Volume 25, Number 2, February 2016
Answers
Revision quizzes
Anne Hodgson
See how you did on the revision questions in the magazine.
Lab page: esterification
1
a
The empirical formula for benzocaine is C9H11NO2
b
Calculating the mass of one mole of benzocaine:
(9 × 12) + (11 × 1) + 14 + (2 × 16) = 165 g
2
To work out which reactant should be used in the yield calculation, we need to work out which
of them is present in excess. From the equation we can see that 1 mole of 4-aminobenzoic acid reacts
with 1 mole of ethanol to give 1 mole of benzocaine. We need to calculate how many moles of each
starting material we had:
The empirical formula of 4-aminobenzoic acid is C7H7NO2, from which we can calculate its molar
mass:
(7 × 12) + (7 × 1) + 14 + (2 × 16) = 137 g
We used 1.30 g of 4-aminobenzoic acid in the reaction, and so we can work out how many moles we
had:
1.30 g
= 0.0095 mol 137 g mol !!
3
–3
3
We used 10 cm of ethanol. From its density (0.79 g cm ) we know that 1 cm has a mass of 0.79 g,
3
so 10 cm will have a mass of 7.9 g.
The molar mass of ethanol (C2H6O) is:
(2 × 12) + (6 × 1) + 16 = 46 g
We can calculate how many moles there are in 7.9 g:
7.9 g
= 0.17 mol 46 g mol !!
From this we can see that the ethanol is in excess (0.17 mol compared to 0.0095 mol of 4aminobenzoic acid). As the 4-aminobenzoic acid is the limiting amount, it is this quantity that we
should use in our yield calculation.
3
The theoretical yield of benzocaine that could be obtained from 1.30 g (0.0095 mol) of 4–1
aminobenzoic acid is 0.0095 mol (which would have a mass of 0.0095 mol × 165 g mol = 1.57 g).
The actual yield of 0.85 g contains
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0.85 g
= 0.0052 mol 165 g mol !!
actual yield
× 100 = percentage yield
theoretical yield
0.0052 mol
×100 = 𝟓𝟒. 𝟕%
0.0095 mol
or
0.85 g
×100 = 𝟓𝟒. 𝟏 %
1.57 g
Note that the difference between these two percentage yields is due to rounding values to two
significant figures at earlier stages of the calculations.
4
The reaction between sodium carbonate and sulfuric acid is:
Na2CO3 + H2SO4 à Na2SO4 + CO2 + H2O
3
3
40 cm (= 0.04 dm ) of 0.5 mol dm
3
0.04 dm × 0.5 mol dm
3
–3
–3
sodium carbonate contain
= 0.02 mol
3
1 cm (0.001 dm ) of concentrated sulfuric acid contains
3
0.001 dm × 18.4 mol dm
–3
= 0.0184 mol
From the equation we can see that 1 mole of sodium carbonate reacts with 1 mole of sulfuric acid, so
3
–3
3
40 cm of 0.5 mol dm sodium carbonate (0.02 mol) is sufficient to neutralise 1 cm of concentrated
sulfuric acid (0.0184 mol).
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25 years of… carbene chemistry
Extra question
4
Work out the oxidation numbers of the Li, Mg, Sn, B and U centres in the structures below:
N
N
N
N
N
Li
N
N
N
N
Mg
N
N
Sn
N
Sn
N
N
N
N
N
N
N
B
N
N
O
U
B
N
N
O O
N
O
N
N
N
Answers
Some examples of NHC complexes with s block, p block and f block elements
Remember that an NHC ligand is electronically neutral and so has no charge.
1
Cr = 0, Hg = +2, Pd = +2 for both
2
Ru = +2 for both
3
Cu = +1
4
Li = +1, Mg = +2, Sn = 0, B = 0, U = +3
Nitro
1
1-bromo-2-methylpropane
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2
There are six positional isomers of trinitrotoluene:
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3
a
CH3C6H2(NO2)3(s) → 1.5 N2(g) + 2.5 H2O(g) + 3.5 CO(g) + 3.5 C(s)
So there are 1.5 + 2.5 + 3.5 = 7.5 moles of gas released when 1 mole of TNT
explodes.
b
C3H5N3O9(l) → 2.5H2O(g) + 3CO2(g) + 1.5 N2(g) + 0.25 O2(g)
So there are 2.5 + 3 + 1.5 + 0.25 = 7.25 moles of gas released when 1 mole of
nitroglycerine explodes.
4
a
The molar mass of TNT (CH3C6H2(NO2)3) is:
(7 × 12) + (5 × 1) +(3 × 14) +(6 × 16) = 227 g
–1
The amount of energy released when 1 mole of TNT explodes is: 4080 J g × 227 g =
–1
926 160 J = 0.9 MJ mol
b
The molar mass of nitroglycerine (C3H5N3O9) is:
(3 × 12) + (5 × 1) + (3 × 14) +(9 × 16) = 227 g
The amount of energy released when 1 mole of nitroglycerine explodes is:
–1
–1
6275 J g × 227 g = 1 424 425 J = 1.4 MJ mol
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