Question

caffeine, a stimulant in coffee and tea has a molar mass of 194.19 g/mol and mass composition 49.48 % c, 5.19 % h, 28.85 % n, and 16.48 % o. what is the molecular formula of caffeine?
a. c4h5n2o
b. c5h5no
c. c8h10n4o2
d. c8h10n2o
e. c4h2n2o

Explanation:

Step1: Assume 100g of caffeine

So, \(m(C) = 49.48g\), \(m(H)=5.19g\), \(m(N) = 28.85g\), \(m(O)=16.48g\)

Step2: Calculate the number of moles of each element

• For \(C\): \(n(C)=\frac{49.48g}{12.01g/mol}\approx4.12mol\)
• For \(H\): \(n(H)=\frac{5.19g}{1.008g/mol}\approx5.15mol\)
• For \(N\): \(n(N)=\frac{28.85g}{14.01g/mol}\approx2.06mol\)
• For \(O\): \(n(O)=\frac{16.48g}{16.00g/mol}\approx1.03mol\)

Step3: Find the mole ratio

Divide each number of moles by the smallest number of moles (\(n(O) = 1.03mol\))

• \(C:\frac{4.12}{1.03}\approx4\)
• \(H:\frac{5.15}{1.03}\approx5\)
• \(N:\frac{2.06}{1.03}\approx2\)
• \(O:\frac{1.03}{1.03}=1\)

The empirical formula is \(C_4H_5N_2O\), molar mass of empirical formula \(M=(4\times12.01 + 5\times1.008+2\times14.01+1\times16.00)g/mol=(48.04 + 5.04+28.02 + 16.00)g/mol = 97.1g/mol\)

Step4: Find the multiple \(n\)

\(n=\frac{194.19g/mol}{97.1g/mol}\approx2\)
Multiply the empirical formula by \(n = 2\)
The molecular formula is \(C_8H_{10}N_4O_2\)

Answer:

C. \(C_8H_{10}N_4O_2\)