Question

Use the data below to generate a calibration curve for an unknown zinc compound. In order to determine the % zinc in the compound, a solution was made with 1.75 g of the compound with a total volume of 50.0 mL. The absorbance of this solution was measured at 1.16. Assuming zinc is the only chemical species contributing to the absorbance, calculate the % zinc in the unknown compound.

Concentration (M)	Absorbance
0.025	0.13
0.070	0.41
0.13	0.87
0.16	1.12
0.20	1.53

Answer 1

Ans. Step1 : Calculation of unknown [Zn²⁺] in solution using standard graph.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 7.928x - 0.1156 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 7.928 units on X-axis (concentration) minus 0.1156

We have, absorbance of unknown, y = 1.16

Putting y = 1.16 in trendline equation-

            1.16 = 7.928x – 0.1156

            Or, 7.928x = 1.16 + 0.1156 = 1.2756

            Or, x = 1.2756 / 7.928

            Hence, x = 0.1609

Hence, [Zn²⁺] in 50.0 mL of the solution of the unknown = 0.1609 M

# Step 2: Calculate Mass of Zn in SOlution

Moles of Zn in 50.0 mL solution = Molarity x Volume of solution in liters

                                                = 0.1609 M x 0.050 L

                                                = 0.008045 mol

Mass of Zn in 50.0 mL solution = Moles of Zn x Molar mass

                                                = 0.008045 mol x (65.39 g/ mol)

                                                = 0.5261 g

Therefore, mass of Zn in 50.0 mL solution = 0.5261 g

Step 3: Calculate % Zn in sample:

Since the 50.0 mL solution is prepared from 1.75 g sample, the total mass of Zn in the sample must be equal to the total mass of Zn in 50.0 mL solution.

So,

            Mass of Zn in sample = 0.5261 g

Now,

            % Zn in sample = (Mass of Zn / Mass of sample) x 100

                                                = (0.5261 g/ 1.75 g) x 100

                                                = 30.02 %