Chapter 27, Problem 27.27P

(a)

Interpretation Introduction

Interpretation:

The value of x when ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}$ undergoes thermogravimetric analysis has to be calculated.

Concept introduction:

Thermogravimetric analysis:

Thermogravimetric analysis is a kind of thermal analysis in that temperature varies over time to determine mass of sample used for thermogravimetric analysis.  This analysis provides various informations such as phase transition, desorption, absorption, chemisorption, decomposition due to thermal, oxidation and reduction.  It is done on thermogravimetric analyzer.  Thermogravimetric analysis is also known as thermal gravimetric analysis (TGA).

Answer to Problem 27.27P

The value of x when ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}$ undergoes thermogravimetric analysis is $\text{0}\text{.2042}$.

Explanation of Solution

To calculate: The value of x when ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}$ undergoes thermogravimetric analysis.

The reaction occurs during thermogravimetric analysis is

${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}\text{+}\text{(3}\text{.5}\text{-}{\text{x)H}}_{\text{2}}{}_{\text{(g)}}\stackrel{{\text{1000}}^{\text{0}}\text{C}}{\to }\text{ 0}{\text{.5Y}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s) }}{\text{+ 2BaO}}_{\text{(s)}}\text{+}{\text{3Cu}}_{\text{(s)}}\text{+}\text{(3}\text{.5}\text{-}{\text{x)H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

The formula mass of ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}$ is $\text{666}\text{.19}\text{-(16}\text{.00)x}$

Number of moles of ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}{}_{\text{(s)}}$ is calculated as follows

$\begin{array}{l}\text{=}\frac{\text{34}\text{.397}\text{mg}}{\text{[66}\text{.19}\text{-}\text{(16}\text{.00)x]mg/mmol}}\\ \\ \text{mmol}\text{of}{\text{o}}_{\text{2}}\text{lost}\text{in}\text{experiment}\text{=}\frac{\text{(34}\text{.397}\text{-}\text{31}\text{.661)mg}}{\text{16}\text{.00}\text{mg/mmol}}\text{=}\text{0}\text{.17100}\text{mmol}\end{array}$

Using the stoichiometry of reaction the value of x is calculated as

$\begin{array}{l}\frac{\text{mmol}\text{of}{\text{O}}_{\text{2}}\text{lost}}{\text{mmol}{\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7-x}}}\text{=}\frac{\text{3}\text{.5}\text{-}\text{x}}{\text{1}}\\ \\ \frac{\text{0}\text{.17100}}{\text{34}\text{.397}\text{/}\text{[666}\text{.19}\text{-}\text{(16}\text{.00)x]}}\text{=}\text{3}\text{.5}\text{-}\text{x}\\ \\ \text{x}\text{=}\text{0}\text{.2042}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The uncertainty in value of x has to be calculated.

Concept introduction:

Thermogravimetric analysis:

Thermogravimetric analysis is a kind of thermal analysis in that temperature varies over time to determine mass of sample used for thermogravimetric analysis.  This analysis provides various informations such as phase transition, desorption, absorption, chemisorption, decomposition due to thermal, oxidation and reduction.  It is done on thermogravimetric analyzer.  Thermogravimetric analysis is also known as thermal gravimetric analysis (TGA).

Answer to Problem 27.27P

The uncertainty in value of x is $\text{0}\text{.204}\text{±}\text{0}\text{.004}$.

Explanation of Solution

To calculate: The uncertainty in value of x.

Given,

$\text{uncertainty}\text{of}\text{mass}\text{=}\text{0}\text{.002}\text{mg}$

Consider the uncertainty of all molecular mass and atomic mass is negligible.

$\begin{array}{l}\text{mmol}\text{of}{\text{O}}_{\text{2}}\text{lost}\text{=}\frac{\text{(34}\text{.397(±0}\text{.002)}\text{-}\text{31}\text{.661(±0}\text{.002))mg}}{\text{16}\text{.00}\text{mg/mmol}}\text{=}\frac{\text{2}\text{.736}\text{(±}\text{0}\text{.0028)}}{\text{16}\text{.00}}\\ \\ \text{=}\text{0}\text{.17100(±0}\text{.102\%)}\\ \\ \text{Relative}\text{error}\text{in}\text{mass}\text{of}\text{reactants}\text{=}\frac{\text{0}\text{.002}}{\text{34}\text{.397}}\text{=}\text{0}\text{.0058}\text{\%}\end{array}$

The uncertainty in x value is determined as

$\begin{array}{l}\frac{\text{0}\text{.17100(±}\text{102\%)}}{\text{34}\text{.397}\text{(±}\text{0}\text{.0058}\text{\%)/[66}\text{.19-(16}\text{.00)x]}}\text{=}\text{3}\text{.5}\text{-}\text{x}\\ \\ \text{0}\text{.17100(±}\text{102\%)}\text{[66}\text{.19-(16}\text{.00)x]}\text{=}\text{(3}\text{.5}\text{-}\text{x)}\text{[34}\text{.397}\text{(±}\text{0}\text{.0058}\text{\%)]}\\ \\ \text{113}\text{.918(±0}\text{.116)}\text{-}\text{[2}\text{.736}\text{(±}\text{0}\text{.00279)]x}\text{=}\text{120}\text{.3895}\text{(±}\text{0}\text{.00698)-[34}\text{.397}\text{(±}\text{0}\text{.002)]x}\\ \\ \text{[31}\text{.66(±}\text{0}\text{.00346)]}\text{x}\text{=}\text{6}\text{.4715}\text{(±}\text{0}\text{.116)}\\ \\ \text{x}\text{=}\text{0}\text{.2044}\text{(±}\text{1}\text{.79}\text{\%)}\text{=}\text{0}\text{.204}\text{±}\text{0}\text{.004}\end{array}$