Question

1. determine the empirical and molecular formulas of caffeine. its molar mass is 194.19 g/mol and its percentage composition is 49.48 % carbon, 5.15 % hydrogen, 28.87 % nitrogen, and 16.49 % oxygen. ti c

Explanation:

Step1: Assume 100g sample

Assume we have a 100 - g sample of caffeine. Then the masses of each element are:

• Mass of \(C = 49.48g\)
• Mass of \(H=5.15g\)
• Mass of \(N = 28.87g\)
• Mass of \(O=16.49g\)

Step2: Calculate the number of moles of each element

Use the formula \(n=\frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass (\(M_C = 12.01g/mol\), \(M_H=1.008g/mol\), \(M_N = 14.01g/mol\), \(M_O = 16.00g/mol\))

• \(n_C=\frac{49.48g}{12.01g/mol}\approx4.12mol\)
• \(n_H=\frac{5.15g}{1.008g/mol}\approx5.11mol\)
• \(n_N=\frac{28.87g}{14.01g/mol}\approx2.06mol\)
• \(n_O=\frac{16.49g}{16.00g/mol}\approx1.03mol\)

Step3: Find the mole ratio

Divide each number of moles by the smallest number of moles (\(n_{min}=1.03mol\))

• \(x_C=\frac{4.12mol}{1.03mol}\approx4\)
• \(x_H=\frac{5.11mol}{1.03mol}\approx5\)
• \(x_N=\frac{2.06mol}{1.03mol}=2\)
• \(x_O=\frac{1.03mol}{1.03mol}=1\)

The empirical formula is \(C_4H_5N_2O\)

Step4: Calculate the empirical formula molar mass

\(M_{empirical}=(4\times12.01 + 5\times1.008+2\times14.01 + 1\times16.00)g/mol=(48.04+5.04 + 28.02+16.00)g/mol = 97.1g/mol\)

Step5: Find the ratio \(n\)

\(n=\frac{M_{molecular}}{M_{empirical}}=\frac{194.19g/mol}{97.1g/mol}\approx2\)

Step6: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n = 2\)
The molecular formula is \(C_8H_{10}N_4O_2\)

Answer:

The empirical formula of caffeine is \(C_4H_5N_2O\) and the molecular formula is \(C_8H_{10}N_4O_2\)