Question

A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of a solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green endpoint. Find the weight percent of${\mathrm{NH}}_2$(FM 17.031) in the cleaner.

Short answer

The weight percent of ${\mathrm{NH}}_2$ in the given sample of window cleaner is calculated as$2.86\%$

Step-by-step solution

Definition of the weight percentage.

Weight percent is one of the parameters that are used to express the concentration of a solution.

It is expressed as,

$\mathrm{The}\mathrm{weight}\mathrm{percent}\mathrm{of}\mathrm{a}\mathrm{solute}=\frac{\mathrm{weight}\mathrm{of}\mathrm{solute}}{\mathrm{weight}\mathrm{of}\mathrm{solution}}\times 100\%$

Calculation of the weight of the cleaner.

The given data is

$$\begin{gathered} \mathrm{Weight}\mathrm{of}\mathrm{sample}=10.231\mathrm{g} \\ \mathrm{Weight}\mathrm{of}\mathrm{water}=39.466\mathrm{g} \\ \mathrm{Weight}\mathrm{of}\mathrm{solution}\mathrm{taken}\mathrm{for}\mathrm{titration}=4.373\mathrm{g} \\ \mathrm{and}, \\ \mathrm{Weight}\mathrm{of}\mathrm{HCl}=14.22\mathrm{mL} \\ \mathrm{Molarity}\mathrm{of}\mathrm{HCl}=0.1063\mathrm{M} \end{gathered}$$

The amount of sample cleaner that is diluted and titrated is equivalent to the amount of cleaner present in 4.373 g of solution.

So, the weight of cleaner in $4.373\mathrm{g}$of solution that is titrated is;

role="math" localid="1663559709519" $$\begin{gathered} =\frac{\mathrm{weight}\mathrm{of}\mathrm{diluted}\mathrm{sample}\mathrm{solution}}{\mathrm{weight}\mathrm{of}\mathrm{water}+\mathrm{weight}\mathrm{of}\mathrm{sample}\mathrm{used}}\times \mathrm{weight}\mathrm{of}\mathrm{sample}\mathrm{used} \\ =\frac{4.373\mathrm{g}}{(39.466\mathrm{g}+10.231\mathrm{g})}\times 10.231\mathrm{g} \\ =0.9003\mathrm{g} \end{gathered}$$

 Determination of the weight along with the equivalence point.

At the equivalent point,

$\mathrm{molHCl}={\mathrm{molNH}}_3$

As we know,

no. of moles = molarity $\times$volume of solution

Number of moles of HCl and ${\mathrm{NH}}_3$are,

$\mathrm{mol}\mathrm{HCI}=\mathrm{mol}{\mathrm{NH}}_3$

$=(0.01422\mathrm{L})\times (0.1063\mathrm{M})$

${\mathrm{molNH}}_3=1.512\mathrm{mmol}$

The formula mass of ammonia is 17.031 and the weight of ammonia present in $10.231\mathrm{g}$of sample cleaner is calculated as follows

localid="1663560572748" $\mathrm{no}.\mathrm{of}\mathrm{moles}=\frac{\mathrm{mass}\mathrm{of}{\mathrm{NH}}_3}{\mathrm{formula}\mathrm{mass}\mathrm{of}\mathrm{NH}3}$

Therefore,

localid="1663560903293" $$\begin{gathered} \mathrm{mass}\mathrm{of}{\mathrm{NH}}_3=\mathrm{Formula}\mathrm{of}{\mathrm{NH}}_3\times \mathrm{no}.\mathrm{of}\mathrm{moles} \\ =17.031\mathrm{g}/\mathrm{mol}\times 1.512\times {10}^{-3}\mathrm{mol} \\ =25.75\times {10}^{-3} \\ =25.75\mathrm{mg} \end{gathered}$$

Now, the weight percent of ${\mathrm{NH}}_3$is,

localid="1663561430961" $$\begin{gathered} \mathrm{Weight}\mathrm{percentof}{\mathrm{NH}}_3=\frac{\mathrm{weight}\mathrm{of}{\mathrm{NH}}_3}{\mathrm{weight}\mathrm{of}\mathrm{cleaner}}\times 100\mathrm{\%} \\ =\frac{25.75\times {10}^{-3}\mathrm{g}}{0.9003\mathrm{g}}\times 100 \\ =2.86\% \end{gathered}$$