Chapter 17
Q. 17.3
Measuring Benzene in Hexane
(a) Pure hexane has negligible ultraviolet absorbance above a wavelength of 200 nm. A solution prepared by dissolving 25.8 mg of benzene (C_{6}H_{6}, FM 78.11) in hexane and diluting to 250.0 mL had an absorption peak at 256 nm and an absorbance of 0.266 in a 1.000-cm cell. Find the molar absorptivity of benzene at this wavelength.
Step-by-Step
Verified Solution
The concentration of benzene is
[C_{6}H_{6}] = \frac{(0.025 8 g)/(78.11 g/mol)}{0.250 0 L} = 1.32_{1} × 10^{−3} M
We find the molar absorptivity from Beer’s law:
Molar absorptivity = ε = \frac{A}{bc} = \frac{(0.266)}{(1.00 cm)(1.32_{1} × 10^{−3} M)} = 201._{3} M^{−1} cm^{−1}
(b) A sample of hexane contaminated with benzene had an absorbance of 0.070 at 256 nm in a cuvet with a 5.000-cm pathlength. Find the concentration of benzene in mg/L.
Using Beer’s law with the molar absorptivity from part (a), we find
[C_{6}H_{6}] = \frac{A}{εb} = \frac{0.070}{201._{3} M^{−1} cm^{−1} (5.00 cm)} = 6.9_{5} × 10^{−5} M
[C_{6}H_{6}] = (6.9_{5} × 10^{−5} \frac{mol}{L})(78.11 × 10^{3} \frac{mg}{mol}) = 5.4 \frac{mg}{L}
Test Yourself 0.10 mM KMnO_{4} has an absorbance maximum of 0.26 near 525 nm in a 1.000-cm cell. Find the molar absorptivity and the concentration of a solution whose absorbance is 0.52 at 525 nm in the same cell. (Answer: 2 600 M^{−1} cm^{−1} , 0.20 mM)
This example illustrates the measurement of molar absorptivity from a single solution. It is better to measure several concentrations to obtain a more reliable absorptivity and to demonstrate that Beer’s law is obeyed.