Ch 18 Sapling

sample cuvette

holds the sample and has a defined pathlength

monochromator

it is typically a grating, prism, or filter that selects a narrow wavelength band of radiation and passes it to the sample

light source

provides electromagnetic radiation in the UV, visible, or IR region of the spectrum

see number 3 for further review

see number 3 for further review

detector

typically a photomultiplier tube that generates an electric signal when struck by photons

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standard Fe3 solution with a concentration of 7.21 × 10-4 M. You treat 14.0 mL of this reference with HNO3 and excess KSCN to form a red complex, and dilute the reference to 55.0 mL. The diluted reference is placed in a cell with a 1.00-cm light path. You then take 10.0 mL of the well water, treat with HNO3 and excess KSCN, and dilute to 100.0 mL. This diluted sample is placed in a variable pathlength cell. The absorbance of the reference and the sample solutions match when the pathlength is 4.45 cm. What is the concentration of iron in the well water? For each solution, the zero is set with a blank.

1. calculate the concentration of the dilute reference solution: Cref = 7.21x10^-4 M x 14.0mL/55.0mL =.000184M 2. Concentration sample = cref x bref/bsample .000184x1cm/4.45cm =.0000412 3. Consider dilution factor C₁V₁ = C₂V₂ (.0000412)(100mL) = C₂(10) =.000412M

Wavelength λ₀ corresponds to the transition from the ground vibrational level of S₀ to the lowest vibrational level in S₁ in absorption and vice versa in emission. If λ₀ represents the same transition, why then does λ₀ for the absorption and emission spectra not overlap in the spectrum above? a. The emission transition is observed at a slightly lower wavelength because additional energy is emitted as a molecule relaxes from the excited state to the ground state. The geometry and solvation of the molecule remains the same during the excitation and emission process. b.After excitation the molecule adopts a different geometry and solvation than the ground state molecule. This change in geometry lowers the energy of the transition between the two states, so when the molecule emits radiation, this transition will be observed at a slightly higher wavelength. c. The absorption transition is observed at a slightly lower wavelength because additional energy is required to promote a molecule from its ground state to the excited state. No additional energy is required during the emission process. The geometry and solvation of the molecule remains the same during the excitation and emission process. d.After excitation the molecule adopts a different geometry and solvation than the ground state molecule. This change in geometry increases the energy of the transition between the two states, so when the molecule emits radiation, this transition will be observed at a slightly lower wavelength.

B

A solution of a specific vitamin has λmax = 247 nm and a concentration of 5.44 × 10-7 M. The absorbance of the solution at 247 nm is A = 0.140. What is the molar absorptivity of the vitamin at 247 nm? (sample pathlength = 1.00 cm)

Beer's Law: A = εlC absorbance = molar absorptivity(length)(concentration) 0.140 = ε(1 cm)(5.44x10⁻⁷) ε = 257352 L/mol cm

For 598-nm visible light, calculate its frequency (ν, Hz), wavenumber (ν, cm-1), and photon energy (J). v = 1/v = E = If a laser were produced at this frequency what color of light would you observe?

C = λ x v 3x10⁸ = 598x10⁻⁹ v v = 5.01x10¹⁴ Hz E = hv E = 6.626x10⁻³⁴ (5.01x10¹⁴) E = 3.32x10⁻¹⁹ J 1/v = 1/598nm 1nm = 1x10⁻⁷ orange light

A certain shade of blue has a frequency of 7.07 × 10^14 Hz. What is the energy of exactly one photon of this light? E = ?

E = hv energy = planck's constant (frequency) energy = 6.626x10^-34(7.07x10^14) E = 4.68x10^-19 J

Absorbance measurements in the range of A = 0.3-2 are considered the most accurate. Why would absorbance measurements of 0.05 and 2.5 be considered inaccurate? Select all that apply. a. The difference in the absorbance of the sample and reference is too small at A = 0.05. b. At A = 2.5, the detector is saturated because too much light is reaching the detector. c. At A = 2.5, a sufficient amount of light does not reach the detector for an accurate measurement. d. If the absorbance is too high (A = 2.5), the signal difference between the sample and the reference is too small for an accurate measurement. e. At A = 2.5, the transmission is too high for an accurate measurement.

a, c

a. What absorbance value corresponds to 40.4% T? b. If a 0.0104 M solution of a certain analyte exhibits 40.4% T at a wavelength of 280 nm, what will the percent transmittance be for a 0.0266 M solution of the same analyte? Assume the pathlengths for each measurement are equal.

a. A = log(P₀/P) = -log(T) A = -log(.404) = .394 b. A₁/C₁ = A₂/C₂ A₁ = .394M C₁ = .0104 A₂ = ? C₂ = .0266 A₂ = 1.0077 A = -logT 1.0077 = -logT 10⁻¹ = T .098 = T 9.8% transmittance

Use the correct term to complete the sentence: halve, double A) If you ______ the wavelength, the electromagnetic radiation energy will double. B)The energy of the electromagnetic radiation will _____ if you halve the wave number. C) When the frequency of the light is doubled, its energy will _______.

a. halve b. halve c. double

A beam of light with an irradiance of P0 passes through a solution of a substance that can absorb some of the light. The pathlength of the solution is b cm, and the molar concentration of the substance is c M. The irradiance of the emerging light is P. Absorbance (A), transmittance (T) and molar absorptivity (ε) are quantities calculated from these values. Identify the expression used for calculating each of these three quantities by dragging the quantity into the box next to the expression. Transmittance = Absorbance = Molar absorptivity = Which of these three quantities is proportional to the concentration?

absorbance is proportional to concentration

The wavelength of some blue light is 470.0 nm. What is the frequency of this blue light?

c = lambda x v speed of light = wavelength x frequency 3x10^8 = 470x10^-9(v) v = 6.38x10^14 Hz

What material is the sample cell commonly made of in the following types of spectroscopy? UV spec a. glass b. NaCl c. quartz d. KBr Visibile spec a. glass b. quartx c. NaCl d. AgCl IR spec a. Quarz b. fused silica c. glass d. KBr

c, a, d