When the high-temperarure supercondactor yttrium barium copper oxide (see Chapter 16 opening and Box 16-3) is heated under flowing ${\mathit{H}}_{\mathbf{2}}$, the solid remaining at $1OOO°C$ is a mixture of $Y_2{O}_3$ , BaO, and Cu . The starting material has the formula

$YB{a}_{2}C{u}_3{O}_{7-1}$ , in which the oxygea stoichiometry varies between 7 and role="math" localid="1663686755590" $6.5\left(x=0to0.5\right)$ .

$$\begin{gathered} YB{a}_{2}C{u}_3{O}_{7-2}\left(s\right)+\left(3.5-x\right)H_2\left(g\right)\stackrel{1000C}{\to } \\ 666.19-16.7 \\ \frac{{\displaystyle \frac{1}{2}}{Y}_2{O}_3\left(s\right)+2BaO\left(s\right)+3Cu\left(s\right)}{YB{a}_{2}C{u}_3{O}_{2}s}+\left(3.5-x\right)H_{2}O\left(g\right) \end{gathered}$$

(a) Thennogruimetric analysis. When $34.397mg$ of $YB{a}_{2}C{u}_3{O}_{-}-x$were subjected to this andysis, 31.661mg of solid remained after heating to $1000°C$ Find the value of x in $YB{a}_{2}C{u}_3{O}_{3-x}$

(b) Propagation of error. Suppowe that the uncertainty in each mass in part (a) is$\pm 0.0012mg$ . Find the uncertainty in the value of x.

• Thermogravimetric analysis (TGA) is a type of thermal analysis in which the mass of a sample is measured over time as the temperature changes. This measurement yields data on physical phenomena such as phase transitions, absorption, adsorption, and desorption.
• The effects of a variable's uncertainty on a function are defined as propagation of error (or propagation of uncertainty). It is a statistical calculation based on calculus that is designed to combine uncertainties from multiple variables in order to provide an accurate measurement of uncertainty.

a) consider the following:

$M\left(YB{a}_{2}C{u}_3{O}_{7-x}\right)=666.19-16x$

First write the expression for the moles of $YB{a}_{2}C{u}_3{O}_{7-x}$ :

$$\begin{gathered} n\left(YB{a}_{2}C{u}_3{O}_{7-x}\right)=\frac{m}{M} \\ n\left(YB{a}_{2}C{u}_3{O}_{7-x}\right)=\frac{34.397mg}{666.19-16xmg/mmol} \end{gathered}$$

Then calculate the moles of oxygen lost during the experiment:

$$\begin{gathered} n\left(O\right)=\frac{m}{M} \\ n\left(O\right)=\frac{\left(34.397-31.661\right)mg}{16mg/mmol} \\ n\left(O\right)=0.171mmol \end{gathered}$$

Next write the following and calculate the value of x :

$$\begin{gathered} \frac{n\left(O\right)}{n\left(YB{a}_{2}C{u}_3{O}_{7-x}\right)}\frac{3.5-x}{1} \\ \frac{0.171}{34.397/\left(666.19-16xmg/mmol\right)}=3.5-x \\ x=0.2042 \end{gathered}$$

Therefore the value of x is 0.2042 .

b) Calculate the moles of oxygen lost:

$$\begin{gathered} n\left(O\right)=\frac{\left(34.397\pm 0.002\right)-\left(31.661\pm 0.002\right)}{16} \\ n\left(O\right)=\frac{2.736\pm 0.0028}{16} \\ n\left(O\right)=0.171\pm 0.102\% \end{gathered}$$

The relative error in the mass would be $0.002/34.397=0.0058\%$

The equation with error would then be:

$$\begin{gathered} \frac{0.171\pm 0.102\%}{34.397\pm 0.0058\%/\left(666.19-16mg/mmol\right)}=3.5-x \\ \left(0.171\pm 0.102\%\right)\left(666.19-16x\right)=\left(3.5-x\right)\left(34.397\pm 0.0058\%\right) \\ \left(113.98\pm 0.116\right)-\left(2.736\pm 0.00279\right)x=\left(120.3895\pm 0.00968\right)-\left(34.397\pm 0.002\right)x \\ \left(31.66\pm 0.00346\right)x=6.4715\pm 0.116=0.2044\pm 1.79\%=0.204\pm 0.004 \end{gathered}$$

Therefore the uncertainty in the value of x is $0.204\pm 0.004$.