\)

Answer:

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