Question

1. a compound contains 75.95% c, 17.72% n, and 6.33% h. what is the empirical formula?

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the masses of \(C\), \(N\), and \(H\) are \(m_{C}=75.95g\), \(m_{N} = 17.72g\), \(m_{H}=6.33g\)

Step2: Calculate the number of moles of each element

The molar mass of \(C\) is \(M_{C}=12.01g/mol\), the molar mass of \(N\) is \(M_{N}=14.01g/mol\), and the molar mass of \(H\) is \(M_{H} = 1.008g/mol\)
The number of moles of \(C\): \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{75.95g}{12.01g/mol}\approx6.324mol\)
The number of moles of \(N\): \(n_{N}=\frac{m_{N}}{M_{N}}=\frac{17.72g}{14.01g/mol}\approx1.265mol\)
The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{6.33g}{1.008g/mol}\approx6.28mol\)

Step3: Divide by the smallest number of moles

Divide the number of moles of each element by the smallest number of moles (\(n_{N}\approx1.265mol\))
For \(C\): \(\frac{n_{C}}{n_{N}}=\frac{6.324}{1.265}\approx5\)
For \(N\): \(\frac{n_{N}}{n_{N}} = 1\)
For \(H\): \(\frac{n_{H}}{n_{N}}=\frac{6.28}{1.265}\approx5\)

Answer:

\(C_{5}NH_{5}\)