Question

If reproducibility is$\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{mg}$, product mass might have been0.1339to 0.1349 g. Find wt\% $\mathbf{Al}\mathbf{(}{\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{6}}\mathbf{NO}{\mathbf{)}}_{\mathbf{3}}$if the product weighed 0.1339 g.

Short answer

$\mathrm{The}\mathrm{weight}\mathrm{percentage}\mathrm{of}\mathrm{aluminum}\mathrm{complex}\mathrm{in}\mathrm{sample}\mathrm{is}30.27\%.$

Step-by-step solution

Concept used

Divide the determined mass of the substance by the total mass of the sample, then multiply the whole thing by 100,the mass percentage of the compound may be calculated.

The formula may be used to compute the mass percent, and the formula can also be used to provide a solution.

$\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{\left(}\mathit{M}\mathit{\right)}\mathit{=}\frac{\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{}\mathit{o}\mathit{f}\mathit{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathit{\left(}\mathit{i}\mathit{n}\mathit{m}\mathit{o}\mathit{l}\mathit{\right)}}{\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{\hspace{0.33em}}\mathit{o}\mathit{f}\mathit{\hspace{0.33em}}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{(}\mathit{i}\mathit{n}\mathit{\hspace{0.33em}}\mathit{L}\mathit{)}}$

Find the weight percent of Al(C9H6NO)3 in the original mixture

The proportion of aluminium complex in the sample by weight.

$x+y=1.0843g$

x is mass of${\mathrm{AIQ}}_3$

y is mass of ${\mathrm{MgQ}}_2$

Anion of Qis8-hydroxyquinoline

One and a half moles of aluminium oxide equals one and a half moles of aluminium.

moles of${\mathrm{Al}}_2{\mathrm{O}}_3=0.5\frac{\mathrm{xg}}{459.43\mathrm{g}/\mathrm{mol}}$

The amount of moles of magnesium equals two moles of magnesium oxide.

$=\frac{\mathrm{yg}}{312.61\mathrm{g}/\mathrm{mol}}$

$0.5\frac{x}{459.43}(101.96)+\frac{y}{312.61}(40.304)=0.1339g$

$0.5\frac{x}{459.43}/(101.96)=$mass of aluminum oxide

$\frac{y}{312.61}(40.304)=$mass of magnesium oxide

The preceding equation becomes $y=1.0843-x$when $y=1.0843-x$is substituted.

$$\begin{gathered} 0.5\left(\frac{x}{459.43}\right)\left(101.96\right)+\left(\frac{1.0843-x}{312.61}\right)\left(40.304\right)=0.1339g \\ x=0.328g \end{gathered}$$

Using the specified weight of the sample, the weight percent of aluminium complex is determined.

role="math" localid="1663666073851" $$\begin{gathered} =\frac{0.3280\mathrm{mgH}}{1.0843\mathrm{mg}\mathrm{sample}}\times 100\% \\ =30.27\% \end{gathered}$$

Thus, the aluminium complex weight in a particular sample is 30.27%.