Chapter 11, Problem 77A

Interpretation Introduction

Interpretation:

The molecular formula of caffeine needs to be determined when the percent composition of caffeine is 49.5% C, 5.20% H, 16.5% O, and 28.9% N and its molar mass is 194.1 g/mol.

Concept Introduction:

The ratio of number of atoms is equal to fractions ratio of their percentage and relative atomic mass and the addition of standard atomic masses of constituent atoms gives molar mass of molecule.

Answer to Problem 77A

The molecular formula of caffeine is ${\text{C}}_8{\text{H}}_{10}{\text{O}}_2{\text{N}}_4$

Explanation of Solution

Step 1:

From the periodic table in the textbook, it is known as:

  $\begin{array}{l}\text{Ar}\left(\text{C}\right)\text{=12}\text{.01 g/mol}\\ \text{Ar}\left(\text{C}\right)\text{=16}\text{.00 g/mol}\\ \text{Ar}\left(\text{N}\right)\text{=14}\text{.01 g/mol}\\ \text{Ar}\left(\text{H}\right)\text{=1}\text{.008 g/mol}\end{array}$

Step 2:

The ratio of number of atoms is equal to fractions ratio of their percentage and relative atomic mass so one has as:

  $\begin{array}{c}\text{N}\left(\text{C}\right)\text{:N}\left(\text{O}\right)\text{:N}\left(\text{N}\right)\text{:N}\left(\text{H}\right)\text{=}\frac{\text{\%C}}{\text{Ar}\left(\text{C}\right)}\text{:}\frac{\text{\%O}}{\text{Ar}\left(\text{O}\right)}\text{:}\frac{\text{\%N}}{\text{Ar}\left(\text{N}\right)}\text{:}\frac{\text{\%H}}{\text{Ar}\left(\text{H}\right)}\\ \text{=}\frac{49.50/100}{12.01}:\frac{16.50/100}{16.00}:\frac{28.90/100}{14.01}:\frac{5.20/100}{1.008}\\ =0.0412:0.0103:0.0206:0.0516\end{array}$

Step 3:

The numbers are divided with smallest one.

  $\begin{array}{c}\text{N}\left(\text{C}\right)\text{:N}\left(\text{O}\right)\text{:N}\left(\text{N}\right)\text{:N}\left(\text{H}\right)\text{= }0.0412:0.0103:0.0206:0.0516/:0.0103\\ =4:1:2:5\end{array}$

Thus, the empirical formula is ${\text{C}}_4{\text{H}}_5{\text{ON}}_2$ .

Step 4:

The molar mass is as:

  $\begin{array}{c}\text{M}\left({\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{ON}}_{\text{2}}\right)\text{=4×Ar}\left(\text{C}\right)\text{+5×Ar}\left(\text{H}\right)\text{+Ar}\left(\text{O}\right)\text{+2×Ar}\left(\text{N}\right)\\ \text{=4×12}\text{.01+5×1}\text{.008+16}\text{.00+2×14}\text{.01}\\ \text{=97}\text{.1 g/mol}\end{array}$

Step 5:

Empirical formula is simplest ratio of number of atoms in molecule and a molecular formula is whole number product of an empirical formula.

  $\begin{array}{c}\frac{\text{M}\left(\text{caffeine}\right)}{\text{M}\left({\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{ON}}_{\text{2}}\right)}=\frac{194.1}{97.1}\\ =2\end{array}$

Molecular formula of caffeine is ${\text{2×C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{ON}}_{\text{2}}{\text{=C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{O}}_{\text{2}}{\text{N}}_{\text{4}}$

Hence the molecular formula is ${\text{C}}_8{\text{H}}_{10}{\text{O}}_2{\text{N}}_4$ .