Q. 7.7 - An Introduction to General, Organic, and Biological Chemistry

Question

 Quinine, ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$, is a component of tonic water and bitter lemon.
a. How many moles of $\mathrm{H}$ are in $1.5$ moles of quinine?
b. How many moles of $\mathrm{C}$ are in $5.0$ moles of quinine?
c. How many moles of $\mathrm{N}$ are in $0.020$ mole of quinine? 

Step-by-Step Solution

Answer

Our required answers are:

1. $36$moles of $\mathrm{H}$
2. $100$moles of $\mathrm{C}$
3. $0.040$moles of $\mathrm{N}$
   

Part(a) Step 1: Given Information

We need to find moles of $\mathrm{H}$ in $1.5$ moles of quinine

Part(a) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$ = $24$ moles of  $\mathrm{H}$

Conversion Factor=$\frac{24 moles \mathrm{H}}{1 mole {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2 }$

Moles of $\mathrm{H}$ in $1.5$ moles of quinine = $$\begin{gathered} \frac{24 mole \mathrm{H}}{1 \mathrm{mole} {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}\times 1.5 \\ =36 \end{gathered}$$

Hence, $36$ moles of $\mathrm{H}$ in $1.5$ mole quinine.

Part(b) Step 1: Given Information

We need to find moles of $\mathrm{C}$ in $5.00$ moles of quinine

Part(b) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$ = $20$ moles of  $\mathrm{C}$

Conversion Factor= $\frac{20 mole \mathrm{C}}{1 mole {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}$

Moles of $\mathrm{C}$  in $5.00$ moles of quinine = $$\begin{gathered} \frac{20 mole \mathrm{C}}{1 mole C_{20}{H}_{24}{N}_2{O}_2}\times 5 mole C_{20}{H}_{24}{N}_2{O}_2 \\ = 100 mole \mathrm{C} \end{gathered}$$

Hence, $100$ moles of $\mathrm{C}$ in $5$ mole quinine.

Part(c) Step 1: Given Information

We need to find moles of $\mathrm{N}$ in $0.020$ moles of quinine

Part(c) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$ =  $2$ moles of  $\mathrm{N}$

Conversion Factor= $\frac{2 mole \mathrm{N}}{ 1 \mathrm{mole} {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}$

Moles of  $\mathrm{N}$ in $0.020$ moles of quinine = $$\begin{gathered} \frac{2 mole \mathrm{N}}{1 \mathrm{mole} {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}\times 0.020 \mathrm{mole} {\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2 \\ =0.040 mole \mathrm{N} \end{gathered}$$

Hence, $0.040$ moles of $\mathrm{N}$ in $0.020$ mole quinine.