STOICHIOMETRY
Greek: Stoicheon = element metron = element measuring
Stoichiometry is the science of measuring the quantitative proportions or mass
ratios in which chemical elements stand to one another
Molar Ratios:
xA + yB → aC + zD
nA = nB
x
y
Stoichiometric relationship:
=n =n
𝐶
a
𝐷
z
Example 1: Consider the following balanced equation and write the
stoichiometric relationship for all reactants and products
2C2 H6(g) + 7O2(g) → 4CO2 (g) + 6H2O (l)
Solution:
nC H (g) = nO (g) = nCO (g) = nH O(g)
2
6
2
2
7
2
4
2
6
1. Mole to Mole conversions: mol → mol
3H2(g) + N2(g) → 2NH3(g)
3 H2 molecules ≡ 1 N2 molecule ≡ 2 NH3 molecules
3 mols H2 ≡ 1 mol N2 ≡ 2 mol NH3
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Example 1.
If 2.0 mol of N2(g) reacts with sufficient H2(g), how many mols of NH3(g) will be
produced?
nN = nNH
Solution:
3
2
1
⇒ 2.0 mols = nNH3
2
1
2
nNH = 2.0 mols x 2 = 4.0 mols
3
Example 2.
If 4.25 mols of H2 (g) reacted with sufficient N2(g), calculate the number of moles
of NH3(g) that would be produced.
2. Mole to mass conversions: mol → grams Example 1:
How many grams of oxygen are produced when 1.50 mols of KClO3(s) are
decomposed according to the balanced equation?
∆
2 KClO3(s)
nKClO = nO
3
2
Therefore
nO = mass of O
2
2
2
3
2KCl(s) + 3O2(g)
⇒ 1.5 mols = nO2
2
3
nO = 1.50 mols × 32 = 2.25 mols
2
⇒ mass of O2
= 2.25 mols × 32.00 g mol-1
Molar mass of O2
= 72.0 g
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3. Mass to mole conversions

∆
2 KClO3(s)
2 KCl(s) + 3 O2(g)
Example: If 80.0 g of O2(g) was produced in the above reaction, calculate the
number of moles of KClO3 decomposed.
Solution:
Stoichiometry from balanced equation is:
nKClO
3
2
=
nO
2
⇒ nKClO3
3
=
nO
2
=
× ²/3
80 g
32.00 g/mol
×
²/3
= 1.67 mols of KClO3 decomposed
Practice Exercise:
Consider the following equation:
2H2(g) + O2(g) → 2H2O
(a) How many grams of H2O are produced when 2.50 moles of O2(g) is
reacted?
(b) If 3.00 moles of H2O is produced, calculate the mass of O2(g) that was
used.
(c) How many grams of H2(g) must be used, given the data in (b) above?
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4. Mass to mass conversions
Example 1
How many grams of Cl2(g) can be liberated from the decomposition of 64.0 g of
AuCl3 in the following reaction:
2AuCl3 (aq) → 2Au(s) + 3Cl2
Solution:
From stoichiometry
nAuCl3 = nAu = nCl
2
2
3
nAuCl = nCl
3
2
nCl = nAuCl
2
3
x 3=
2
64.0 g
303.32 g/mol
x
2
3
3
2
mass of Cl2 (g) = 0.211 mols x 70.9 g mol
=
2
=
0.211 mols
-1
22.40 g
Practice Exercises
1.
Calculate the mass of AuCl3 that can be produced from 100 g of Cl2 in the
following reaction:
2Au(s) + 3Cl2 → 2 AuCl3(aq)
2.
Calculate the mass of AgCl(s) that can be produced by reacting 200.0 g of
AlCl3 and sufficient AgNO3, using the following reaction:
3AgNO3(aq) + AlCl3(aq) → 3AgCl(s) + Al(NO3)3 (aq)
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3.
Using the following reaction:
2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)
Calculate the mass of PbI2(s) by reacting 30.0 g of KI with excess Pb(NO3)2 .
4.
How many grams of Na(s) are required to react completely with 75.0 g of
Cl2(g) using the following equation:
2Na(s) + Cl2(g) → NaCl (unbalanced)
5.
A component of acid rain is sulfuric acid which forms when SO2(g), a
pollutant reacts with oxygen and rain water according to the following
reaction:
2SO2 (g) + O2 (g) + H2O(ℓ) → H2SO4 (aq)
Assuming that there is plenty of O2 (g) and H2O(ℓ), how much H2SO4 in
3
kilograms forms from 2.6 x 10 kg of SO2 (g)?
6.
Limiting Reagents
The limiting reagent (or reactant) is the reactant that is completely consumed in
a chemical reaction.
The maximum amount of product formed depends on how much of this (limiting)
reactant was originally present.
Excess reagents are present in quantities greater than necessary to react with the
quantity of the limiting reagent.
Theoretical yield - the amount of product that can be made in a chemical reaction
based on the amount of limiting reagent.
Actual or (experimental) yield – the amount of product actually produced by a
chemical reaction.
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Percentage Yield =
Actual Yield
Theoretical Yield
x 100
Consider a recipe to bake pancakes:
1 cup flour + 2 eggs + ½ tsp baking powder → 5 pancakes
Suppose we have:
3 cups flour + 10 eggs + 4 tsp baking powder → ? pancakes
We can make:
3 cups flour → 15 pancakes
10 eggs → 25 pancakes
4 tsp baking powder → 40 pancakes
Flour is the limiting reagent as it produces the least amount of pancakes
Practice Exercise 1.
Consider the following reaction:
Ti(s) + 2Cl2 (g) → TiCl4(s)
If we begin with 1.8 mol of Ti and 3.2 mol of Cl2, what is the limiting reagent and
calculate the theoretical yield of TiCl4 in moles?
Solution:
Given: 1.8 mol Ti
Find: limiting reagent
3.2 mol Cl2
Theoretical yield
Stoichiometry:
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nTiCl
4
=
1
nCl
2
2
nTiCl
4
=
1
=
3.2 mols
= 1.6 mols
2
nTi
= 1.8 mols
1
Since smaller amount of mols of TiCl4 is produced from Cl2,
Therefore Cl2 is the limiting reagent while Ti is the excess reagent.
Therefore theoretical yield of TiCl4(s) = 1.6 mols
Practice Exercise 1.
Consider the following reaction:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
If we begin with 0.552 mol of aluminium and 0.887 mol of chlorine, what is the
limiting reagent and the theoretical yield?
7.
Limiting Reagent, Theoretical Yield, and Percent Yield from Initial masses
of Reactants
Consider the following reaction:
2Na(s) + Cl2(g) → 2NaCl(s)
If we begin with 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant and
theoretical yield?
nNa =
53.2g
22.99g/mol
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= 2.31 mols,
nCl = 65.8g
2
= 0.928 mols
70.90g/mol
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nNaCl = nNa =
2
2.31
2
nNaCl =
= 2.31 mols
2
nCl
2
2
⇒
nNaCl = 0.928 mols x 2 = 1.856 mols
1
Therefore the limiting reagent is Cl
Theoretical yield (calculated from Cl)
-1
Mass of NaCl = 1.856 mols x 58.44 g mol = 108.5 g NaCl
Suppose when the synthesis was carried out, the actual yield of NaCl was
found to be 86.4 g. What is the percent yield?
Percentage Yield = Actual Yield
x 100
Theoretical Yield
= 86.4 g x 100
108.5 g
= 80.0 %
Practice Exercise
1.
Ammonia can be synthesized by the Haber Process according to the
following reaction:
3H2(g) + N2(g) → 2NH3(g)
(a) What is the maximum amount of ammonia in grams that can be
synthesized from 25.2g of N2(g) and 8.42g of H2 (g)?
(b) What is the maximum amount of ammonia in grams that can be
synthesized from 5.22g of H2(g) and 31.5 of N2(g)?
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2.
Consider the following reaction:
Cu2O(s) + C(s) → 2Cu(s) + CO2(g)
When 11.5 g of C are allowed to react with 114.5 g of Cu2O(s), 87.4 g of Cu
are obtained. Find the limiting reagent, theoretical yield and percent yield.
COMBUSTION REACTIONS
A combustion reaction is one in which the elements in a compound react with
molecular oxygen to form the oxides of those elements. For example C in a
carbon-containing compound will be converted to CO2 and if there is
hydrogen it will be converted to H2O
Notice that by accurately measuring the mass of CO2 obtained by combustion
of the carbon-containing compound, the mass of carbon in the original sample
can be calculated.
Similarly, by measuring the mass of H2O formed in the reaction, the mass of
hydrogen in the original sample can be calculated. These calculations assume
that all the carbon in the sample is captured in the CO2 and that all the
hydrogen is captured in the H2O
EXAMPLE 1: Combustion reaction involving C,H and O only
Vitamin C is a compound that contains the elements C. H and O. Complete
combustion of a sample of mass 0.2000 g of vitamin C produced 0.2998 g of
CO2 and 0.08185 g of H2O. Determine the empirical formula of vitamin C.
-1
Molar masses (in g mol ) :
CO2 = 44.01
H2O = 18.02 ; C = 12.01 H = 1.008 O = 16.00
CxHyOz
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+ O2 →
x
CO2
+
y
H2O
Page 9
2
All the C is converted to CO2 and all the H is converted to H2O.
Solution:
Step1:
Determine the mass of C in CO2
x 0.2998 g CO2 = 0.08181 g of C
12.01 g mol−
144.01 g mol−1
Step 2:
Calculate the mass of water hydrogen in water
2 x 1.008 g mol−
118.02 g mol−1
x 0.08185 g H2O = 0.009157g of H
(note: there are 2 mols of hydrogen in water)
Step 3: Calculate the mass of O which is obtained by difference
Mass of O = Mass of sample – (mass of C + mass of H)
= 0.2000 g – (0.08181 g + 0.00915 g)
= 0.1090 g of O
Step 4: Convert to moles
mass
Atomic mass
C
Moles
0.006812
H
0.009084
O
0.006814
÷ by smallest number of moles
0.006812
0.006812
1
0.009084
0.006812
1.33
0.006814
0.006812
1
convert to whole numbers by multiplying by 3
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1x3
3
1.33 x 3
1x3
4
3
Therefore the empirical formula of vitamin C is C3H4O3
EXAMPLE 2.
Combustion reaction involving C,H,O and N only
The compound caffeine contains the elements C, H, N and O. Combustion analysis
of a 1.500 g sample of caffeine produces 2.737g of CO2 and 0.6814 g of H2O. A
separate further analysis of another sample of mass 2.500 g of caffeine produces
0.8677 g of NH3 .
Determine the empirical formula of caffeine.
-1
Molar masses (in g mol ): H = 1.008; C = 12.01; N = 14.0 ; O = 16.00 ;
CO2 = 44.01; H2O = 18.0 ; NH3 = 17.04.
Important features of this problem are that:
1.
You are analysing for 4 elements – C, H, N and O.
2.
The analysis is performed on two samples (for example sample 1 and 2)
which have different masses.
NOTE:
In the one sample you analyse for carbon and hydrogen.
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In the other you analyse for nitrogen. Remember that the oxygen is always
obtained by difference.
Solution:
In sample 1 - 1.500g of caffeine
Step1:
Determine the mass of C in CO2
12.01 g mol−
x 2.737 g CO2 =
0.7469 g of C
144.01 g mol−
Step 2:
Calculate the mass of water hydrogen in water
2 x 1.008 g mol− x 0.06814 g H2O = 0.07623 g of H
118.02 g mol−1
In sample 2 - 2.500 g of caffeine
Step 3.
Determine the mass of N in NH3
14.01 g mol−1
x 0.8677 g NH3 = 0.7134 g of N
117.04 g mol−
Step 4:
Calculate % N in sample 2 = % N in sample 1
Note: The % composition of a pure compound is constant. If you know the percentage of N in
sample 1 then the percentage of nitrogen in sample 2 is exactly the same.
Therefore % N in sample 1 = 0.7134 x 100 = 28.54 %
2.500 g
Step 5. You must determine the masses of all the elements in the compound in a
common sample therefore find the mass of N in sample 1
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i.e.
Step 6.
28.54 g
100 g
X 1.500 g = 0.4280 g mass of N
The mass of O in sample 1 is obtained by difference
Mass of O = Mass of sample 1 – (mass of C + mass of H + mass of N)
= 1.500 g – (0.7469 g + 0.07623 g + 0.4280 g)
= 0.2489 g
Determine the empirical formula
C
0.7469
H
0.07623
N
0.4280
O
0.2489g
Atomic mass (g mol )
12.01
1.008
14.01
16.00
moles:
0.06219
0.07563
0.03057
0.01555
Mass (g)
-1
Divide by the smallest number of mol = 0.01555
4
4.9
2
1
~5
∴ Empirical formula is = C4H5N2O
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PRACTICE EXERCISES
1.
An 0.1888g sample of a hydrocarbon produces 0.6260g of CO2 and
0.1602g of H2O in combustion analysis. Its molecular weight is found to
be 106 u.
For this hydrocarbon determine
2.
(a) its mass percent composition
Ans. 90.47% C. 9.50% H
(b) its empirical formula
Ans. C4H5
(c) its molecular formula
Ans. C8H10
Dimethylhydrazine is a carbon–hydrogen-nitrogen containing
compound. Combustion analysis of a 0.312g sample of the compound
produces 0.458 g of CO2. From a separate 0.525 g sample the nitrogen
content is converted to 0.244 g N2 .
(a) What is the empirical formula of dimethylhydrazine?
Ans. CH4N
(b) If the molecular weight was found to be 60.02 what is the molecular
formula of the compound?
Ans. C2H8N2
3. A 1.35 g sample of a substance containing C. H. N and O is burned in air
to produce 0.810 g of H2O and 1.32 g of CO2. In a separate analysis of the
same substance all of the N in a sample of mass 0.735 g is converted to
0.284g NH3. Determine the empirical formula of the substance.
Ans. CH3NO
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