Question

A 3.03-g petroleum specimen was decomposed by wet ashing and subsequently diluted to 500 mL in a volumetric flask. Cobalt was determined by treating 25.00-mL aliquots of this diluted solution as follows:

Assume that the Co(II)-ligand chelate obeys Beer’s law, and calculate the percentage of cobalt in the original sample.

Short answer

The percent of Co present in the sample of the petroleum is 0.015 %.

Step-by-step solution

Given Information

The percentage of the amount of cobalt present in the petroleum solution is to be calculated.

Given data:

Explanation

The expression for the relation between molar absorptivity, volume of the solution and molar analytical concentration (c) of the solution is:

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

Where,

• A_x is the absorbance of the unknown solution.

• is the molar absorptivity of the solution.

• c_x is the molar analytical concentration of the unkonown solution.

• V_x is the volume of unknown solution.

• V_t is the total volume of the solution.

• b is the path length of the cell.

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

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

Where,

• A_x+s is the absorbance of the unknown solution added to standard solution.

• $\epsilon$is the molar absorptivity of the solution.

• c_x is the molar analytical concentration of the unkonown solution.

• V_x is the volume of unknown solution.

• c_s is the molar analytical concentration of standard solution.

• V_s is the volume of the standard solution.

• V_t is the total volume of the solution.

• b is the path length of the cell.

Divide the equation (I) by equation (II) and rearrange for c_x.

$c_x=\frac{A_x{c}_x{V}_s}{(A_{x+s}-A_x)V_x}$…… (3)

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

$\%\mathrm{Co}=\left(\frac{{\mathrm{V}}_{\mathrm{t}}.\mathrm{c}}{\mathrm{m}}\right)\times 100\%$ …… (4)

Where,

• V_t is the total volume of the solution.

• c is the concentration of the Co in the solution.

• m is the amount of Co present in the sample.

The value of A_x is 0.212.

The value of A_x+s is 0.399.

The value of c_s is 4.00ppm.

The value of V_x is 25.00mL.

The value of V_s is 5.00 mL.

Substitute the values in equation (3).

$$\begin{gathered} {\mathrm{c}}_{\mathrm{x}}=\frac{{\mathrm{A}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{x}}{\mathrm{V}}_{\mathrm{s}}}{({\mathrm{A}}_{\mathrm{x}+\mathrm{s}}-{\mathrm{A}}_{\mathrm{x}}){\mathrm{V}}_{\mathrm{x}}} \\ {\mathrm{c}}_{\mathrm{x}}=\frac{(0.212)(4.00\mathrm{ppm})(5.00\mathrm{ml})}{\left(\right(0.399)-(0.212\left)\right)(25.00\mathrm{ml})} \\ {\mathrm{c}}_{\mathrm{x}}=\frac{4.24}{4.675}\mathrm{ppm} \\ =0.907\mathrm{ppm} \end{gathered}$$

Since 1 ppm is equivalent to 1 mg/mL. Therefore, the concentration of the Co can be determined as follows.

$$\begin{gathered} \mathrm{c}=0.907\mathrm{ppm}\left(\frac{1\mathrm{mg}/\mathrm{L}}{1\mathrm{ppm}}\right)\left(\frac{1\mathrm{g}}{1000\mathrm{mg}}\right)\left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right) \\ =0.907\times {10}^{-6}\mathrm{g}/\mathrm{mL} \end{gathered}$$

The value of V_t is 500.0 mL.

The value of c is 0.907 * 10^-6g/mL.

The value of m is 30.3 g.

Substitute the values in equation (4).

localid="1649068605738" $$\begin{gathered} \%\mathrm{Co}=\left(\frac{{\mathrm{V}}_{\mathrm{t}}.\mathrm{c}}{\mathrm{m}}\right)\times 100\% \\ \%\mathrm{Co}=\left(\frac{(500.0\mathrm{mL})(0.907\times {10}^{-6}\mathrm{g}/\mathrm{mL})}{(3.03\mathrm{g})}\right)\times 100\% \\ =0.0149\% \\ \approx 0.015\% \end{gathered}$$

The percent of Co present in the sample of the petroleum is 0.015 %.