Problems on spectrophotometry Lab 8 Week 9 Problems on spectrophotometry E=h E = h c/ wavenumber = 1/ , cm-1 . Frequency (, nu) is the number of waves in 1 second (it is expressed in Hz = 1 cycle/s) Frequency (, nu) = C / Wavenumber ( , nu par), gives number of waves per unite distance (it is expressed in cm-1). wavenumber = 1/ , cm-1 . E = Energy in joules (J) or ergs (erg) = Frequency (Hz) h = Plank’s constant (6.625 X 10-34 J.Sec. or 6.625 X 10-27 erg. s) = Wavelength (cm) = C/ c = Speed of light (3 x 1010 cm/s in vacuum = Different units of length are used to express wavelengths and their amplitudes 1 m = 102 cm = 103 mm = 106 = 109 nm = 1010 o. Beer’s - Lambert’s Law: A=abC Units of absorptivivity 1- Absorptivity (a): A =abc ( a is expressed in L.g-1cm-1) where concentration (c) is as gram / Liter. A = ε bc 2- Molar absorptivity (ε , Epsilon): (ε is expressed in L.mol-1cm-1) where conc. (c) expressed as molar 3- A1% 1cm A one percent one centimeter 1% A1cm = 1% A bc A= 1cm It is expressed in g% -1 cm -1 if c is expressed in g/100 mL 1% A1cm units = ε x 10 / Mol.Wt. 10 a Transformation between units • 1cm = 10 mm= 10000 m = 10000000 nm = 100000000 A • 1 cm = 10 3 m = 107 nm = 10 8 A To convert g/L → g/ ml : we have to multiply by 10 3 To convert g/ ml → g/L : : we have to multiply by 10 - 3 Transformation between units Different units of length are used to express wavelengths and their amplitudes 1 m = 102 cm = 103 mm = 106 = 109 nm = 1010 o. gm mg 𝜇g ng 1 103 106 109 m 1 cm 102 L 1 𝜇g 106 mm 103 ml 103 A∘ 1010 nm 109 𝜇l 106 Examples of problems with answers 1- Calculate frequency and wavenumber of light beams with a wavelength 200 nm? C = velocity of light C = 3 x 10 10 cm/sec. • = c/ = 3 x 10 10 / 2 x 10 -7 =1.5 x 10 17 cycle /sec.( or Hertz) • ύ = 1/ = 1/ 2 x 10 -7 = 50000 cm -1 • -------------------------------------------------------------------------------------------2- Calculate the wavelength and frequency of light beams with wavenumber of 5X 10 5 cm -1 = 1/ ύ = 1/ 5 x 10 5 = 2 x 10 -6 cm • ύ = 1/ • = c/ = 3 x 10 10 / 2 x 10 -6 = 1.5 x 10 16 Hz • 3 Express the wavelength 2500 A as micrometers and nm. 2500 A = 2500 / 10 = 250 nm = 2500 x 10 -4 = 0.25 m -------------------------------------------------------------------------------------------4 What is the energy of photons with wavelength equal to 100 nm. E = h = hC / = 6.625 x 10 -27 x 3 x 10 10 / 1 x 10 -5 = 1.987 x 10 -11 erg. 5- A molecule absorb energy of 3.3125 x 10 wavelength in nm of its spectrum ? -12 erg. what is the E = hC / 3.3125 x 10 -12 = 6.625 x 10 -27 x 3 x 10 10 / = 6 x 10 -5 cm = 6 x 10 -5 x 10 7 = 600 nm • • • (1cm = 10 mm= 10000 m = 10000000 nm) • 1 cm = 10 7 nm 6- What is the wavelength and wavenumber of photons having a frequency = 5 x 10 16 Hertz. • = c / = 3 x 10 6 / 5 x 10 16 = 6 x 10 -7 cm • ῡ= 1/ = 1 / 6 x 10 -7 = 1666666.667 cm -1 --------------------------------------------------------------------------------• 7- What is the wavelength and wavenumber of photons having an energy = 5 x 10 -15 erg. • E = hC / = hc / E • = 6.625 x 10 -27 x 10 10 / 5 x 10 -15 = 0.03975 cm • ύ = 1/ = 1 / 0.03975 = 25.157 cm -1 8-Convert the wavelength 4000 A , 250 nm, 35 m and 5.5 mm into frequencies and wavenumber. • 400 A = 4 x 10 -5 cm • ῡ = 1/ λ = 1/ 4 x 10 -5 = 25 000 cm -1 • γ = c/ λ = 3 x 10 10 / 4 x 10 -5 = 7.5 x 10 14 cycle/ sec. or Hz. 9-Find the absorbance of the following solution, in all cases 1 cm cells are used: • Solution A of = 1500 and C= 0.0001 M. • A = bc = 1500 x 1x 0.0001 = 0.15 • Solution B of A1% = 200 and C = 0.003 g% • A = A1% b c = 200 x 1 x 0.003 = 0.6 • Solution C of a = 30 and c = 0.015 g/ L • A = abc = 30 x 1 x 0.015 = 0.45 • Solution E of =5000 and c = 0. 001 g% (Mwt – 100) • A 1% = x 10 / Mwt = 5000 x 10 / 100 =500 • A = A 1% x b x c = 500 x 1 x 0.001 = 0.5 10- A drug of Mwt . 250, its absorbace is 0.8 in 1 cm cell at 550 nm and its concentration is 40 g/ml .Calculate its , A1%. And absorptivity. 40 g/ml → x 10 -3 = 0.04g/L A=abc 0.8 = a x 1 x 0.04 absorptivity or a = 0.8 / 0.04 = 20 L. g -1. cm -1 = a x Mwt = 20 x 250 = 5000 L. mol-1 cm -1 • A1% = 10 a = 10 x 20 = 200 • 11- Drug of molecular weight 350 has molar absorptivity =1400 at 26 0 nm. Calculate its absorptivity and A 1% 1 cm at that wavelength. If this drug shows an absorption = 0.7 in 1 cm cell, find its concentration in g/ml. • = a x Mwt • a = / Mwt = 1400 / 350 = 4 • A 1% 1 cm = 10 a = 10 x4 = 40 • A = abc • 0.7 = 4 x 1 x c • c = 0.7 / 4 = 0.175 g/L = 0.175 x 10 6 / 10 3 • = 175 g / ml • 12- A drug has A1% of 100 at 300 nm. Calculate its molar absorptivity at that wavelength if its Mwt is 280 . Then if an aquous solution of this drug exhibits an absorbance of 0.4 in 0.5 cm cell. Find its molar concentration. = a x Mwt = A1% x Mwt / 10 = 100 x 280 / 10 = 2800. A = bc 0.4 = 2800 x 0.5 x c c = 0.4 / 2800 x 0.5 = 0.0002857 M • 13- A compound of Mwt. 450, its absorbnce is 0.33 in 2 cm cell at 400 nm and its conc. is 20 g/ml. calculate its , A1%. • C = 20 g/ml = 20 x 10 -3 g/L •A = abc • 0.33 = a x 2 x 20 x 10 -3 • a = 0.33 / 2x 20 x 10 -3 = 8.25 L.g-1 cm -1 • = a x Mwt = 8.25 x 450 = 3712.5 L.mol -1 cm -1 • A1%.= 10 a = 10 x 8.25 = 82.5