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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.
Atomic Mass Constant 1 amu = 𝑚(12
C)
12 = 1.660 538 × 10−27 kg
Relative Atomic Mass 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.
It is defined by the means of Atomic Mass Constant – 𝐴𝑟 (Li) = 𝑚(Li)
amu
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)
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) = 14.0067; 𝐴𝑟 (H) = 1.008)
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) = 32.065; 𝐴𝑟 (O) = 16.00; 𝐴𝑟 (H) = 1.008
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) = 1.008; 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 this case we have to multiply the coefficients by three to obtain the whole numbers.
The stoichiometric formula of the compound is C9H13Br3
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Chemical equation based calculations
There is chemical reaction of compounds A and B, compounds C and D arise
a A + b B c C + d D,
the letters a, b, c, and d represents molar coefficients of particular compounds in the chemical
equation. Based on the equation is possible to calculate masses of all particular compounds
in the reaction using general formulas
𝑚(A) = 𝑎 ·𝑛 · 𝑀𝑟 (A), 𝑚(B) = 𝑏 ·𝑛 · 𝑀𝑟 (B), 𝑚(C) = 𝑐 ·𝑛 · 𝑀𝑟 (C), 𝑚(D) = 𝑑 ·𝑛 · 𝑀𝑟 (D)
where 𝑛 is amount of substance of one equivalent in the particular preparation. To be able
to calculate the masses is necessary to know mass of at least one compound.
Exercises
1. Calculate number of atoms, which contains 1 mg of pure silver.
𝐴𝑟 (Ag) = 107.87
2. 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?
3. Calculate relative atomic mass of a gas, if 25 litres of the gas weights 31.258 g at
standard conditions.
4. Calculate content of vanadium and oxygen (in mass %) in the vanadium(V) oxide.
𝐴𝑟 (V) = 50.94; 𝐴𝑟 (O) = 15.9994
5. Calculate content of all particular elements as mass percents in the compound, which
molecular formula is C23H29N3O4. For the result use four decimal places.
𝐴𝑟 (C) = 12.01; 𝐴𝑟 (H) = 1.008; 𝐴𝑟 (N) = 14.0067; 𝐴𝑟 (O) = 15.9994
6. 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
7. Determine stoichiometric formula of a compound, which contains 11.9847 % of aluminium,
2.6866 % of hydrogen, 63.9635 % of oxygen and 21.3653 % of sulphur.
𝐴𝑟 (Al) = 26.98; 𝐴𝑟 (H) = 1.008; 𝐴𝑟 (O) = 15.9994, 𝐴𝑟 (S) = 32.065
8. 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 stoichiometric and summary formula of the
compound of 𝑀𝑟 = 60.
𝐴𝑟 (H) = 1.008; 𝐴𝑟 (C) = 12.01; 𝐴𝑟 (O) = 15.9994
9. 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
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10. Calculate amount of water, which can be evaporated from 250 g of sodium carbonate
decahydrate.
𝑀𝑟 (Na2CO3 ·10 H2O) = 286.141; 𝑀𝑟 (Na2CO3) = 105.99
11. 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
12. 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) = 14.0067; 𝐴𝑟 (O) = 15.9994
13. 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
14. 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) = 65.38; 𝑀𝑟 (H2SO4) = 98.08; 𝑀𝑟 (ZnSO4 ·7 H2O) = 287.55
15. 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
16. Iodine can be prepared by the reaction of potassium iodide with potassium chlorate
and sulfuric acid. Calculate masses of potassium iodide, potassium chlorate and volume
of 15% sulfuric acid solution (𝜌 = 1.11 g · cm−3) needed for preparation of 25 g of
iodine. Assume 75% yield of the process.
6 KI + KClO3 + 3 H2SO4 KCl + 3 I2 + 3 K2SO4 + 3 H2O
𝑀𝑟 (KI) = 166.00; 𝑀𝑟 (KClO3) = 122.55; 𝑀𝑟 (H2SO4) = 98.08; 𝐴𝑟 (I) = 126.904
17. 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
18. 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 (KOH) = 56.11
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Results
1. 9.27 × 10−6 mol; 5.583 × 1018 atoms of Ag
2. 𝐴𝑟 = 195.092; platinum
3. 28.02 is relative molecular mass (N2) 14.0098, nitrogen
4. 𝑀𝑟 (V2O5) = 181.877; 56.01588 % of vanadium; 43.9841 % of oxygen
5. 𝑀𝑟 (C23H29N3O4) = 411.4797; 67.13089 % of carbon; 7.10412 % of hydrogen; 10.21195 %
of nitrogen; 15.55304 % of oxygen
6. Fe3O4
7. Al2H12O18S3
8. C2H4O2
9. 48.2004 % of iron; 433.804 kg of iron
10. 157.397 g of water
11. FeCl2 ·4 H2O
12. N2O3; dinitrogen trioxide
13. 504.27 kg of CaO; 201.51 m3 of CO2
14. 87.963 g of ZnSO4 ·7 H2O; 6.855 L of H2
15. 1829.45 mL of 10% H2SO4
16. 43.603 g of KI; 5.365 g of KClO3; 77.36 mL of 15% H2SO4
17. 500.76 g of 10% KOH; 55.44 g of KCl; 18.23 g of KClO3
18. 14.98 L of Cl2