Quantitative analysis by isotope dilution. In isotope dilution, a known amount of an

Chapter 21, Problem 21-32

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Quantitative analysis by isotope dilution. In isotope dilution, a known amount of an unusual isotope (called the spike) is added to an unknown as an internal standard for quantitative analysis. After the mixture has been homogenized, some of the element of interest must be isolated. The ratio of the isotopes is then measured. From this ratio, the quantity of the element in the original unknown can be calculated. Natural vanadium has atom fractions 51V 0.997 5 and 50V 0.002 5. The atom fraction is defined as A spike enriched in 50V has atom fractions 51V 0.639 1 and 50V 0.360 9. (a) Let isotope A be 51V and isotope B be 50V. Let Ax be the atom fraction of isotope A ( atoms of A/[atoms of A atoms of B]) in an unknown. Let Bx be the atom fraction of B in an unknown. Let As and Bs be the corresponding atom fractions in a spike. Let Cx be the total concentration of all isotopes of vanadium (mol/g) in the unknown and let Cs be the concentration in the spike. Let mx be the mass of unknown and ms be the mass of spike. After mx grams of unknown are mixed with ms grams of spike, the ratio of isotopes in the mixture is found to be R. Show that (A) (b) Solve Equation A for Cx to show that (B) (c) A 0.401 67-g sample of crude oil containing an unknown concentration of natural vanadium was mixed with 0.419 46 g of spike containing 2.243 5 mol V/g enriched with 50V (atom fractions: 51V 0.639 1, 50V 0.360 9).48 After dissolution and equilibration of the oil and the spike, some of the vanadium was isolated by ion-exchange chromatography. The measured isotope ratio in the isolated vanadium was R 51V/50V 10.545. Find the concentration of vanadium (mol/g) in the crude oil. (d) Examine the calculation in part (c) and express the answer with the correct number of significant figures. 21-