Question

1. 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 sample

If we assume a 100 - g sample of caffeine, then the masses of each element are:

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

Step2: Calculate the number of moles of each element

The molar mass of \(C\) is \(M(C)=12.01g/mol\), the molar mass of \(H\) is \(M(H)=1.008g/mol\), the molar mass of \(N\) is \(M(N) = 14.01g/mol\), and the molar mass of \(O\) is \(M(O)=16.00g/mol\)

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

Step3: Find the mole ratio

Divide each number of moles by the smallest number of moles (\(n(O)\approx1.03mol\))

• \(x=\frac{n(C)}{n(O)}\approx\frac{4.12mol}{1.03mol} = 4\)
• \(y=\frac{n(H)}{n(O)}\approx\frac{5.15mol}{1.03mol}=5\)
• \(z=\frac{n(N)}{n(O)}\approx\frac{2.06mol}{1.03mol}=2\)
• \(w = 1\)

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

Step4: Calculate the ratio of molar masses

Let \(n=\frac{M_{molecular}}{M_{empirical}}\), where \(M_{molecular}=194.19g/mol\) and \(M_{empirical}\approx97.1g/mol\)
\(n=\frac{194.19g/mol}{97.1g/mol}\approx2\)

Step5: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n = 2\)
The molecular formula is \(C_{4\times2}H_{5\times2}N_{2\times2}O_{1\times2}=C_8H_{10}N_4O_2\)

Answer:

C. \(C_8H_{10}N_4O_2\)