Question #dbc04

1 Answer
Jan 2, 2017

Answer:

WARNING! Very long answer! (a) The balanced equation is #"C"_2"H"_6"(g)" + "4Cl"_2"(g)" → "C"_2"H"_2"Cl"_4"(l)" + "4HCl(g)"#. (b) You need to use 7.01 g of ethane.

Explanation:

(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.