Question

Caffeine ( Mol. Wt. $ = 194$ ) contains $28.9\% $ of nitrogen. How many numbers of nitrogen atoms are present in one molecule of caffeine?
(A). $2$
(B). $5$
(C). $3$
(D). $4$

Answer 1

Hint:
The $1$ mole of a substance contains Avogadro’s number, that is $6.022 \times {10^{23}}$ molecules or atoms. Moreover, $1$ mole is also equal to the gram molecular mass/atomic mass of that substance.

Complete step by step answer:
As we know that a mole is defined as the amount of substance which has mass equal to gram molar mass. The molar weight of caffeine given is $194g$ . This means that $1$ mole of caffeine has a weight of $194g$ . We also know that a mole is equal to the Avogadro number $\left( {6.022 \times {{10}^{23}}} \right)$ of molecules or atoms.
This implies that
$6.022 \times {10^{23}}$ molecules of caffeine contain mole ${\text{ = 1}}$
$\therefore 1$ molecule of caffeine contain mole ${\text{ = 1mole}} \times \dfrac{1}{{6.022 \times {{10}^{23}}}}$
We have already discussed that $1$ mole of caffeine has a mass of $194g$ . So, the
mass of $1$ molecule will be ${\text{ = 194g}} \times \dfrac{1}{{6.022 \times {{10}^{23}}}}$
 $ = \dfrac{{194 \times {{10}^{ - 23}}}}{{6.022}}grams$
Caffeine contains $28.9\% $ of nitrogen of mass, so the amount of nitrogen present in $1$ molecule of caffeine = $\operatorname{mass} of 1 \operatorname{molecule} of caffeine{\text{ }} \times {\text{28}}{\text{.9% }}$
$ = \dfrac{{194 \times {{10}^{23}}}}{{6.022}}grams \times \dfrac{{28.9}}{{100}}$
$ = \dfrac{{28.9 \times 194 \times {{10}^{ - 23}}}}{{6.022 \times 100}}g$
which is mass of nitrogen present in $1$ molecule of caffeine
Number of moles of nitrogen present in $1$ molecule of caffeine $ = \dfrac{\text{Given mass of nitrogen in 1 molecule of caffeine}}{\text{molar weight of Nitrogen}}$
We have already find the weight of nitrogen in $1$ molecule of caffeine which is $\dfrac{{28.9 \times 194 \times {{10}^{ - 23}}}}{{6.022 \times 100}}grams$
And we know that the molar weight of nitrogen is $14$ gram. Putting values, we get
No of moles of nitrogen present in $1$ molecule of caffeine = $\dfrac{{28.9 \times 194 \times {{10}^{23}}}}{{\dfrac{{6.022 \times 100}}{{14}}}}$
$ = \dfrac{{28.9 \times 194 \times {{10}^{23}}}}{{6.022 \times 14 \times 100}}$ moles
Moreover we know
$1$ mole of nitrogen contains atom $ = 6.022 \times {10^{23}}$ atoms
This implies that
$\dfrac{{28.9 \times 194 \times {{10}^{23}}}}{{6.022 \times 14 \times 100}}$ moles of nitrogen contain atom = $6.022 \times {10^{23}} \times \left[ {\dfrac{{28.9 \times 194 \times {{12}^{23}}}}{{6.022 \times 14 \times 100}}} \right]$
$ \approx 4atoms$
This implies that $1$ molecule of caffeine contains $4$ atoms of nitrogen.
Hence option (D) is the correct answer.

Note:The above question can be directly solved by using the formula :
$\text{No . of atom of N} = \dfrac{\text{mass of N in one molecule of caffeine}}{\text{molar mass of Nitrogen}}$
-Since mass of $N$ in one molecule of caffeine is $ = \dfrac{{28.9}}{{100}} \times 194$
-And the molecular weight of nitrogen is $14$.