Question

A 0.4740-g pesticide sample was decomposed by wet ashing and then diluted to 200.0 mL in a volumetric flask. The analysis was completed by treating aliquots of this solution as indicated.

Calculate the percentage of copper in the sample.

Short answer

The percent of ${\mathrm{Cu}}^{2+}$present in the sample of the pesticide is$0.0795\%$.

Step-by-step solution

Step 1. Given information

A 0.4740-g pesticide sample was decomposed by wet ashing and then diluted to 200.0 mL in a volumetric flask. The analysis was completed by treating aliquots of this solution as indicated.

Step 2. Relationship between the absorbance, molar absorptivity, molar analytical concentration, and the volume of solution 

The concentration of the ${\mathrm{Cu}}^{2+}$present in the pesticide is given by establishing the relation between the unknown concentration and absorbance to the known concentration absorbance and the volume of the solution.

The relationship between the absorbance, molar absorptivity, molar analytical concentration of the solution, and the volume of solution is given by.

$A_x=\epsilon b{c}_x\frac{V_x}{V_t}........\left(1\right)$

The expression for the relation between molar absorptivity, the volume of the solution, and molar analytical concentration of the solution for the standard solution added to the unknown solution is:

$A_{x+s}=\epsilon b\frac{c_x{V}_x+c_s{V}_s}{V_t}........\left(2\right)$

Step 3. Expression for the percentage of copper 

Divide the equation (1) by equation (2) and rearrange for $c_x$.

localid="1647016726187" $c_x=\frac{A_x{c}_s{V}_s}{\left(A_{x+s}-A_x\right)V_x}.........\left(3\right)$

The expression for the percentage of copper present in the sample is:

localid="1647017414719" $\%{\mathrm{Cu}}^{2+}=\frac{\left(V_t.c\right)}{m}\times 100..........\left(4\right)$

Step 4. Concentration of Cu2+

The value of $A_x$is 0.671.

The value of $A_{x+s}$is 0.849.

The value of $c_x$is 2.50 ppm.

The value of $V_x$is 5.00 mL.

The value of is 1.00 mL.

Substitute the values in equation (3).

$$\begin{gathered} c_x=\frac{A_x{c}_s{V}_s}{\left(A_{x+s}-A_x\right)V_x} \\ =\frac{0.670\times 2.50\mathrm{ppm}\times 1.00\mathrm{mL}}{\left(0.849-0.671\right)5.00\mathrm{mL}} \\ =\frac{1.6775\mathrm{pmm}}{0.89} \\ =1.8848\mathrm{ppm} \end{gathered}$$

Step 5. Conversion of ppm to g/mL

The concentration of the ${\mathrm{Cu}}^{2+}$can be determined as follows:

$$\begin{gathered} c=1.8848\mathrm{ppm}\times \frac{1\mathrm{mg}/\mathrm{L}}{1\mathrm{ppm}}\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1\mathrm{L}}{1000\mathrm{mL}} \\ =1.8848\times {10}^{-6}\mathrm{g}/\mathrm{mL} \end{gathered}$$

Step 6. Percentage of copper in the sample 

The value of $V_t$is 200 mL.

The value of $c$is$1.8848\times {10}^{-6}\mathrm{g}/\mathrm{mL}$.

The value of $m$is 0.4740 g.

Substitute the values in equation (4) as follows:

$$\begin{gathered} \%{\mathrm{Cu}}^{2+}=\frac{\left(V_t.c\right)}{m}\times 100 \\ =\frac{\left(200\mathrm{mL}.\times 1.8848\times {10}^{-6}\mathrm{g}/\mathrm{mL}\right)}{0.4740\mathrm{g}}\times 100 \\ =0.0795\% \end{gathered}$$