(30 marks) A 0.40167g sample of crude oil containing an unknown concentration of
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Question
(30 marks) A 0.40167g sample of crude oil containing an unknown concentration of natural vanadium was mixed with 0.41946g of spike containing 2.2435umol V/g enriched with 5°V. After dissolution and equilibration of the oil and the spike, some of the vanadium was isolated by ion exchange resin. The measured isotope ratio in the extracted vanadium was 10.545. Find the concentration of vanadium (umol/g) in the crude oil. Natural abundances of vanadium: 51V = 0.9975 and 50V = 0.0025 Abundance in the vanadium spike: 51V-06391 and 50V = 0.3609 The ratio of isotopes in the mixture can be defined as R= A is 5IV, B is 50V, x is unknown, s is spike, m is mass in g and C is the concentration in umol/sg.Explanation / Answer
given :
R=10.545
Ax=51V in crude oil=0.9975
Cx=concentration of V in crude oil=unknown
mx=mass of V in crude oil=0.40167 g
As=51 V in spike=0.6391
Cs=2.2435 umol V/g
ms=0.41946 g
Bx=50V in crude oil=0.0025
Bs=50V in spike=0.3609
Putting these values in the equation for ratio of isotopes,R=(AxCxmx+AsCsms)/(BxCxmx+BsCsms)
10.545=[0.9975*Cx*(0.40167 g)+0.6391* (2.2435 umol V/g)*(0.41946 g)]/[0.0025*Cx*0.40167g +0.3609*(2.2435 umol V/g)*(0.41946 g)]
or, 10.545=[0.40066*Cx+0.60143]/[0.0010042*Cx+0.33963]
or, 0.39008*Cx=2.97997
Cx=2.97997/0.39008=7.63938 umol/g
Cx=concentration of vanadium in crude oil=7.63938 umol/g
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