Chapter 16, Problem 16.37P

(a)

Interpretation Introduction

Interpretation:

Average oxidation numbers of Bismuth and Copper in the superconductor and the oxygen stoichiometry coefficient has to be found.

Concept introduction:

Oxidation number: Oxidation number is defined as degree of oxidation (ie. loss of an electrons) of an atom within the chemical compound.

There is slightly difference between oxidation state and oxidation number.  In oxidation state, there is electronegativity of an atom in a bond should be considered.  However, when describing oxidation number, electronegativity does not considered.

Answer to Problem 16.37P

Average oxidation number of Bi and copper are found to be ${3.2000}^{+}$ and ${2.2001}^{+}$ respectively.

Explanation of Solution

To calculate: Average oxidation number of Bi and copper

Experiment A:

Amount of initial copper ion is $\text{0}\text{.2000}\text{mmol}$ and final amount is $\text{0}\text{.1085}\text{mmol}$.  So, consumption of $102.3\text{mg}$ of superconductor is $\text{0}\text{.0915}\text{mmol}{\text{Cu}}^{\text{+}}$.

${\text{2×mmol Bi}}^{\text{5+}}{\text{+ mmolCu}}^{\text{3+}}\text{in 102}\text{.3mg of super conductor = 0}\text{.0915}{\text{mmol Cu}}^{\text{+}}$

Experiment B:

Amount of initial ${\text{Fe}}^{\text{2+}}$ ion is $0.1000\text{mmol}$ and final amount is $\text{0}\text{.0577 mmol}$.  So, consumption of $94.6\text{mg}$ of superconductor is $\text{0}\text{.0423}\text{mmol}{\text{Fe}}^{\text{2}}{}^{\text{+}}$.

${\text{2×mmol Bi}}^{\text{5+}}{\text{+ mmolCu}}^{\text{3+}}\text{in 94}\text{.6 mg of super conductor = 0}\text{.0423}{\text{mmol Fe}}^2{}^{\text{+}}$

Normalizing to one gram of super conductor = $0.0423$

Experiment A: $\text{2}\left({\text{mmolBi}}^{\text{5+}}\right){\text{+ mmolCu}}^{\text{3+}}\text{in1g of superconductor = 0}\text{.89443}$

Experiment B: $\text{2}\left({\text{mmolBi}}^{\text{5+}}\right)\text{in1g of superconductor = 0}\text{.44715}$

It is easier not to get missing the arithmetic if that oxidized Bismuth is ${\text{Bi}}^{\text{4+}}$ and equate 1 mol of ${\text{Bi}}^{\text{5+}}$ to two moles of Bismuth.

So, rewrite these previous equation as

$\begin{array}{l}{\text{mmol Bi}}^{\text{4+}}{\text{+ mmolCu}}^{\text{3+}}\text{in 1}\text{.0 g of super conductor = 0}\text{.89443 }\mathrm{......}\left(1\right)\\ \\ {\text{mmol Bi}}^{\text{4+}}\text{in 1}\text{.0 g of super conductor = 0}\text{.44715}\mathrm{......}\left(2\right)\end{array}$

Equation (2) has to be subtracted from equation (1), we get

${\text{mmol Cu}}^{\text{3+}}\text{in 1}\text{.0 g of super conductor = 0}\text{.44728}\mathrm{......}\left(3\right)$

Equation (2) and (3) shows that the relationship in the formula of the superconductor is $\text{b/c = 0}\text{.44715/0}\text{.44728 = 0}\text{.9997}$.

One gram of superconductor contains $\text{0}\text{.44728}\text{mmol}$ ${\text{Cu}}^{\text{3+}}$, we say that

  $\begin{array}{l}\frac{\text{mol}{\text{Cu}}^{\text{3+}}}{\text{mol}\text{solid}}\text{= 2c}\\ \\ \frac{\text{mol}{\text{cu}}^{\text{3+}}\text{/mol solid}}{\text{gram solid/mol solid}}\text{ =}\frac{\text{2c}}{\text{760}\text{.37+15}\text{.9994}\left(\text{8+b+c}\right)}\\ \\ \frac{\text{mol}{\text{cu}}^{\text{3+}}}{\text{gram solid}}\text{=}\frac{\text{2c}}{\text{760}\text{.37+15}\text{.9994}\left(\text{8+b+c}\right)}\text{= 4}{\text{.4728×10}}^{\text{-4}}\mathrm{......}\left(\text{4}\right)\end{array}$

Substituting   $\text{b = 0}\text{.9997c}$ in the denominator of equation (4) to solve c.

$\begin{array}{l}\frac{\text{2c}}{\text{760}\text{.37+15}\text{.9994}\left(\text{8+1}\text{.9994c}\right)}\text{= 4}{\text{.4728×10}}^{\text{-4}}\\ \\ \text{c = 0}\text{.2001}\\ \\ \text{b = 0}\text{.9997c = 0}\text{.2000}\end{array}$

Average oxidation number of Bi is $3.2000+$ and copper is found to be ${2.2001}^{+}$ and the formula of the compound is ${\text{Bi}}_{\text{2}}{\text{Sr}}_{\text{2}}{\text{CaCu}}_{\text{2}}{\text{O}}_{\text{8}\text{.4001}}$.  Since the Oxygen stoichiometry derived at the starting of the solution is $\text{x = 8 + b + c}$.

(b)

Interpretation Introduction

Interpretation:

The uncertainties in the oxidation number and x when the quantities in Experiment A are $\text{102}\text{.3}\left(\text{±0}\text{.2}\right)\text{mg}$ and $0.1085\left(\text{±0}\text{.007}\right)\text{mmol}$ and the quantities in experiment B are $94.6\left(\text{±0}\text{.2}\right)\text{mg}$ and $0.0577\left(\text{±0}\text{.007}\right)\text{mmol}$.

Explanation of Solution

To found: The uncertainties in the oxidation number and x when given quantities in experiment A and experiment B

Experiment A: Consumption of $\text{102}\text{.3}\left(\text{±0}\text{.2}\right)\text{mg}$ compound is $\text{0}\text{.0915}\left(\text{±0}\text{.0007}\right){\text{mmol Cu}}^{\text{+}}$

Experiment B: Consumption of $94.6\left(\text{±0}\text{.2}\right)\text{mg}$ compound is $\text{0}\text{.0423}\left(\text{±0}\text{.0007}\right){\text{mmol Fe}}^2{}^{\text{+}}$

Normalising to one gram of superconductor gives,

Experiment A:

$\begin{array}{l}{\text{mmolBi}}^{\text{4+}}{\text{+ mmolCu}}^{\text{3+}}\text{in1g of superconductor}\\ \\ \text{=}\frac{\text{0}\text{.0915}\left(\text{±0}\text{.0007}\right)}{\text{0}\text{.1023}\left(\text{±0}\text{.0002}\right)}\text{= 0}\text{.89443}\left(\text{±0}\text{.00706}\right)\frac{\text{mmol}}{\text{gram}}\end{array}$

Experiment B:

$\begin{array}{l}{\text{mmolBi}}^{\text{4+}}{\text{+ mmolCu}}^{\text{3+}}\text{in1g of superconductor}\\ \\ \text{=}\frac{\text{0}\text{.0423}\left(\text{±0}\text{.0007}\right)}{\text{0}\text{.0946}\left(\text{±0}\text{.0002}\right)}\text{= 0}\text{.44715}\left(\text{±0}\text{.00746}\right)\frac{\text{mmol}}{\text{gram}}\\ \\ \frac{{\text{m molCu}}^{\text{3+}}}{\text{g sperconductor}}\text{= 0}\text{.89443}\left(\text{±0}\text{.00746}\right)\\ \\ \frac{\text{b}}{\text{c}}\text{=}\frac{\text{0}\text{.44715}\left(\text{±0}\text{.00746}\right)}{\text{0}\text{.44728}\left(\text{0}\text{.01027}\right)}\text{= 0}\text{.9997}\left(\text{±0}\text{.0284}\right)\\ \\ \frac{\text{2c}}{\text{760}\text{.37+15}\text{.9994}\left(\text{8+}\left[\text{1}\text{.9997}\left(\text{±0}\text{.0284}\right)\text{c}\right]\right)}\text{= 4}\text{.4728}\left(\text{±0}\text{.1027}\right){\text{×10}}^{\text{-4}}\\ \\ \left[\text{4471}\text{.47}\left(\text{±102}\text{.7}\right)\right]\text{c=888}\text{.365+}\left[\text{31}\text{.9994}\left(\text{±0}\text{.445}\right)\right]\text{c}\\ \\ \text{c = 0}\text{.2001}\left(\pm 0.0033\right)\end{array}$

The relative uncertainty in b given as

$\begin{array}{l}\text{Uncertainty in b =}\frac{\text{0}\text{.00746/0}\text{.44715}}{\text{0}\text{.01027/0}\text{.44728}}\left(\text{uncertainty in c}\right)\\ \\ \text{=}\frac{\text{0}\text{.00746/0}\text{.44715}}{\text{0}\text{.01027/0}\text{.44728}}\left(\text{±}\text{0}\text{.0046}\right)\text{= ±0}\text{.0033}\\ \\ \text{b = 0}\text{.200}\left(\text{±0}\text{.0033}\right)\end{array}$

The average oxidation number of Bi is given as ${\text{Bi}}^{\text{+3}\text{.200}\left(\text{±0}\text{.0033}\right)}\text{and}{\text{Cu}}^{\text{+2}\text{.2001}\left(\text{±0}\text{.0046}\right)}$ and the formula of the compound is ${\text{Bi}}_{\text{2}}{\text{Sr}}_{\text{2}}{\text{CaCu}}_{\text{2}}{\text{O}}_{\text{8}\text{.4001}\left(\pm 0.0057\right)}$.