Question

A particular \[100\] octane aviation gasoline used \[{\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\] of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]of density \[{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}}\]per litre of product. This compound is made as follows:
\[{\rm{4}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,\,{\rm{ + }}\,\,{\rm{4NaPb}} \to {{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ + }}\,\,{\rm{4NaCl}}\,\,{\rm{ + }}\,\,{\rm{3Pb}}\]

How many grams of ethyl chloride is needed to make enough tetraethyl lead for \[{\rm{1}}\,\,{\rm{L}}\] of gasoline? [Atomic mass of \[{\rm{Pb}}\,\,{\rm{ = }}\,\,{\rm{207}}\]]

Answer 1

Hint: Density of tetraethyllead is required to calculate first followed by the number of moles tetraethyllead. Number of moles of tetraethyllead is calculated by using its molar mass. The molar ratio from the balanced molecular equation is used to find the moles of ethyl chloride and then its mass in grams is calculated with the help of its molar mass.

Complete Step by Step Solution:
Density of a substance is defined as the mass per unit volume.
Mathematically, \[{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\]
As per given data,
Volume of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,{\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\]
Density of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}}\]
Find the mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]as shown below:
\[\begin{array}{l}{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{density}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ \times }}\,\,{\rm{volume}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}} \times {\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,{\rm{g}}\end{array}\]
Molar mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]is known to be \[{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
The relationship between moles (n), mass (m), and molar mass (M) is as shown below:
\[{\rm{moles}}\,{\rm{(n) = }}\dfrac{{{\rm{mass(m)}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}}}\]
Find the moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] as shown below:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{1}}{\rm{.66}}\,\,{\rm{g}}}}{{{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\end{array}\]
The balanced molecular equation is given as:
\[{\rm{4}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,\,{\rm{ + }}\,\,{\rm{4NaPb}} \to {{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ + }}\,\,{\rm{4NaCl}}\,\,{\rm{ + }}\,\,{\rm{3Pb}}\]
From the above equation, it can be seen that \[4\]moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]reacts to form \[{\rm{1}}\]moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\].
So, molar ratio of \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,:{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = 4\,\,:\,\,1\]
Find the moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from the calculated moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] and their molar ratio as:
\[{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\,\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb \times }}\dfrac{{{\rm{4}}\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}}}{{{\rm{1}}\,\,{\rm{mol}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ = \,0.0204\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]

Molar mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is known to be \[{\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
Lastly, find the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from its calculated moles by using the molar mass as:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\, = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} \times {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,0.0204\,\,{\rm{mol}} \times {\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,1.32\,\,{\rm{g}}\end{array}\]
Hence, the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is found to be \[1.32\,\,{\rm{g}}\]

Therefore, the answer is \[\,1.32\,\,{\rm{g}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]

Note: The calculation of relative amounts of substances in chemical reactions is known as stoichiometry (derived from the Greek stoicheion meaning element and meterin meaning measures). The key to weight relationship in chemical change is the mole concept. Coefficients placed in front of formulae in a balanced chemical equation represent the ratio by moles in which substances are consumed and produced by the chemical change. Since the mass of a mole of a substance is directly related to its formula or molecular weight, a balanced chemical equation represents a ratio by weight also.