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Stoichiometry and related calculations
Basic constants
Avogadro’s Number NA = 6.022 14 × 1023 represents the number of atoms, molecules, or
ions in one mole of a substance.
Atomic Mass Constant 1 amu = 1.660 538 × 10−27 kg – the unified atomic mass unit or
dalton (symbol: Da) is the standard unit that is used for indicating mass on an atomic
or molecular scale (atomic mass). One unified atomic mass unit is approximately the
mass of one nucleon (either a single proton or neutron) and is numerically equivalent to
1 g/mol.
Molar volume Vm = 22.413 83 dm3 · mol−1 – the volume, occupied by a 1 mol of any gas
at the Standard Temperature and Pressure (273.15 K and 101.325 kPa)
Basic formulas
m = n · Mr n =
m
Mr
m = ρ · V V =
m
ρ
Vgas = n · Vm n =
Vgas
Vm
where: m is mass; Mr is relative molecular weight; n is the mass of the substance (mol);
V is the volume of a liquid; ρ is relative density (usually in g · cm−3); Vgas is volume of
a gas; Vm is molar volume (dm3 · mol−1)
Avogadro’s Number
1. The atom of an element has mass of 3.2395 × 10−25 kg. Calculate its relative atomic
mass Ar. What is the name of the element?
2. Calculate relative atomic mass of nitrogen, if 25 litres of nitrogen weights 31.258 g at
standard conditions
Elemental analysis
Example 1. Calculate molar and mass ratio of elements in ammonia
Ammonia – NH3 – Molar ratio of nitrogen and hydrogen in the molecule is 1 : 3. Based
on this and the relative atomic masses we can calculate the mass ratio of particular elements.
(Ar(N) = 14.01; Ar(H) = 1.0079)
The mass ratio can be calculated as Ar(N) : 3 × Ar(H) = 14.01 : 3.0237. We usually
calculate mass composition of a compound in mass percents.
w%(element) =
Ar(element) × n
Mr(compound)
× 100 where n is a molar coefficient
In case of the ammonia it is:
w%N =
14.01
17.04
× 100 = 82.25 % w%H =
1.0079 × 3
17.04
× 100 = 17.75 %
Ammonia contains 82.25 % of nitrogen and 17.75 % of hydrogen.
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Example 2. We have a compound, which contains potassium, aluminium, sulfur, oxygen and
hydrogen. The composition of the compound is showed below.
K 8.2418 %
Al 5.6877 %
S 13.5190 %
O 67.4530 %
H 5.0993 %
Ar(K) = 39.10; Ar(Al) = 26.98; Ar(S) = 32.07; Ar(O) = 15.9994; Ar(H) = 1.0079
Calculate its stoichiometric formula.
We need calculate molar ratio of particular elements For this purpose, we can assume,
that we have 100 g of a compound.
n =
m
Ar
n(K) =
8.2418
39.098
= 0.2107985 mol
n(Al) =
5.6877
26.982
= 0.210796 mol n(S) =
13.519
32.065
= 0.421612 mol
n(O) =
67.453
15.9994
= 4.21597 mol n(H) =
5.0993
1.0079
= 5.05933 mol
As the next step divide all obtained amounts of substances by the lowest number. We
obtain:
K 1.0
Al 1.0
S 2.0
O 20.0
H 24.0
The stoichiometric formula of the compound is KAlS2O20H24
Example 3. We have a compound, which contains carbon, hydrogen and bromine. The
composition of the compound is showed below.
C 29.9509 %
H 3.6306 %
Br 66.4185 %
Ar(C) = 12.01; Ar(H) = 1.0079; Ar(Br) = 79.904
Calculate its stoichiometric formula.
According the procedure, described previously we obtain the molar ratios of the particular
elements
C 3.000
H 4.333
Br 1.000
In thic case we have to multiply the coeficients to obtain the whole numbers. The stoichiometric
formula of the compound is C9H13Br3
Tasks
1. Calculate content of vanadium and oxygen (in mass %) in the vanadium(V) oxide.
Ar(V) = 50.94; Ar(O) = 15.9994
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2. Determine stoichiometric formula of an oxide, which contains 72.3591 % of iron and
27.6409 % of oxygen.
Ar(Fe) = 55.845; Ar(O) = 15.9994
3. Determine stoichiometric formula of a compound, which contains 25 % of hydrogen and
75 % of carbon.
Ar(H) = 1.0079; Ar(C) = 12.01
4. In an analysator, 21 mg of an organic compound was burned and 30.75 mg of CO2 and
12.60 mg of H2O arise. Determine stochimetric and summary formula of the compound
of Mr = 60.
Ar(H) = 1.0079; Ar(C) = 12.01; Ar(O) = 15.9994
5. Calculate content of the iron (in mass %) in FeCO3. Calculate mass of iron, which can
be obtained from 1000 kg of the compound with 10 % of impurities.
Ar(Fe) = 55.845; Mr(FeCO3) = 115.86
6. Calculate amount of water, which can be evaporated from 250 g of sodium carbonate
decahydrate.
Mr(Na2CO3 ·10 H2O) = 286.141
7. A hydrate of iron(II) chloride was dried to constant mass. The starting mass of the
hydrate was 25 g, the final mass was 15.9385 g. Determine composition of this hydrate.
Mr(FeCl2) = 126.751; Mr(H2O) = 18.016
8. In 15 g of an oxide of nitrogen was found 5.52811 g of nitrogen. Determine stoichiometric
formula of the oxide and write the name of this compound.
Ar(N) = 14.01; Ar(O) = 15.9994
Calculation based on chemical equations
1. Calculate mass of calcium oxide and volume of carbon dioxide (at the standard conditions),
which can be obtained by thermal decomposition of 900 kg of pure calcium
carbonate.
Mr(CaCO3) = 100.09; Mr(CaO) = 56.08; Mr(CO2) = 44.01
CaCO3 CaO + CO2
2. Zinc reacts with a diluted sulfuric acid and gives zinc(II) sulfate and hydrogen.
Zn + H2SO4 ZnSO4 + H2
Calculate mass of Zinc(II) sulfate heptahydrate, which arises from 20 g of the zinc.
Calculate volume of hydrogen (at the standard conditions), arising by this reaction.
Ar(Zn) = 65.39; Mr(H2SO4) = 98.08; Mr(ZnSO4 ·7 H2O) = 287.55
3. Based on the previous reaction calculate volume of 10% sulfuric acid, needed for reaction
with 130 g of the zinc.
Relative density of the 10% solution of H2SO4 ρ = 1.066 g · cm−3
4. The gaseous ammonia reacts with 500 g of 15% solution of sulfuric acid and ammonium
sulfate was obtained. Calculate volume (at the standard conditions) of gaseous ammonia,
needed for this reaction.
2 NH3 + H2SO4 (NH4)2SO4
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Calculate volumes of ammonia (at the standard conditions) and the sulfuric acid solu-
tion
(ρ = 1.11 g · cm−3), needed for preparation of 60 g of the ammonium sulfate.
Mr(H2SO4) = 98.08; Mr(NH3) = 17.03; Mr((NH4)2SO4) = 132.14
5. By reaction of gaseous chlorine with aqueous solution of potassium hydroxide arise
potassium chloride and potassium chlorate. Calculate mass of the 10% solution of
KOH, needed for reaction with 10 L of chlorine (at standard conditions). Calculate
mass of both salts, arising by this reaction.
3 Cl2 + 6 KOH 5 KCl + KClO3 + 3 H2O
Mr(Cl2) = 70.90; Mr(KOH) = 56.11; Mr(KCl) = 74.55; Mr(KClO3) = 122.55
6. Based on the previous reaction calculate volume of chlorine (at the standard conditions),
which can react with 500 g of 15% solution of KOH.
Mr(H2SO4) = 98.08