Question

a compound is analyzed as 32.0% c, 42.6% o, 18.7% n, and 6.7% h. what is its empirical formula? c₂h₇o₂n c₂h₅o₂n c₃h₇o₃n c₃h₆o₄n₂

Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of the compound. Then the masses are:

• \(m(C)=32.0g\)
• \(m(O) = 42.6g\)
• \(m(N)=18.7g\)
• \(m(H)=6.7g\)

Step2: Calculate the number of moles of each element

Use the formula \(n=\frac{m}{M}\), where \(M\) is the molar mass (\(M(C) = 12.01g/mol\), \(M(O)=16.00g/mol\), \(M(N)=14.01g/mol\), \(M(H) = 1.008g/mol\))

• \(n(C)=\frac{32.0g}{12.01g/mol}\approx2.66mol\)
• \(n(O)=\frac{42.6g}{16.00g/mol}=2.66mol\)
• \(n(N)=\frac{18.7g}{14.01g/mol}\approx1.33mol\)
• \(n(H)=\frac{6.7g}{1.008g/mol}\approx6.65mol\)

Step3: Divide by the smallest number of moles

The smallest number of moles is \(n(N)\approx1.33mol\)

• \(x_{C}=\frac{2.66mol}{1.33mol}=2\)
• \(x_{O}=\frac{2.66mol}{1.33mol}=2\)
• \(x_{N}=\frac{1.33mol}{1.33mol}=1\)
• \(x_{H}=\frac{6.65mol}{1.33mol}\approx5\)

Answer:

\(C_{2}H_{5}O_{2}N\)