Chapter 22, Problem 22.43P

(a)

Interpretation Introduction

Interpretation:

Form the given information, the ratio of isotopes in the mixture is R has to found.

Explanation of Solution

Let’s take the term $A_x{C}_x{m}_x,$ which applies to the unknown.

$\begin{array}{l}{A}_x{C}_x{m}_x\\ \text{=}\left(\frac{\text{μmol}\text{isotope}\text{A}}{\text{μmol}\text{isotope}\text{A + μmol}\text{isotope}\text{B}}\right)\left(\frac{\text{μmol}}{\text{g}\text{unknown}}\right)\left(\text{g}\text{unknown}\right)\\ =\left(\frac{\text{μmol}\text{isotope}\text{A}}{\text{μmol}\text{isotope}\text{A + μmol}\text{isotope}\text{B}}\right)\left[\text{μmol}\text{V}\right]\\ =\left(\frac{\text{μmol}\text{isotope}\text{A}}{\text{μmol}\text{isotope}\text{A + μmol}\text{isotope}\text{B}}\right)\left(\text{μmol}\text{isotope}\text{A+μmol}\text{isotope}\text{B}\right)\\ \\ =\text{μmol}\text{isotope}\text{A}\text{in}\text{the}\text{unknown}\text{.}\end{array}$

In the same way, $B_x{C}_x{m}_x=\text{μmol}\text{isotope}\text{B in the unknown,}$

$A_s{C}_{s}m{s}_x=\text{μmol}\text{isotope A in the spike, }$ and $B_s{C}_s{m}_s=\text{μmol}\text{isotope B in the unknown}\text{.}$

Now, mix the unknown and spike, then we will get the isotope ratio as follows.

$\begin{array}{l}R=\frac{\text{μmol A}}{\text{μmol B}}=\frac{\text{μmol A in unknown + μmol A in spike}}{\text{μmol B in unknown + μmol B in spike}}\\ \\ R=\frac{A_x{C}_x{m}_x+A_s{C}_s{m}_s}{B_x{C}_x{m}_x+B_s{C}_s{m}_s}\end{array}$

Therefore, the ratio of isotopes in the mixture is found to be as $R=\frac{A_x{C}_x{m}_x+A_s{C}_s{m}_s}{B_x{C}_x{m}_x+B_s{C}_s{m}_s}.$

(b)

Interpretation Introduction

Interpretation:

By using the given equation A is to show that $C_x$ as given has to be solved.

Explanation of Solution

The given equation A is $R=\frac{A_x{C}_x{m}_x+A_s{C}_s{m}_s}{B_x{C}_x{m}_x+B_s{C}_s{m}_s}.$

In order to obtain the value or equation for $C_x$, we have to cross-multiply the equation A and doing some simple mathematic calculation we can get the final answer.

$\begin{array}{l}R\left(B_x{C}_x{m}_x+B_s{C}_s{m}_s\right)=A_x{C}_x{m}_x+A_s{C}_s{m}_s\\ R{B}_x{C}_x{m}_x+R{B}_s{C}_s{m}_s=A_x{C}_x{m}_x+A_s{C}_s{m}_s\\ R{B}_x{C}_x{m}_x-A_x{C}_x{m}_x=A_s{C}_s{m}_s-R{B}_s{C}_s{m}_s\\ C_x=\frac{A_s{C}_s{m}_s-R{B}_s{C}_s{m}_s}{R{B}_x{m}_x-A_x{m}_x}\\ C_x=\left(\frac{C_s{m}_s}{m_x}\right)\left(\frac{A_s-R{B}_s}{R{B}_x-A_x}\right)\end{array}$

Hence, we derived and proved as $C_x=\left(\frac{C_s{m}_s}{m_x}\right)\left(\frac{A_s-R{B}_s}{R{B}_x-A_x}\right).$

(c)

Interpretation Introduction

Interpretation:

The concentration of vanadium in the crude oil has to be found.

Answer to Problem 22.43P

The concentration of vanadium in the crude oil is $\text{7}\text{.6394 μmol}\text{V/g}\text{.}$

Explanation of Solution

The given information is ${\text{A = }}^{\text{51}}\text{V}\text{and}{\text{B = }}^{\text{50}}\text{V}$

Atom fractions in unknown is: $A_x=0.9975and{B}_x=0.0025$

Atom fractions in spike is: $A_s=0.6391and{B}_x=0.3609$

Now, let’s substitute these values in the above obtained formula of $C_x$ and the calculation can be done as follows.

$C_x=\left(\frac{C_s{m}_s}{m_x}\right)\left(\frac{A_s-R{B}_s}{R{B}_x-A_x}\right)$

$\begin{array}{l}{C}_x=\left(\frac{\left(\text{2}\text{.243μmolV/g}\right)\left(0.41946g\right)}{0.40167g}\right)\left(\frac{06391-\left(10.545\right)\left(0.3609\right)}{\left(10.545\right)\left(0.0025\right)-0.9975}\right)\\ \\ C_x\text{=}\text{7}\text{.6394}\text{μmol}\text{V/g}\end{array}$

Therefore, the concentration of vanadium in the crude oil is $C_x\text{=}\text{7}\text{.6394}\text{μmol}\text{V/g}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The calculation in the part (c) and the answer with the correct number of significant figures has to be examined and expressed.

Explanation of Solution

The part (c) calculation is

$\begin{array}{l}{C}_x=\left(\frac{\left(\text{2}\text{.243μmolV/g}\right)\left(0.41946g\right)}{0.40167g}\right)\left(\frac{06391-\left(10.545\right)\left(0.3609\right)}{\left(10.545\right)\left(0.0025\right)-0.9975}\right)\\ \end{array}$

To examine this we can do some simple mathematical calculation as follows to get the final answer.

$\begin{array}{l}{C}_x=\left(\frac{\left(\text{2}\text{.243μmolV/g}\right)\left(0.41946g\right)}{0.40167g}\right)\left(\frac{06391-3.8057}{0.02636-0.9975}\right)\\ \end{array}$

$\begin{array}{l}{C}_x=\left(2.3429\right)\left(\frac{-3.166}{-0.9711}\right)\\ \\ C_x=\text{7}\text{.639 μmol}\text{V/g}\\ \end{array}$

In this way, we have to examine and express with the correct number of significant figures for the part (c) answer.