To analyze the chromium in an unknown sample, 1.5623 g of the sample was pretrea

Question

To analyze the chromium in an unknown sample, 1.5623 g of the sample was pretreated and diluted to the exact 100 mL mark in a volumetric flask. Two 20.0 mL samples of this solution were taken. In the first sample, a Cr-EDTA complex is developed and sufficient water is added to give a total volume of 100.0 mL. In the second sample, 10.0 mL of Cr(III) standard solution at a concentration of 12 mM was added before developing the Cr-EDTA complex. Sufficient water is also added to give a total volume of 100.0 mL. The absorbance of each is measured in a 1.0 cm cuvette at 450 nm and is found to be 0.11 for the first solution and 0.22 for the second solution. Calculate the % chromium (atomic wt 52.0 g/mol) in the original sample.

Explanation / Answer

1. Let's gather the information we have:

mass of sample = 1.5623g

initial volume = 100mL (V1)

1.a Now, we took 20mL from V1 and we prepared the Cr-EDTA complex which was completed to another 100mL (V2) and the absorbance for this is A = 0.11

Let's remember the A formula:

A = bEC

In this case, C (concentration) is exclusively related to the concentration of Cr in the sample, remember concentration is given in mol/L units, let's re-write this equation using the value of absorbance (A = 0.11) and the volume (V2 = 0.1L), b = 1.0cm and as well as the moles of Cr (nCr) from the sample:

0.11 = E(nCr/0.1L)

1.b In this case, we did the same as in 1.a but we added 10mL (0.01L) of a standard of concentration 12mM, we need to calculate how many extra moles of Cr are added to the sample:

M = n/L

mM = mmol/L

12mM = mmol/0.01L

Solve for mmol

mmol = 0.12 mmol = 0.00012 mol (1 mol = 1000 mmol)

Now let's write the A = bEC equation using the absorbance for this solution (A = 0.22) and let's express the concentration as the sum of moles of Cr from the sample (nCr) plus the moles of Cr from the standard (nE = 0.00012 mol).

0.22 = E(nCr+0.00012/0.1L)

2. Let's gather the 2 equations we have:

0.11 = E(nCr/0.1L)

0.22 = E(nCr+0.00012/0.1L)

3. We have 2 equations and 2 variables (nCr, E), let's solve the system of equations then:

0.011 = nCr*E

0.022 = nCr*E + 0.00012E

Use the first equation and insert it into the second one:

0.022 = 0.011 + 0.00012E

Solve for E:

0.011 = 0.00012E

E = 91.67

Now that we have E, we can insert it into the first equation and solve for nCr:

0.011 = 91.67nCr

nCr = 0.00012 mol

So, the concentration of Cr in the diluted sample is:

C = 0.00012mol/0.1L

C = 0.0012M

4. Using C1V1 = C2V2 we can calculate the concentration in the first solution:

C1*(20mL) = 0.0012M*100mL

C1 = 0.006M

5. That is the concentration of Cr in mol/L in the very first solution, which means we have 0.006moles of Cr per liter of solution, but since we have 100mL of that solution, we only have 0.0006moles of Cr.

6. We need to convert those moles to grams using the atomic weight = 52g/mol

m = 0.0006mol*(52g/mol)

m= 0.0312g

7. To calculate %Cr we just divide the mass of Cr we got in step 6 by the total mass of sample (1.5623g) and multiply times 100.

%Cr = 0.0312g/1.5623g x 100

%Cr = 2.0%

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