A mixture of X and Y was analyzed using UV/vis absorption spectroscopy at four d

Question

A mixture of X and Y was analyzed using UV/vis absorption spectroscopy at four
different wavelengths to determine the concentrations of the two unknowns. The pathlength of
the cell was 1.00 cm. The following data were obtained:
Atot(327 nm) = 0.1304 eX(327 nm) = 517 M-1cm-1 eY(327 nm) = 279 M-1cm-1
Atot(403 nm) = 0.0695 eX(403 nm) = 85.0 M-1cm-1 eY(403 nm) = 738 M-1cm-1
Atot(478 nm) = 0.1386 eX(478 nm) = 246 M-1cm-1 eY(478 nm) = 1236 M-1cm-1
Atot(551 nm) = 0.2289 eX(551 nm) = 928 M-1cm-1 eY(551 nm) = 426 M-1cm-1
Determine the concentrations of [X] and [Y].

Explanation / Answer

Absorbance = e*C*l

where e = molar absorbtivity (M^(-1) cm^(-1))

C : concentration (M)

l: path length (1 cm)

A = e*C (as l = 1 cm)

C = A/e (M)

At 327 nm

[X] = 0.1304/517 = 2.52 * 10^(-4) M

[Y] = 0.1304/279 = 4.67*10^(-4) M

At 403 nm

[X] = 0.0695/85 = 8.17 * 10^(-4) M

[Y] = 0.0695/738 = 0.94*10^(-4) M

At 478 nm

[X] = 0.1386/246 = 5.63 * 10^(-4) M

[Y] = 0.1386/1236 = 1.21*10^(-4) M

At 551 nm

[X] = 0.2289/928 = 2.46 * 10^(-4) M

[Y] = 0.2289/426 = 5.37*10^(-4) M

Thus [X] = (2.52 + 8.17 + 5.63+2.46)/4 *10^(-4) M = 4.695 *10^(-4) M

[Y] = (4.67 + 0.94 + 1.21+5.37)/4 *10^(-4) M = 3.048*10^(-4) M

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