Thermogravimetric analysis and propagation of error. 16 Crystals of deuteratedpotassium dihydrogen phosphate,$\mathbf{K}{\left(D_x{H}_{1-x}\right)}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$are used in optics as a light valve, as a light deflector, and for frequency doubling of lasers. The optical properties are sensitive to the fraction of deuterium in the material. A publication states that deuterium content can be determined by measuring the mass lost by dehydration of the crystal after slow heating to$\mathbf{450}\mathbf{°}\mathbf{C}$ in a Pt crucible under${\mathbf{N}}_{\mathbf{2}}$

(a) Let$\mathbf{\alpha }$ be the mass of product divided by the mass of reactant:

$\mathbf{\alpha }\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{KPO}}_{\mathbf{3}}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{K}{\left(D_x{H}_{1-x}\right)}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}}$

Show that the coefficient xin $\mathbf{K}{\left(D_x{H}_{1-x}\right)}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$is related to $\mathbf{\alpha }$by the equation

$\mathbf{x}\mathbf{=}\frac{\mathbf{58}\mathbf{.}\mathbf{6670}}{\mathbf{\alpha }}\mathbf{-}\mathbf{67}\mathbf{.}\mathbf{6183}$

What would be the value of$\mathbf{\alpha }$if starting material were 100% deuterated?

(b) One crystal measured in triplicate gave an average value of$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{8567}}_{\mathbf{7}}$.Find x for this crystal.

(c) From equation B.1 in Appendix B, show that the uncertainty in$\mathbf{x}\left(e_x\right)$ is related to the uncertainty in$\mathbf{\alpha }\left(e_{\alpha }\right)$by the equation

${\mathbf{e}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{58}\mathbf{.}\mathbf{6670}{\mathbf{e}}_{\mathbf{\alpha }}}{{\mathbf{\alpha }}^{\mathbf{2}}}$

(d) The uncertainty in deuterium: hydrogen stoichiometry is$\mathbf{e\_}\left\{x\right\}$The authors estimate that their uncertainty in$\mathbf{\alpha }$ is${\mathbf{e}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0001}\mathbf{.}$ From${\mathbf{e}}_{\mathbf{\alpha }}$ compute$\mathbf{e\_}\left\{x\right\}$. Write the stoichiometry in the form$\mathbf{x}\mathbf{\pm }{\mathbf{e}}_{\mathbf{x}}$ If${\mathbf{e}}_{\mathbf{x}}$were 0.001 (which is perfectly reasonable), what would${\mathbf{e}}_{\mathbf{x}}$be?

Thermogravimetric analysis is a type of thermal analysis in which the temperature fluctuates over time to determine the mass of a sample.

Thermogravimetric analvsis is a kind of thermogravimetric analvsis. This analvsis delivers a variety of data.

(a)

During thermogravimetric analysis, the value of alpha when all the reactants are 100 percent deuterated.

During thermogravimetric analysis, a reaction happens.

$\mathrm{K}{\left({\mathrm{D}}_{\mathrm{x}}{\mathrm{H}}_{1-\mathrm{x}}\right)}_2{\mathrm{PO}}_{4\left(\mathrm{s}\right)}\stackrel{450°\mathrm{C}}{\to }{\mathrm{KPO}}_{3\left(\mathrm{s}\right)}+{\left({\mathrm{D}}_{\mathrm{x}}{\mathrm{H}}_{1-\mathrm{x}}\right)}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

The weight of the product divided by the weight of the reactant is the value of

$$\begin{gathered} \alpha =\frac{118.0703}{136.0853+2.01255x} \\ \alpha =\frac{118.0703}{136.0853+2.01255x}\left(136.0853\right)\alpha +\left(2.01255\right)\alpha x \\ =118.0703\left(2.01255\right)\alpha x \\ =118.0703-\left(136.0853\right)\alpha \end{gathered}$$

While deviding the equation by $\left(2.01255\alpha \right)$

$$\begin{gathered} nx=\frac{118.0703}{\left(2.01255\right)\alpha }-\frac{\left(136.0853\right)a}{\left(2.01255\right)\alpha }nx \\ =\frac{58.6670}{\alpha }-67.6183 \\ =0.854976 \end{gathered}$$

Thus, the value of $\alpha$when all the reactants are 100% deuterated during therm ogravimetric analysis is 0.854976.

(b)

Thermogravimetric analysis is a type of thermal analysis in which the temperature of the sample is measured over time to calculate its mass. This analysis includes information on phase transitions, absorption, chemisorption, thermal breakdown, oxidation, and reduction, among other things. The analysis is carried out using a thermogravimetric analyzer. Thermogravimetric analysis, also known as thermal gravimetric analysis, is a type of thermogravimetric analysis (TGA).

For a crystal measured in triplicate with specified$\alpha$the value of x is 0.85632.

To figure out: The x value for a crystal measured in triplicate with a specified alpha value.

Given,

$$\begin{gathered} \alpha =0.85677x=\frac{58.6670}{\alpha }-67.6183 \\ x=\frac{58.6670}{0.85677}-67.6183 \\ =0.85632 \end{gathered}$$

When this crystal is measured in triplicate, the x value is computed as 0.85632.

(c)

The degree of uncertainty in a given equation must be specified.

The level of uncertainty in the provided equation.

The function$x=f\left(\alpha \right)e_x=\sqrt{{\left(\frac{\left(\partial F\right)}{\left(da\right)}\right)}^2{e}_{\alpha }^2}$

When the$\mathrm{x}=\frac{58.6670}{\mathrm{\alpha }}-67.6183\frac{\left(\partial \mathrm{F}\right)}{\left(\mathrm{da}\right)}=-\frac{58.6670}{{\mathrm{\alpha }}^2}$

The function$\mathrm{x}=\mathrm{f}\left(\mathrm{\alpha }\right)$ by inserting the aforementioned value in the above equation.

$$\begin{gathered} \mathrm{x}=\mathrm{f}\left(\mathrm{\alpha }\right) \\ {\mathrm{e}}_{\mathrm{x}}=\sqrt{{\left(-\frac{58.6670}{{\mathrm{\alpha }}^2}\right)}^2} \\ {\mathrm{e}}_{\mathrm{\alpha }}^2=\frac{58.6670{\mathrm{e}}_{\mathrm{a}}}{{\mathrm{\sigma }}^2} \end{gathered}$$

The relation is true and proved.

(d)

The uncertainty of$e_a$, has to be determined based on the supplied uncertainty of$e_x$ is$0.856\pm 0.008$

From the given uncertainty of$e_{\alpha }$ the uncertainty of$e_x$ is$0.856\pm 0.008$

The uncertainty of$e_x$ based on the supplied uncertainty of$e_{\alpha }$

$e_a=0.0001$

When the$e_a=0.0001$

$$\begin{gathered} e_x=\frac{\left(58.6670\right)\left(0.0001\right)}{{\left(0.85677\right)}^2} \\ =0.008 \end{gathered}$$

D:His

$x\pm e_x=0.856\pm 0.008$

When is

$$\begin{gathered} e_x=\frac{\left(58.6670\right)\left(0.0001\right)}{{\left(0.85677\right)}^2} \\ =0.008 \end{gathered}$$

D:His is a ratio $0.86\pm 0.08$.