2.0184 g of a compound of the general formula Cr x (NH 3 ) y Cl z was prepared u
ID: 912188 • Letter: 2
Question
2.0184 g of a compound of the general formula Crx(NH3)yClz was prepared using 3.5246 g CrCl3 * 6H2O and 10.00 mL of 15 M NH3. It was then analyzed as follows:
The concentration of Cr+3 was determined spectrophotometrically. The chromium (III) ion is purple, with a maximum light absorption at 570 nm. A stock solution (Solution 1) of 0.100 M Cr+3 was used to make the following dilutes:
1.2532 g of the unknown chromium compound was dissolved in enough water to make 100 mL of solution. Absorbance readings are given below:
Calculate the concentration of Cr+3 in solutions 2 and 3. Then make a graph of absorbance vs. concetration using the three known (standard) solutions. From the graph, find the concetration of the chromium compound. calculate the moles of Cr+3, grams of Cr+3, and % Cr+3 in the unkown compound. Attach the graph and show your work.
Needed:
_____moles/L Cr+3
_____moles Cr+3
_____g Cr+3
_____% Cr+3
Solution 2: 10.00 mL stock solution diluted to a total of 100 mL with water Solution 3: 20.00 mL stock solution diluted to a total of 100 mL with waterExplanation / Answer
Solution 2 is a 1:10 dilution of solution one. Its concentration is
0.1M /10 = 0.0100 M
Solution 3 is a 1:5 dilution of solution 2. Its concentration is
0.01M / 5 = 0.00200
Solution no.
Concentration, C
Absorbance,
A/C
1
0.100
1.850
18.50
2
0.0100
0.187
18.7
3
0.00200
0.0374*
18.7
Unknown sol.
1.122
18.6**
*correct the value. One 0 is missing
** average value
Draw a plot for A vs. C
Cunknown = 1.122 / 18.6 = 0.060 M
In 100 mL solution (and also in 1.2532 g of the unknown chromium compound) there are
0.060 mol/L x 0.100 L = 0.0060 mol Cr3+
0.0060 mol Cr3+ x 60.0 g Cr/mol = 0.360 g Cr3+
100 x 0.360 g/ 1.2532 g = 28.7 %
Solution no.
Concentration, C
Absorbance,
A/C
1
0.100
1.850
18.50
2
0.0100
0.187
18.7
3
0.00200
0.0374*
18.7
Unknown sol.
1.122
18.6**
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