Question

Given the following data, calculate the percent by mass Fe²⁺ in the unknown sample. Assume the experimental data provided is obtained using the same procedure you will be using in the lab.

Unknown sample mass = 0.1700

Absorbance (average) of unknown sample = 0.343

Beer’s Law Data:        Slope = 0.2143 L/mg Fe²⁺

                                    y-intercept = -0.002

Additional Information (Lab procedure):

1. Obtain and clean three 100-ml volumetric flasks

2. Weigh 0.15-0.20g of the unknown accurately into a clean 100ml beaker

3. Add about 30ml distilled water and 1ml concentrated sufuric acid. Swirl to dissolve sample and quantatively transfer into a 100ml volumetric flask. Make up the flask to the mark with distilled water and mix thoroughly

4. Pipet 10.00ml of this solution into the second volumetric flask and make up to the mark

5. Pipet 10.00ml of this diluted solution ( from step 4) into the third volumetric flask. Add 10ml saturated sodium acetate and 10ml 10% hydroxylamine hydrochloride, and wait five minutes

6. Add 10ml 0.1% 1,10 phenanthroline to the flask. Wait 10 minutes, then make up the flask to the mark with distilled water

Answer 1

Ans. The trendline equation for the graph “y = 0.2143x - 0.002” is in form of y = mx + c. where, c = y-intercept     , m = slope

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 0.2143x - 0.002 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 0.2143 units on X-axis (concentration) minus 0.002.

Given, absorbance of unknown, y = 0.343

Putting y = 0.343 in trendline equation-

            0.343 = 0.2143x - 0.002

            Or, 0.343 + 0.002 = 0.2143x

            Or, x = 0.345 / 0.2143 = 1.6099

Hence, [Fe²⁺] in final (3^rd flask) = 1.6099 mg/ L

# From flask 3 to flask 2:

10.0 mL (V1) of flask 2 solution is diluted to 100.0 mL (V2) to give a solution of [Fe²⁺] = 1.6099 mg/ L (C2, calculated using spectrophotometer) in flask 3.

Now, using C1V1 (flask 2) = C2V2 (flask 3)

            Or, C1 x 10.0 mL = (1.6099 mg/ L) x 100.0 mL

            Or, C1 = (1.6099 mg/ L) x 100.0 mL / 10.0 mL = 16.099 mg/ L

Hence, [Fe²⁺] in 2^nd flask = 16.099 mg/ L

# From flask 2 to flask 1:

10.0 mL (V1) of flask 1 solution is diluted to 100.0 mL (V2) to give a solution of [Fe²⁺] = 16.099 mg/ L (C2) in flask 2

Again using C1V1 (flask 1) = C2V2 (flask 2)

            C1 = (16.099 mg/ L) x 100.0 mL / 10.0 mL = 160.99 mg/ L

Hence, [Fe²⁺] in 1^st flask = 160.99 mg/ L

# Total Fe²⁺ content of flask 1:

Total Fe²⁺ in flask 1 = [Fe²⁺] in flask 1 x Volume of flask 1 in liters

                                    = (160.99 mg/ L) x 0.100 L                                      ; [100 mL = 0.100 L]

                                    = 16.099 mg

Therefore, total Fe²⁺ content in flask 1 = 16.099 mg

# Since flask 1 solution is prepared from 0.1700 g of sample, the total Fe²⁺ content of the flask is equal to the total Fe²⁺content in the sample.

Therefore, total Fe²⁺ content in sample = 16.099 mg = 0.016099 g

Now,

            % Fe in sample = (Mass of Fe²⁺ in sample / Mass of sample) x 100

                                                = (0.016099 g / 0.1700 g) x 100

                                                = 9.47 %