Answers to Problem set for 2/14/07
In the chlor-alkali process, concentrated sodium chloride solution (brine) is electrolyzed at high
current density to produce chlorine gas, hydrogen gas and lye (sodium hydroxide) according to
the following chemical reaction equation:
2 NaCl + 2 H2O → Cl2 + H2 + 2 NaOH
Depending on which is currently fetching a higher market price, either the chlorine or the lye can
be considered a by-product of this industrial process.
How much salt (sodium chloride) is needed to produce a thousand kilograms (one “tonne”) of
lye? (Remember, the salt is in water solution, so there’s plenty of water!)
The “mole” functions as a way to convert chemical quantities (like “2 NaOH”) into quantities that
can be weighed out in bulk. The chemical equation above tells us that, to produce two NaOH, we
need two NaCl. That means that for every NaOH we want to produce, we need one NaCl; in other
words, the “mole ratio” is one-to-one.
The “formula weight” of any compound is found by adding up the atomic weights of the atoms in
its formula. Sodium “Na” has an atomic weight of 23.0; chlorine “Cl” an atomic weight of 35.5;
so “NaCl” has a formula weight of 58.5.
“NaOH” has a formula weight of 23.0 + 16.0 + 1.0 = 40.0.
This means that one mole of NaCl weighs 58.5 g, and one mole of NaOH weighs 40.0 g. But…
we could just as easily use kg, or pounds, or ounces, or grains, or hundredweights – the key is
that, to have the same number of molecules, we need a mass ratio of 58.5 to 40.0.
Therefore, if we want 1000 kg of NaOH (“40.0”), how much NaCl (“58.5”) do we need?
1000 kg
1000 kg
x
=
⇒x=
× 58.5 = 1460 kg
40
58.5
40
Thus we need at least 1460 kg of NaCl to produce 1000 kg of NaOH by the chlor-alkali process.
We could set this up another way:
moles NaCl = moles NaOH
moles NaOH = 1000 kg ÷ 40 grams per mole
moles NaCl = x kg ÷ 58.5 grams per mole
x kg ÷ 58.5 grams per mole = 1000 kg ÷ 40 grams per mole
units cancel, we multiply both sides by 58.5 and we get
x = 1000 ÷ 40 × 58.5 = 1462 kg (rounded to the nearest kg)
How much chlorine and hydrogen are produced as by-products?
We set this up using moles, explicitly. We do this because the reaction equation tells us that we
get ONE H2 and one Cl2 for every TWO NaCl or NaOH.
Note that H2 has a formula weight of 2.0 and Cl2 a formula weight of 71.0 (2 × 35.5).
moles NaOH = 2 × moles H2
(because we make 2 NaOH for every 1 H2)
moles NaOH = 1000 kg ÷ 40 grams per mole
moles H2 = x kg ÷ 2.0 grams per mole
substituting the quantities for “moles NaOH” and “moles H2”, we get
2 × (x kg ÷ 2.0 grams per mole) = 1000 kg ÷ 40 grams per mole
units cancel, 2 ÷ 2.0 = 1, and we get
x = 1000 ÷ 40 = 25 kg (rounded to the nearest kg)
We do the same for Cl2:
moles NaOH = 2 × moles Cl2
(because we make 2 NaOH for every 1 H2)
moles NaOH = 1000 kg ÷ 40 grams per mole
moles Cl2 = x kg ÷ 71.0 grams per mole
substituting the quantities for “moles NaOH” and “moles Cl2”, we get
2 × (x kg ÷ 71.0 grams per mole) = 1000 kg ÷ 40 grams per mole
units cancel, and we get
2 × x = 1000 ÷ 40 × 71 = 1775 kg
x = 888 kg (rounded to the nearest kg)
In real chemistry, one very often doesn’t get the theoretical amount of a desired product. Usually
this is because of side reactions that either draw off the raw materials into other, unwanted
products or because of further, unwanted reactions that the desired product undergoes.
Valuable precursors for biodiesel and synthetic lubricants are obtained by the ozonolysis of
monounsaturated fatty acids such as oleic acid, which are obtained from natural fats and oils.
When oleic acid (C18H34O2) is treated with ozone (O3), it generates two useful products: pelargic
acid (C9H18O2) and azelaic acid (C9H16O4), which can be further processed into a synthetic
lubricant with excellent heat stability, pelargonic azeleate.
3 C18H34O2 + 4 O3 → 3 C9H18O2 + 3 C9H16O4
Suppose that 200 kg of oleic acid is treated with an excess of ozone (that is, more ozone than
required to complete the reaction).
The important quantities are the formula weights of oleic acid (18× 12.01 + 34 × 1.08 + 2 × 16.00
= 284.90), pelargic acid (9 × 12.01 + 18 × 1.08 + 2 × 16.00 = 159.53) and azelaic acid (9 × 12.01
+ 16 × 1.08 + 4 × 16.00 = 189.37). The reaction equation tells us that each mole of oleic acid
produces one mole of pelargic acid and one mole of azelaic acid.
We are using 200 kg of oleic acid.
moles oleic acid = 200 kg ÷ 284.90 grams per mole
How many kg of pelargic acid do we expect to get from this reaction?
moles pelargic acid = x kg ÷ 159.53 grams per mole
moles pelargic acid = moles oleic acid
substituting into this equation, we get
x kg ÷ 159.53 grams per mole = 200 kg ÷ 284.90 grams per mole
allowing units to cancel and solving for x, we get
x kg = 200 ÷ 284.90 × 159.53 = 112 kg (rounded to the nearest kg)
How many kg of azelaic acid do we expect to get from this reaction?
moles azelaic acid = x kg ÷ 189.37 grams per mole
moles azelaic acid = moles oleic acid
substituting into this equation, we get
x kg ÷ 189.37 grams per mole = 200 kg ÷ 284.90 grams per mole
allowing units to cancel and solving for x, we get
x kg = 200 ÷ 284.90 × 189.37 = 133 kg (rounded to the nearest kg)
When the product is collected and purified, there are 100 kg of pelargic acid. What percentage of
the theoretical amount of pelargic acid was obtained?
We expected to obtain 112 kg, but only got 100 kg. The “yield” was therefore
100 ÷ 112 × 100% = 89%
When the product is collected and purified, there are 100 kg of azelaic acid. What percentage of
the theoretical amount of azelaic acid was obtained?
We expected to obtain 133 kg, but only got 100 kg. The “yield” was therefore
100 ÷ 133 × 100% = 75%