Q. 7.9 - An Introduction to General, Organic, and Biological Chemistry

Question

Naproxen, found in Aleve, is used to treat the pain and inflammation caused by arthritis. Naproxen has a formula of ${\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3$.
a. How many moles of $\mathrm{C}$ are present in $2.30$ moles of naproxen?
b. How many moles of $\mathrm{H}$ are present in $0.444$ mole of naproxen?
c. How many moles of $\mathrm{O}$ are present in $0.0765$ mole of naproxen? 

Step-by-Step Solution

Answer

Our required answers are:

1. $32.2$ moles of data-custom-editor="chemistry" $\mathrm{C}$ present in $2.30$ moles naproxen .
2.  $6.216$ moles of data-custom-editor="chemistry" $\mathrm{H}$ present in $0.444$ moles naproxen .
3.   $0.229$ moles of data-custom-editor="chemistry" $\mathrm{O}$ present in $0.0765$ moles naproxen .

Part(a) Step 1: Given Information

We need to find moles of $\mathrm{C}$ in $2.30$ moles of naproxen  .

Part(a) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3$ = $14$ moles of $\mathrm{C}$

Conversion Factor= $\frac{14 mole \mathrm{C}}{1 mole {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}$

Moles of $\mathrm{C}$ in $2.30$ moles of  naproxen = $$\begin{gathered} \frac{14 mole \mathrm{C}}{1 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}\times 2.30 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3 \\ =32.2 \mathrm{mole} \mathrm{C} \end{gathered}$$

Hence,  $32.2$ moles of $\mathrm{C}$ present in $2.30$moles naproxen .

Part(b) Step 1: Given Information

We need to find moles of $\mathrm{H}$ in $0.444$ moles of  naproxen   .

Part(b) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{C}}_{14}{\mathrm{H}}_{{14}_{}}{\mathrm{O}}_3$ = $14$ moles of $\mathrm{H}$ 

Conversion Factor= $\frac{14 mole \mathrm{H}}{1 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}$

Moles of $\mathrm{H}$ in $0.444$ moles of naproxen = $$\begin{gathered} \frac{14 mole \mathrm{H}}{1 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}\times 0.444 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3 \\ =6.216 \mathrm{mole} H \end{gathered}$$

Hence,  $6.216$ moles of $\mathrm{H}$ present in $0.444$ moles naproxen .

Part(c) Step 1: Given Information

We need to find moles of $\mathrm{O}$ in $0.0765$ moles of naproxen  .

Part(c) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3$ = $3$ moles of  $\mathrm{O}$

Conversion Factor= $\frac{3 mole \mathrm{O}}{1 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}$

Moles of $\mathrm{O}$ in $0.0765$ moles of naproxen = $$\begin{gathered} \frac{3 mole \mathrm{O}}{1 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3}\times 0.0765 \mathrm{mole} {\mathrm{C}}_{14}{\mathrm{H}}_{14}{\mathrm{O}}_3 \\ =0.229 \mathrm{mole} \mathrm{O} \end{gathered}$$

Hence,  $0.229$ moles of $\mathrm{O}$ present in $0.0765$ moles naproxen .