From: Sam W Date: Thu, 15 Mar 2018 20:43:48 +0000 (+0000) Subject: Completed calculations section for copper solution (excluding evaluating X-Git-Url: https://git.dalvak.com/public/?a=commitdiff_plain;h=e70f508d12e3cad86e37923ac3d73f23f785b45c;p=chemistry%2Funiversity-chemistry-lab-reports.git Completed calculations section for copper solution (excluding evaluating random error in exchange percentage) in project. --- diff --git a/project/project.pdf b/project/project.pdf index d641a59..6c4efb1 100644 Binary files a/project/project.pdf and b/project/project.pdf differ diff --git a/project/project.tex b/project/project.tex index dd63933..f52a882 100644 --- a/project/project.tex +++ b/project/project.tex @@ -26,6 +26,11 @@ \setcounter{secnumdepth}{4} %Label paragraphs as subsubsubsections. \setcounter{tocdepth}{4} %Treat paragraphs as sections in table of contents. +%Allow manual adding of very large brackets. +\makeatletter +\newcommand{\vast}{\bBigg@{5}} +\makeatother + %Document Headings. \begin{document} \title{Investigation on the Effect of the Cation Counterion used on the Ion Exchange Efficiency with HZSM-5} @@ -130,14 +135,48 @@ After one week the zeolite had settled in the bottom of the solutions. The clear \end{tabular} \end{table} +\section{Calculations} + +\subsection{Calculation of Maximum Theoretical Number of Ion Exchanges} +The \ce{SiO2}/\ce{Al2O3} ratio in the zeolite used was $23$. In this ratio there are two \ce{Al} atoms per \ce{Si}, so $\ce{Si}/\ce{Al} = \frac{23}{2} = 11.5$. + +Using the unit cell general formula (equation \ref{eq:unit-cell}) letting the \ce{Si}/\ce{Al} ratio be $r$ and with $n$ being the number of aluminium atoms per unit cell: +\begin{gather*} + r = \frac{\text{Number of \ce{Si} per unit cell}}{\text{Number of \ce{Al} per unit cell}} = \frac{96 - n}{n} \\ + n r + n = 96 \\ + \therefore n = \frac{96}{r + 1} +\end{gather*} + +Hence for $r = 11.5$ there are $n = \frac{96}{11.5 + 1} = 7.68$ \ce{Al} per unit cell. Letting $q$ be the cation charge and $x$ be the number of water molecules for unit cell: + +\begin{align*} + Mr_{\text{unit cell}} = \frac{7.68}{q} Mr_{\text{cation}} &+ (11.5(26.982)+ (96-7.68)(28.085) + 192(15.999) \\ + &+ x(2(1.008) + 15.999)) \si{\gram\per\mole} \\ + = \frac{7.68}{q} Mr_{\text{cation}} &+ \SI{5759.4692}{\gram\per\mole} + x(\SI{450.375}{\gram\per\mole}) +\end{align*} + +Thus for HZSM-5 where the cation is \ce{H+} and $x \approx 25$~\autocite{donder06}. +\begin{equation}\label{eq:hzsm-5-mr} +\begin{split} + Mr_{\text{HZSM-5 unit cell}} &= \frac{7.68}{1} \times \SI{1.008}{\gram\per\mole} + (5759.49692 + 25(450.375)) \text{ \si{\gram\per\mole}} \\ + &= \SI{6217.6134}{\gram\per\mole} +\end{split} +\end{equation} + +Let: $q$ be the cation charge; $n_{\text{max. cation}}$ be the theoretical maximum amount of cation which can be exchanged and $n_{\text{cation}}$, $m_{\text{cation}}$ and $Mr_{\text{cation}}$ be the actual amount, mass and $Mr$ of the cation exchanged respectively. -%\begin{align*} -% m_{\text{\ce{CuSO4.5H2O}}} &= \SI{0.5014}{\gram} \\ -% n_{\text{\ce{CuSO4.5H2O}}} &= \frac{\SI{0.5014}{\gram}}{(63.546 + 32.066 + 4(15.999) + 5(2(1.008) + 15.999)) \text{ \si{\gram\per\mole}}} \\ -% &= \frac{\SI{0.5014}{\gram}}{\SI{249.677}{\gram\per\mole}} = \SI{2.008e-3}{\mole} \\ -% [\ce{CuSO4}] &= \SI{0.04016}{\mole\per\deci\metre\cubed} \\ -%\end{align*} +\begin{align} + n_{\text{HZSM-5 unit cell}} &= \frac{m_{\text{HZSM-5}}}{Mr_{\text{HZSM-5 unit cell}}} \nonumber \\ + n_{\text{max. cation}} &= \frac{7.68}{q} n_{\text{HZSM-5 unit cell}} \nonumber \\ + &= \frac{7.68}{q} \frac{m_{\text{HZSM-5}}}{Mr_{\text{HZSM-5 unit cell}}} \nonumber \\ +% + \si{\percent} \text{ Exchanged} &= \frac{n_{\text{cation}}}{n_{\text{max. cation}}} \times \SI{100}{\percent} \nonumber \\ + \label{eq:cation-percent-exchanged} + &= \frac{q Mr_{\text{HZSM-5 unit cell}} n_{\text{cation}}}{7.68 m_{\text{HZSM-5}}} \times \SI{100}{\percent} +\end{align} +\subsection{Calculations for Copper Solution} +\subsubsection{Determination of Molar Extinction Coefficient} Let $V_{\ce{Cu}_\text{std.}}$ be the volume and $[\ce{CuSO4}]_\text{std.}$ be the concentration of the standard \ce{Cu^{2+}} solution. \begin{align} n_{\ce{CuSO4}} &= \frac{m_{\ce{CuSO4.5H2O}}}{Mr_{\ce{CuSO4.5H2O}}} \nonumber \\ @@ -171,8 +210,7 @@ This hence gives: &= \frac{0.484 \times \SI{50.00e-3}{\deci\metre\cubed} \times \SI{249.677}{\gram\per\mole}}{\SI{1.0}{\centi\metre} \times \SI{0.5014}{\gram}} = \SI{12.05}{\deci\metre\cubed\per\mole\per\centi\metre} \end{align} -%TODO: Look up (and compare to) literature value. - +\subsubsection{Determination of Percentage of \ce{Cu^2+} Exchanged Compared to the Theoretical Maximum} By rearranging the Beer-Lambert Law (equation \ref{eq:beer-lambert}) for concentration: \begin{equation} \label{eq:beer-lambert-c} @@ -199,78 +237,87 @@ Using equations \ref{eq:[cuso4]-std} and \ref{eq:n_cu-prod-final} to determine t \begin{align} n_{\ce{Cu}_\text{ex.}} &= [\ce{CuSO4}]V_{\ce{Cu}_\text{react.}} - n_{\ce{Cu}_\text{prod.}} \nonumber\\ % \label{ - &= \frac{m_{\ce{CuSO4.5H2O}} V_{\ce{Cu}_\text{react.}}}{V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} - \frac{A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} m_{\ce{CuSO4.5H2O}}}{A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} + &= \frac{m_{\ce{CuSO4.5H2O}} V_{\ce{Cu}_\text{react.}}}{V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} - \frac{A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} m_{\ce{CuSO4.5H2O}}}{A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} \nonumber \\ + \label{eq:cu-exchanged} + &= \frac{m_{\ce{CuSO4.5H2O}} \left(A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{react.}} - A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} \right)}{A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} \end{align} -Substitutiong equation ... into equation \ref{eq:cation-percent-exchanged} and setting $q = 2$ hence gives: - -%TODO: Continue from here. +Substituting equation \ref{eq:cu-exchanged} into equation \ref{eq:cation-percent-exchanged} and setting $q = 2$ hence gives: -\begin{equation} \label{eq:cu-exchanged} - \si{\percent} \text{ \ce{Cu^{2+}} Exchanged} = \frac{2 Mr_{\text{HZSM-5 unit cell}} A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} m_{\ce{CuSO4.5H2O}}}{7.68 m_{\text{HZSM-5}} A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} \times \SI{100}{\percent} +\begin{equation} \label{eq:cu-percent-exchanged} + \si{\percent} \text{ \ce{Cu^{2+}} Exchanged} = \frac{2 Mr_{\text{HZSM-5 unit cell}} m_{\ce{CuSO4.5H2O}} \left(A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{react.}} - A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} \right)}{7.68 m_{\text{HZSM-5}} A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{std.}} Mr_{\ce{CuSO4.5H2O}}} \times \SI{100}{\percent} \end{equation} -Using \ref{eq:cu-exchanged} with: +Using \ref{eq:cu-percent-exchanged} with: -%TODO: Add uncertainties to these values. +%TODO: Look up uncertainty in absorbance value for spectrophotometer and that for relative atomic masses. \begin{align*} Mr_{\text{HZSM-5 unit cell}} &= \SI{6217.6134}{\gram\per\mole} \text{ from equation \ref{eq:hzsm-5-mr}} \\ + m_{\ce{CuSO4.5H2O}} &= \SI{0.5014 \pm 0.00005}{\gram} \text{ from table \ref{tbl:masses}} \\ + A_{\ce{Cu}_\text{std.}} &= \num{0.484} \text{ from table \ref{tbl:absorbance}} \\ + V_{\ce{Cu}_\text{react.}} &= \SI{20.00 \pm 0.06 e-3}{\deci\metre\cubed} \\ A_{\ce{Cu}_\text{prod.}} &= \num{0.110} \text{ from table \ref{tbl:absorbance}} \\ V_{\ce{Cu}_\text{prod.}} &= \SI{100.00 \pm 0.20 e-3}{\deci\metre\cubed} \\ - m_{\ce{CuSO4.5H2O}} &= \SI{0.5014 \pm 0.00005}{\gram} \text{ from table \ref{tbl:masses}} \\ m_{\text{HZSM-5}} &= \SI{0.4810 \pm 0.00005}{\gram} \text{ from table \ref{tbl:masses}} \\ - A_{\ce{Cu}_\text{std.}} &= \num{0.484} \text{ from table \ref{tbl:absorbance}} \\ V_{\ce{Cu}_\text{std.}} &= \SI{50.00 \pm 0.06 e-3}{\deci\metre\cubed} \\ Mr_{\ce{CuSO4.5H2O}} &= \SI{249.577}{\gram\per\mole} \text{ from equation \ref{eq:molar-extinction-calc}} \end{align*} -%TODO: Put numbers above into calculation and write below: -\begin{displaymath} - \si{\percent} \text{ \ce{Cu^{2+}} Exchanged} = \frac{2 \times \SI{6217.6134}{\gram\per\mole} \times \SI{0.00}{\mole} }{7.68 \times \SI{0.4810}{\gram}} \times \SI{100}{\percent} = \SI{0.00}{\percent} -\end{displaymath} - -%TODO: Error propagation. -%TODO: Look up uncertainty in absorbance value for spectrophotometer. - -\subsection{Calculations} - -\subsubsection{Calculation of Maximum Theoretical Number of Ion Exchanges} -The \ce{SiO2}/\ce{Al2O3} ratio in the zeolite used was $23$. In this ratio there are two \ce{Al} atoms per \ce{Si}, so $\ce{Si}/\ce{Al} = \frac{23}{2} = 11.5$. - -Using the unit cell general formula (equation \ref{eq:unit-cell}) letting the \ce{Si}/\ce{Al} ratio be $r$ and with $n$ being the number of aluminium atoms per unit cell: -\begin{gather*} - r = \frac{\text{Number of \ce{Si} per unit cell}}{\text{Number of \ce{Al} per unit cell}} = \frac{96 - n}{n} \\ - n r + n = 96 \\ - \therefore n = \frac{96}{r + 1} -\end{gather*} - -Hence for $r = 11.5$ there are $n = \frac{96}{11.5 + 1} = 7.68$ \ce{Al} per unit cell. Letting $q$ be the cation charge and $x$ be the number of water molecules for unit cell: - \begin{align*} - Mr_{\text{unit cell}} = \frac{7.68}{q} Mr_{\text{cation}} &+ (11.5(26.982)+ (96-7.68)(28.085) + 192(15.999) \\ - &+ x(2(1.008) + 15.999)) \si{\gram\per\mole} \\ - = \frac{7.68}{q} Mr_{\text{cation}} &+ \SI{5759.4692}{\gram\per\mole} + x(\SI{450.375}{\gram\per\mole}) + \begin{split} + \si{\percent} \text{ \ce{Cu^{2+}} Exchanged} &= \frac{2 \times \SI{6217.6134}{\gram\per\mole} \times \SI{0.50140}{\gram}\left(0.484 \times 20.00 - 0.110 \times 100.00\right)\num{e-3} \text{ \si{\deci\metre\cubed}}}{7.68 \times \SI{0.4810}{\gram} \times 0.484 \times \SI{50.00e-3}{\deci\metre\cubed} \times \SI{249.577}{\gram\per\mole}} \\ + &\text{ } \times \SI{100}{\percent} + \end{split} \\ + &= \SI{-18}{\percent} \end{align*} -Thus for HZSM-5 where the cation is \ce{H+} and $x \approx 25$~\autocite{donder06}. -\begin{equation}\label{eq:hzsm-5-mr} +\subsubsection{Error Propagation} +Let the percentage of \ce{Cu^2+} exchanged be $v_{\ce{Cu}}$ in the error propagation below: + +\begin{equation} +\label{eq:cu-error-propagation} \begin{split} - Mr_{\text{HZSM-5 unit cell}} &= \frac{7.68}{1} \times \SI{1.008}{\gram\per\mole} + (5759.49692 + 25(450.375)) \text{ \si{\gram\per\mole}} \\ - &= \SI{6217.6134}{\gram\per\mole} + \delta v_{\ce{Cu}} = &v_{\ce{Cu}} + \vast( %Manually putting in large brackets for square root. + \left(\frac{\delta Mr_{\text{HZSM-5 unit cell}}}{Mr_{\text{HZSM-5 unit cell}}}\right)^2 + + \left(\frac{\delta m_{\ce{CuSO4.5H2O}}}{m_{\ce{CuSO4.5H2O}}}\right)^2 \\ + %New Line + &+ \frac{ + A_{\ce{Cu}_\text{std.}}^2 V_{\ce{Cu}_\text{react.}}^2 + \left( + \left(\frac{\delta A_{\ce{Cu}_\text{std.}}}{A_{\ce{Cu}_\text{std.}}}\right)^2 + + \left(\frac{\delta V_{\ce{Cu}_\text{react.}}}{V_{\ce{Cu}_\text{react.}}}\right)^2 + \right) + + A_{\ce{Cu}_\text{std.}}^2 V_{\ce{Cu}_\text{react.}}^2 + \left( + \left(\frac{\delta A_{\ce{Cu}_\text{prod.}}}{A_{\ce{Cu}_\text{prod.}}}\right)^2 + + \left(\frac{\delta V_{\ce{Cu}_\text{prod.}}}{V_{\ce{Cu}_\text{prod.}}}\right)^2 + \right) + } + { + \left( + A_{\ce{Cu}_\text{std.}} V_{\ce{Cu}_\text{react.}} - A_{\ce{Cu}_\text{prod.}} V_{\ce{Cu}_\text{prod.}} + \right)^2 + } \\ + %New Line. + &+ \left(\frac{\delta m_{\text{HZSM-5}}}{m_{\text{HZSM-5}}}\right)^2 + + \left(\frac{\delta A_{\ce{Cu}_\text{std.}}}{A_{\ce{Cu}_\text{std.}}}\right)^2 + + \left(\frac{\delta V_{\ce{Cu}_\text{std.}}}{V_{\ce{Cu}_\text{std.}}}\right)^2 + + \left(\frac{\delta Mr_{\ce{CuSO4.5H2O}}}{Mr_{\ce{CuSO4.5H2O}}}\right)^2 + \vast)^{1/2} \end{split} \end{equation} -Let: $q$ be the cation charge; $n_{\text{max. cation}}$ be the theoretical maximum amount of cation which can be exchanged and $n_{\text{cation}}$, $m_{\text{cation}}$ and $Mr_{\text{cation}}$ be the actual amount, mass and $Mr$ of the cation exchanged respectively. +%TODO: Substitute uncertainties into equation. +Substituting values into equation \ref{eq:cu-error-propagation} thus yields: -\begin{align} - n_{\text{HZSM-5 unit cell}} &= \frac{m_{\text{HZSM-5}}}{Mr_{\text{HZSM-5 unit cell}}} \nonumber \\ - n_{\text{max. cation}} &= \frac{7.68}{q} n_{\text{HZSM-5 unit cell}} \nonumber \\ - &= \frac{7.68}{q} \frac{m_{\text{HZSM-5}}}{Mr_{\text{HZSM-5 unit cell}}} \nonumber \\ -% - \si{\percent} \text{ Exchanged} &= \frac{n_{\text{cation}}}{n_{\text{max. cation}}} \times \SI{100}{\percent} \nonumber \\ - \label{eq:cation-percent-exchanged} - &= \frac{q Mr_{\text{HZSM-5 unit cell}} n_{\text{cation}}}{7.68 m_{\text{HZSM-5}}} \times \SI{100}{\percent} -\end{align} +\begin{displaymath} + \delta v_{\ce{Cu}} = \pm \SI{0.00}{\percent} +\end{displaymath} + +So the percentage of \ce{Cu^2+} exchanged is \SI{-18 \pm 0.00}{\percent}. + +\subsection{Calculation of Ion-Exchange Efficiency for Zinc Solution} %Over 100% exchange is possible e.g. due to formation of oxide species phyllosilicate outside zeolite e.t.c. influence on cobalt salt precursers on cobalt speciation and catalytic properties of H-ZSM-5 modified ... mhamdi @@ -278,14 +325,6 @@ Let: $q$ be the cation charge; $n_{\text{max. cation}}$ be the theoretical maxim %TODO: Subsubsub section package? %\subsubsubsection{Copper} -\paragraph{Copper} - - - -\subsection{Copper} - -For zeolite: - \section{Analysis} %TODO: Compared molar extinction coefficient value to literature value.