Q. 7.112 - An Introduction to General, Organic, and Biological Chemistry

Question

A toothpaste contains $0.240 \%$ by mass sodium fluoride used to prevent dental caries and $5.0 \%$ by mass ${\mathrm{KNO}}_3$, which decreases pain sensitivity. One tube contains $119 g$ of toothpaste
$\mathrm{NaF}$ and ${\mathrm{KNO}}_3$ are components found in toothpastes.

1. How many moles of $\mathrm{NaF}$ are in the tube of toothpaste?
2. How many fluoride ions, ${\mathrm{F}}^{-}$, are in the tube of toothpaste?
3. How many grams of sodium ion, ${\mathrm{Na}}^{+}$, are in $1.50 g$ of toothpaste?
4. How many ${\mathrm{KNO}}_3$ formula units are in the tube of toothpaste? 

Step-by-Step Solution

Answer

Our required answers are:

1. $0.0068$ moles of $\mathrm{NaF}$
   
2. $4.09\times {10}^{21}$ ions
3. $1.17\times {10}^{22} g$ of ${\mathrm{Na}}^{+}$ions 
4. $3.14\times {10}^{21}$ formula units of ${\mathrm{KNO}}_3$ present in toothpaste

Part (a) Step 1: Given Information

We are given a toothpaste tube of $119 g$ containing $0.240 \%$ by mass of $\mathrm{NaF}$.

We need to find moles of $\mathrm{NaF}$ in toothpaste.

Part (a) Step 2: Explanation

Amount of toothpaste=$119 g$

Composition of $\mathrm{NaF}$ in toothpaste= $0.240 \% = \frac{0.240\times 119 g}{100}=0.2856 g$

Number of moles =$\frac{Weight of subs\tan ce}{Molecular mass of subs\tan ce}$

So, number of moles in $\mathrm{NaF}$ = $\frac{0.2856 \cancel{g}}{42 \cancel{g}}=0.068$

Hence, $0.0068$ moles of $\mathrm{NaF}$ is present in toothpaste.

Part (b) Step 1: Given Information

We are given a toothpaste tube of $119 g$ containing $0.240 \%$ by mass of $\mathrm{NaF}$.

We need to find how many fluoride ions are in toothpaste.

Part (b) Step 2: Explanation

• Moles of $\mathrm{NaF}=0.0068$(Calculated in part (a) step $2$)
• Number of ions= Number of moles* Avogadro’s number
• Number of ${\mathrm{F}}^{-}$ ions in toothpaste = $0.0068\times (6.022\times {10}^{23})= 4.09\times {10}^{21}$

Hence, in one toothpaste tube $4.09\times {10}^{21}$ fluoride ions are present.

Part (c) Step 1: Given Information

We are given a toothpaste tube of $119 g$containing  $0.240 \%$ by mass of $\mathrm{NaF}$.

We need to find how many grams of ${\mathrm{Na}}^{+}$ ions are in toothpaste.

Part (c) Step 2: Explanation

• Composition of $\mathrm{NaF} = \frac{0.240\times 1.50 g}{100}=0.036 \mathrm{g}$

• Moles of $\mathrm{NaF}$ in $1.50 g$of toothpaste = $\frac{0.036 \cancel{g}}{42 \cancel{g}}=0.00085$
• Number of ions of $\mathrm{Na}$=$0.00085\times 6.022\times {10}^{23}=5.1\times {10}^{20}$

Hence, Grams of sodium ion= $5.1\times {10}^{20}\times 23 g=1.17\times {10}^{22} g$

Part (d) Step 1: Given Information

We are given that in toothpaste tube of $119 g$, $5 \%$ by mass of ${\mathrm{KNO}}_3$ is present.

We need to find formula unit of ${\mathrm{KNO}}_3$ present in toothpaste.

Part (d) Step 2: Explanation

• Composition by mass of ${\mathrm{KNO}}_3$ = $\frac{5\times 119}{100}=5.95 g$
• Moles of ${\mathrm{KNO}}_3= \frac{5.95 \cancel{g}}{101 \cancel{g}}= 0.058$
• Number of formula units = Number of moles * Avogadro’s number 

         $$\begin{gathered} =0.058\times 6.022\times {10}^{23} \\ =0.034\times {10}^{23} \\ =3.4\times {10}^{21} \end{gathered}$$

Hence, number of formula units of ${\mathrm{KNO}}_3$ is $3.4\times {10}^{21}$.