CHEMISTRY 31

Fall, 2017 - Dixon

Homework Set 2.2 Solutions

Ch. 7: 18, 26 Now not covering this material

Ch. 18: 1, 3, 4, 11, 18

1. Fill in the blanks.

a) If you double the frequency of electromagnetic radiation, you double the energy.

b) If you double the wavelength, you halve the energy.

c) If you double the wavenumber, you double the energy.

3. Calculate the frequency (Hz), wavenumber (cm^-1), and energy (J/photon and J/[mole of photons]) of visible light with a wavelength of 562 nm.

frequency: ν = c/λ = 3.00 x 10⁸ m s^-1/(562 nm*1x10^-9 m/nm) = 5.33 x 10¹⁴ Hz.

wavenumber: = 1/λ = 1/(5.62 x 10^-5 cm) = 1.78 x 10⁴ cm^-1

Energy: E = hc/λ = 6.63 x 10^-34J*s*3.00 x 10⁸ m s^-1/(562 nm*1x10^-9 m/nm) = 3.54 x 10^-19 J

E = N*E_photon = 6.02 x 10²³*3.54 x 10^-19 J = 213 kJ/mole

4. Which molecular processes correspond to the energies of microwave, infrared, visible, and ultraviolet photons?

microwave: molecular rotation; infrared: molecular vibration; visible and ultraviolet: transitions in electronic state

11. Why is it most accurate to measure absorbances in the range of A = 0.3 to 2?

At low A values, there is little difference between I and I_o so small errors affecting either value become important. At very high values of A, I is small and subject to larger percent errors.

18. A compound with molecular mass 292.16 g/mol was dissolved in a 5 mL volumetric flask. A 1.00 mL aliquot was withdrawn, placed in a 10 mL volumetric flask, and diluted to the mark. The absorbance at 340 nm was 0.427 in a 1.000 cm cuvet. The molar absorptivity at 340 nm is e₃₄₀ = 6130 M^-1 cm^-1.

a) Calculate the concentration of compound in the cuvet.

A = ebC or C = A/eb = 0.427/[(6130 M^-1 cm^-1)( 1.000 cm)] = 6.97 x 10^-5 M

b) What was the concentration of compound in the 5 mL flask?

C_cuvetteV_cuvette = C_5mLflaskV_5mLflask or C_5mLflask = (6.97 x 10^-5 M)(10 mL)/(1 mL) = 6.97 x 10^-4 M

c) How many milligrams of compound were used to make the 5 mL solution?

Mass = (5 mL)(6.97 x 10^-4 mmol/L)(292.16 mg/mmol) = 1.02 mg

Ch. 23 3, 8a, 18, 20, 23

3. If you wish to extract aqueous acetic acid into hexane, is it more effective to adjust the aqueous phase to pH 3 or pH 8?

pH 3. Under acidic conditions, acetic acid will be in the HA form that can partition between both phases while the A^- form will only exist in the aqueous phase.

8. Solute S has a partition coefficient of 4.0 between water (phase 1) and chloroform (phase 2) in Equation 23-1.

a) Calculate the concentration of S in chloroform if [S(aq)] is 0.020 M.

K = 4.0 = [S(CHCl₃)]/[S(aq)]

[S(CHCl₃)] = K[S(aq)] = 4.0(0.020 M) = 0.080 M

18. The partition coefficient for a solute in chromatography is K = C_s/C_m, where C_s is the concentration in the stationary phase and C_m is the concentration in the mobile phase. The larger the partition coefficient, the longer it takes a solute to be eluted. Explain why.

A large partition coefficient means that the analyte is attracted to the stationary phase relative to the mobile phase. Thus, it will spend a greater fraction of its time in the stationary phase. Since the analyte only moves along the column when it is in the mobile phase, analytes with larger partition coefficients will take longer to exit the column.

20. (a) A chromatography column with a length of 10.3 cm and inner diameter of 4.61 mm is packed with a stationary phase that occupies 61.0% of the volume. If the volume flow rate is 1.13 mL/min, find the linear flow rate in cm/min.

V_m = (1 – 0.610)∙(Volume of cylinder) (assuming everything not stationary phase is mobile phase)

V(Cylinder) = pd²L/4 = p(0.461 cm)²(10.3 cm)/4 = 1.72 cm³

V_m = 0.390∙1.72 mL = 0.670 mL

Time to pass through column = 0.670 mL/1.13 mL/min = 0.593 min

u_x = 10.3 cm/0.593 min = 17.4 cm/min

b) How long does it take for solvent (which is the same as unretained solute) to pass through the column?

0.593 min (see above calculation)

c) Find the retention time for a solute with a retention factor of 10.0.

k = 10.0 = (t_r – t_m)/t_m 10.0∙t_m = t_r – t_m

t_r = 10.0∙t_m + t_m = 11(0.593 min) = 6.52 min.

23. Solvent passes through a column in 3.0 min but solute requires 9.0 min.

a) Calculate the retention factor, k.

k = (9.0 min – 3.0 min)/3.0 min = 2.0

b) What fraction of time is the solute in the mobile phase in the column?

In mobile phase 3 min (time to pass through) vs. 9.0 total time required

So 3.0 min/9.0 min = 1/3 (or 0.33)

c) The volume of stationary phase is 1/10 of the volume of the mobile phase in the column (V_s = 0.10 V_m). Find the partition coefficient, K, for the system.

K = k(V_m/V_s) = (2.0)(10) = 20