1. [spectroscopy and chemical
1. [spectroscopy and chemical dyes] A compound with molecular mass292.16 g/mol was dissolvedin a 5 mL volumetric flask. A 1.00 mLaliquot was withdrawn, placed in a 150.0 mL volumetricflask, anddiluted to the mark. The absorbance at 340 nm was 0.427 in a 1.00cm cuvette. The molar absorptivity at 340 nm is equal to 6130 M-1cm-1.
a. Calculate the concentration of the compound in thecuvette.
b. What was the concentration of the compound in the 5 mLflask?
c. How many milligrams of compound were used to make the 5 mLsolution?
Answer:
a) We can calculate the concentration of the compound in thecuvette by using the Beer-Lambert Law
A=ebc
Where A is absorbance = 0.427
e is the molar absorptivity = 6130 M-1 cm-1.
b is the path length of the sample =1.00 cm
c is the concentration of the compound in solution
A=ebc
c=A/eb
c = 0.427/ (6130 M-1 cm-1. x 1.00cm)
c = 6.97e-5M
The concentration of the compound in the cuvette =6.97e-5M
b. The concentration of the compound in the 5 mL flask-
Only 1.00 ml of the stock solution was diluted up to 10 ml
C1 V1=C2V2
C1 (1.00mL)=(6.966e-5)(10.00ml)
C1= 6.97e-4 M
The concentration of the compound in the 5 mL flask is6.97e-4 M
c.
Calculate number of moles- n = cv
C= 6.966e-4M
V = 5 ml = 0.005 litres
We could calculate the moles
0.005 litres x 6.966e-4 mol/litre = 3.483e-6 moles
Now use the molar mass to calculate the grams used
Molar mass = 292.16 g/mol
3.483e-6 moles x 292.16 g/mol = 1.0176e-3 grams
= 0.00102 grams
= 1.02 milligrams
1.02 milligrams of compound were used to make the 5 mlsolution
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