(a) Write the balanced equation
This is really a three-part question:
i. Find the empirical formula.
ii. Find the molecular formula.
iii. Write the equation.
i. Find the empirical formula.
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of #"C"# to #"H"# to #"Cl"#.
Your compound contains 14.3 % #"C"# and 84.5 % #"Cl"#.
Then #"% H = 100 % - % C - % Cl" = "100 % - 14.3 % - 84.5 %" = 1.2 %#
Assume that we have 100 g of sample.
Then it contains 14.3 g of #"C"#, 1.2 g of #"H"#, and 84.5 g of #"Cl"#.
#"Moles of C" = 14.3 color(red)(cancel(color(black)("g c"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "1.191 mol H"#
#"Moles of H" = 1.2 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "1.19 mol H"#
#"Moles of Cl" = 84.5 color(red)(cancel(color(black)("g O"))) × "1 mol Cl"/(35.45color(red)(cancel(color(black)( "g Cl")))) = "2.384 mol Cl"#
From this point on, I like to summarize the calculations in a table.
#mathbf"Element"color(white)(m) mathbf"Mass/g"color(white)(X) mathbf"Moles"color(white)(Xll)mathbf "Ratio"color(white)(m)mathbf"Integers"#
#color(white)(m)"C" color(white)(XXXmll)14.3color(white)(Xmm)1.191
color(white)(mml)1.001color(white)(mmml)1#
#color(white)(m)"H"color(white)(mmmmml)1.2 color(white)(mmll)1.19 color(white)(mmm) 1color(white)(mmmmml)1#
#color(white)(m)"Cl"color(white)(mmmml)84.59color(white)(mml)2.384color(white)(mml)2.003color(white)(mmml)2#
The empirical formula of #"Y"# is #"CHCl"_2#.
ii. Calculate the molecular formula
The empirical formula mass of #"CHCl"_2# is 83.92 u.
The molecular mass is 168 u.
The molecular mass must be an integral multiple of the empirical formula mass.
#"MM"/"EFM" = (168 color(red)(cancel(color(black)("u"))))/(83.92 color(red)(cancel(color(black)("u")))) = 2.00 ≈ 2#
∴ The molecular formula is twice the empirical formula.
#"Molecular formula" = ("CHCl"_2)_2 = "C"_2"H"_2"Cl"_4#
Compound #"Y"# is #"C"_2"H"_2"Cl"_4#
iii. Write the balanced equation
The unbalanced equation must be
#"C"_2"H"_6"(g)" + "Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "HCl(g)"#
The balanced equation is
#"C"_2"H"_6"(g)" + "4Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "4HCl(g)"#
(b) Calculate the mass of ethane required
#"57.0 % yield" = "22.3 g Y"#
∴ #"Theoretical yield" = 100 color(red)(cancel(color(black)("% yield"))) × "22.3 g Y"/(57.0 color(red)(cancel(color(black)("% yield")))) = "39.12 g Y"#
#"Mass of C"_2"H"_6 = 39.12 color(red)(cancel(color(black)("g Y"))) × (1 color(red)(cancel(color(black)("mol Y"))))/(167.85 color(red)(cancel(color(black)("g Y")))) × (1 color(red)(cancel(color(black)("mol Y"))))/(1 color(red)(cancel(color(black)("mol C"_2"H"_6)))) × ("30.07 g C"_2"H"_6)/(1 color(red)(cancel(color(black)("mol C"_2"H"_6)))) = 7.01"g C"_2"H"_6#
The mass of ethane required is 7.01 g.
