\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}
\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 \\
&= \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}
\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
%TODO: Subsubsub section package?
%\subsubsubsection{Copper}
-\paragraph{Copper}
-
-
-
-\subsection{Copper}
-
-For zeolite:
-
\section{Analysis}
%TODO: Compared molar extinction coefficient value to literature value.
projects / chemistry / university-chemistry-lab-reports.git / commitdiff