Maris Stella High School
Mole Concept & Chemical Calculation
Name : ________________________(
)
Class: 3
Date : _______
Topic 7.2 : The Mole Concept & Chemical Calculation From Equation
Syllabus Objectives :
(a) Define relative atomic mass, A
r
(b) Define relative molecular mass, M and calculate relative molecular mass
r,
(c)
(d)
(e)
(g)
(and relative formula mass) as the sum of relative atomic masses
Calculate the percentage mass of an element in a compound when given
appropriate information.
Calculate number of moles of atoms and molecules and molar mass
Calculate molar gas volume
Calculate stoichiometric reacting masses and volumes of gases (one mole
3
of gas occupies 24 dm at room temperature and pressure); calculations
involving the idea of limiting reactants may be set .
(The gas laws and the calculations of gaseous volumes at different temperatures
and pressures are not required.)
3
3
(h) Apply the concept of solution concentration (in mol/dm or g/dm ) to process
The results of volumetric experiments and to solve simple problems.
(Appropriate guidance will be provided where unfamiliar reactions such as
redox are involved.
(Calculations on % yield and % purity are NOT required. )
First 40 years after Atomic Theory (1820 –1850), there was chaos in Chemistry.
Chemists were groping about trying to solve the problem of comparing the masses of
atoms, they were making little progress. It took 100 years to find the mass of a
hydrogen atom.
Mass of one hydrogen atom = 0.000 000 000 000 000 000 000 00167g
Atoms are so small that it seems inconvenient to express their absolute masses in
grams. Therefore, it is preferable to express their atomic masses by comparing theirs
with that of a standard atom.
It was then agreed to take Carbon-12, the commonest isotope of carbon, as a
standard, and compare the masses of other atoms with it. The carbon-12 atom is
taken to have a mass of exactly 12 units.
1.
Relative Atomic Mass
Definition:
The relative atomic mass, Ar , of an element is defined as the average mass of
an atom of the element compared with 1 of the mass of an atom of carbon-12.
12
Note: * Symbol for the relative atomic mass is Ar and Ar has no unit.
The values of the Ar of atoms of the elements are given in the Periodic Table.
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Relationship between Nucleon number and Relative atomic mass, Ar
♣ The mass of an atom is due mainly to the total number of protons and neutrons in
its nucleus (Nucleon number). Thus, we would expect the Ar of the element to be
the same as its nucleon number. This is true for elements which have no isotopes.
♣ For most elements, the nucleon number and the Ar are not the same due to the
existence of isotopes.
♣ The nucleon number is always a whole number (since we can’t have fractions of
protons or neutrons in the nucleus), but the Ar values may not be whole numbers.
If the Ar is not a whole number and is different from nucleon number, we can
conclude that the element has naturally occurring isotopes.
Question 1
Chlorine has a relative atomic mass of 35.5. How can we get a mass number that is not
a whole number?
Reason: Most naturally occurring samples of an element contain a mixture of isotopes.
A normal sample of chlorine contains 75% of chlorine-35 and 25% of chlorine- 37.
Thus, the average mass of a chlorine atom = [(75/100 x 35) + (25/100 x 37) ] = 35.5 ,
which is not a whole number.
Question 2 [ “O” Level, Nov 2001 Sect. A 3b(ii) ]
10
11
Naturally occurring boron contains atoms represented by the symbols 5 B and 5 B .
Suggest why the relative mass of naturally occurring boron is not a whole number.
Reason : This is because there are different isotopes of boron. The relative atomic
mass is the average of the atomic mass of all the different isotopes of boron.
Relative Molecular Mass
♣ Many elements and compounds exist as molecules.
♣ The mass of a molecule is measured in terms of its relative molecular mass, Mr
Definition: The relative molecular mass, Mr, of a substance is defined as the
average mass of a molecule of the substance compared with 1 of
the mass of a carbon-12 atom.
12
♣ Relative molecular mass is often abbreviated to molecular mass.
♣ Symbol is Mr and Mr has no units.
♣ It is calculated as the sum of the atomic masses of all the atoms in the formula.
Example 1 : Ar of Oxygen (O) is 16, thus Mr of Oxygen (O2) = 16 x 2 = 32
Example 2 : Ar of Hydrogen(H) is 1; Ar of Oxygen(O) is16,
Thus, Mr of H2O = (2x1) + 16 = 18
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Questions
1. Calculate the relative molecular mass of the following:
(a) Carbon dioxide (CO2 )
(b)
(d)
Sugar (C12H22O11)
(c)
Fullarene ( C60 )
Ethanoic acid ( CH3COOH)
2.
Define relative atomic mass. [ Nov 2001 B6a ]
3.
Define relative molecular mass. [ Nov‘04 A3a, Nov’06 BQ10, Nov’09 A3b,
Nov 2010,B11b(ii) ]
Relative Formula Mass
♣ Compounds can also exist as ions, such as ionic compounds , sodium chloride
containing Na+ and Cl− ions and silver chloride containing Ag+ and Cl− ions.
♣ Since ionic compounds do not contain molecules, the sum of the Ar of the atoms in
the formula is called the relative formula mass (also given the symbol Mr ). The
calculation of relative formula mass is the same as above, since the charge does not
affect the overall mass.
Example 1 : Ar of Na = 23 and Ar of Cl = 35.5
Thus, relative formula mass/Mr of NaCl = 23 + 35.5 = 58.5
Example 2 : Ar of Ag = 108 and Ar of Cl = 35.5
Thus Mr of AgCl = 108 + 35.5 = 143.5
Questions
Calculate the relative molecular mass of the following:
(a) Zinc chloride (ZnCl2 )
(b) Aluminium sulfate ( Al2(SO4)3 )
(c) Lead(II) nitrate (Pb(NO3)2)
(f) Hydrated copper(II) sulfate(CuSO4.5H2O )
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2.
Mole Concept & Chemical Calculation
Calculating percentage by mass composition of element in compound
2.1 The percentage composition of a compound can be calculated from
• its formula, and
• the relative atomic masses of its elements
The percentage composition
of an element in a compound
=
No. of atoms of the element x Ar x 100%
Mr of the compound
Example 1
Calculate the percentage composition of (a) sulfur; (b) oxygen, in sulfuric acid.
Mr of H2SO4 = ( 2 x Ar of H ) + Ar of S + ( 4 x Ar of O)
= (2x1) + (32 x 1) + (16 x 4 ) = 2 + 32 + 64 = 98
(a) % of S in H2SO4
= 1 x Ar of S
x 100 %
Mr of H2SO4
= 32 x 100 %
98
= 32.7 %
(b) % of O in H2SO4
= 4 x Ar of O x 100%
Mr of H2SO4
= 64 x 100 %
98
= 65.3%
Example 2
(a)
Calculate the percentage by mass of sodium in Na2CO3.10H2O.
[Ar of Na = 23, Ar of O = 16, Ar of C = 12, Ar of H = 1, Mr of H2O = (2 + 16) ]
Mr of Na2CO3.10H2O = (23x2) + (12x1) + (16x3 ) + 10(2 + 16)
= 46 + 12 +
48 + 180
= 286
% of Na = 2 x 23 x 100
286
=16.083
= 16.1% (3 sig fig)
(b)
Calculate the percentage by mass of water of crystallization in Na2CO3.10H2O.
% of H2O
=
10 x 18 x 100
286
= 62.9 %
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Question
1.
Calculate the percentage by mass of germanium in germanium (IV) oxide,GeO2.
[ “O” Level 2005 A5c(i)]
2.
Calculate the percentage by mass of carbon in ethanol, C2H5OH. [Nov ’08 B11c]
2.2 Calculating the mass of an Element in a Compound
The mass of an element = No. of atoms x Ar of the element x mass of sample
in a compound
Mr of compound
Example: Calculate the mass of oxygen in 10g of sodium carbonate crystals,
Na2CO3.10H2O.
Solution : Mass of oxygen = 13 x 16 x 10 g
286
= 7.27g (3 sig. fig.)
Questions
1.
Calculate the mass of water in 10g of sodium carbonate crystals, Na2CO3.10H2O.
2.
What is the mass of aluminium in 204 g of aluminium oxide, Al2O3 ?
3.
An ore contains 32% iron(III) oxide(Fe2O3). Calculate the mass of iron(Fe) in
1000g of the ore.
[Nov 1998 P3 B8b]
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3.
The Mole Concept
3.1
Avogadro number (Avogadro constant)
The mole is the standard method in chemistry for communicating how much of
a substance is present.
A ‘mole’ is the amount of substance which contains 6x1023 of particles of that
substance. This number, 6 x 1023 , is known as the Avogadro number.
The particles may be atoms, molecules, ions or electrons.
e.g 1 mole of copper contains 6x1023 copper atoms.
1 mole of carbon dioxide contains 6x1023 CO2 molecules.
1 mole of sodium ions contains 6x1023 sodium ions.
1 mole of electrons contains 6x1023 electrons.
No. of moles of substance = no. of particles of substance
6 x 1023
3.2
Molar Mass
Molar mass is the mass of one mole of any substance. It is the relative
atomic mass (Ar) or relative molecular mass (Mr) expressed in grams.
(i) Molar Mass of Atoms
e.g. One mole of carbon atoms has a mass of 12 g.
No. of atoms in 1 mole = 6 x 1023
e.g. One mole of aluminium atoms has a mass of 27 g.
No. of atoms in 1 mole = 6 x 1023
(Ar of C = 12)
(Ar of Al = 27)
No. of moles of atoms in = mass of the sample (g)
a sample of an element
molar mass of element (g/mol) or
simply,
No. of moles
=
mass
Ar
Note: all masses measured are in grams.
E.g. 1
No of moles in 46g sodium = 46
23
= 2
(Ar of sodium = 23)
E.g. 2
Mass of 4 moles of calcium = 4 x 40 g = 160g
(Ar of calcium = 40)
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(ii)
Mole Concept & Chemical Calculation
Molar mass of molecules
E.g. One mole of water has a mass of 18g.
[since Mr of H2O = (2x1)+ 16 = 18 ]
No. of water molecules in one mole is 6 x 1023 .
E.g. One mole of carbon dioxide has a mass of 44g
[ since Mr of CO2 = 12 + (16x2) = 44 ]
No. of carbon dioxide molecules in one mole is 6 x 1023
No. of moles of molecules in = mass of sample (g)
a sample of a substance
molar mass of substance(g/mol)
or simply ,
No. of moles = mass
Mr
E.g. 1
No. of moles in 80g oxygen = 80 = 2.5
32
( since Mr of O2 = 16x2 = 32)
E.g. 2
Mass of 2 moles of hydrogen = 2 x 2 g = 4 g
E.g. 3
No. of Hydrogen molecules in 2 moles = 2 x 6 x 1023 = 1.2 x 1024
( since Mr of H2 = 2 x1 = 2)
Questions :
1. Calculate the mass of one mole (molar mass) of the following substances :
2.
a) Cl2 : __________________
b) C3H8 : _________________
c)
d) Zn(NO3)2 : _________________
Na2CO3 : _________________
Calculate the mass of the following :
(a) 2 moles of oxygen atoms
3
(b) 4 moles of iron(III) oxide
How many chlorine molecules are there in 35.5 g of gaseous chlorine?
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4. Molar Volume of a Gas
♣ Avogadro’s Law : Equal volumes of all gases under the same conditions of
temperature and pressure contain the same number of molecules.
Hence, 1 mole of any gas always has the same volume.
♣ One mole of all gases at room temperature and pressure (r.t.p)
occupies a volume of 24 dm3, known as the molar volume.
i.e. The molar volume for all gases at r.t.p = 24 dm3 or 24000 cm3
No. of moles of gas = Volume of gas in dm3 =
Molar volume (dm3/mol)
Volume
24
Since 1 dm3 is 1000 cm3 , we can also calculate no. of moles of gas as below,
if volume of gas is given in cm3 :
No. of moles of gas = Volume
24000
E.g.1
(when volume of gas is in cm3)
Calculate the number of moles of CO2 when 12 dm3 of it was evolved in a
reaction.
Solution : No. of moles of CO2 = 12 =
24
E.g. 2
0.5
Calculate the mass of oxygen if 4 dm3 of the gas was liberated at r.t.p.
Solution : No. of moles of O2 = 4 = 1 (not final answer, can be in fraction)
24 6
Mass of O2 = mole x Mr = 1 x 32 g = 5.33g ( final ans in 3 s.f. )
6
Questions
A sample of carbon monoxide has a volume of 6 dm3 at r.t.p. Calculate
(a)
the no. of moles of gas in the sample,
(b)
the no. of gas molecules in the sample,
(c)
the mass of the sample.
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5.
Mole Concept & Chemical Calculation
Concentration of Solutions
The concentration of a solution tells you the amount of solute in 1 dm3 of solution.
It can be measured in mol/dm3 or in g/dm3.
Note : 1 dm3 = 1 litre = 1000 cm3
Mol/dm3 is commonly used because it is convenient to express concentrations of
substances in it. It is sometimes referred as “Molarity” of a solution represented
by the symbol “M”.
Molarity, M , means
mol/dm3
A molar solution (1M ) or 1 mol/dm3, contains 1 mole of the substance in 1 dm3 of
the solution .
2 M solution or 2 mol/dm3, contains 2 moles of the substance in 1 dm3 of solution.
0.05 M solution or 0.05 mol/dm3, contains 0.05 moles of the substance in 1 dm3
of solution.
In summary :
(a)
Concentration in mol/dm3 = No of moles of solute
Volume of solution in dm3
Hence, in a volume of a solution :
No. of moles = volume of solution(dm3) x concentration in mol/dm3
(b)
Concentration in g/dm3 =
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Volume of solution in dm3
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Questions
1.
Complete the following table :
Solution
No. of moles
of substance
in a volume
of the
solution
Mass of
substance in
a volume of
the solution
A volume of
the solution
Sodium
hydroxide
(NaOH)
0.2
0.2 x (23
+16 + 1)
= 8g
50 cm3
= 50
1000
= 0.05 dm3
1 dm3
Sulfuric
acid
(H2SO4)
4.9g
Nitric acid
(HNO3)
2 x 25
1000
= 0.05
Potassium
hydroxide
(KOH)
Concentration
of solution in
mol/dm3
= no. of mole
vol of solution
Concentration
of solution in
g/dm3
= mass
vol of solution
4.9
1
= 4.9g/dm3
25 cm3
2.0 mol/dm3
500 cm3
0.4 mol/dm3
3
3
(Apply The Mole Concept : No. of mole = concentration in mol/dm x vol in dm ,
No of mole = mass/Mr and Mass = No. of mole x Mr )
2(a) Calculate the relative molecular mass of sodium hydroxide, NaOH.
(b) Calculate the mass of sodium hydroxide in
(i) 1000 cm3 of 0.2 mol/dm3 sodium hydroxide solution,
(ii)
3.
20 cm3 of 2.0 mol/dm3 sodium hydroxide solution.
[Nov 2002P3A3]
Calculate the mass of sodium carbonate required to make 250cm3 of a
2.0 mol/dm3 solution of sodium carbonate.
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6.
Mole Concept & Chemical Calculation
Summary Exercise on The Mole Concept
Carry out the following calculations, starting with the sample of carbon dioxide,
CO2, with a mass of 11g. Write your answers and working in the boxes.
(Source of Question: Science in Focus- Chemistry for GCE ‘O’ Level Theory
Workbook by JGR Briggs ,Longman)
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7. Chemical Calculations From Equations
Three Basic Methods to do chemical calculations:
Method 1 - Using mole ratio (when masses in g, volume of gas, concentration are given)
Method 2 - Using mass ratio ( where masses are not given in grams, but, in kg or tonnes)
Method 3 - Using volume ratio (if only volumes of gases are involved)
Method 1 – Using Mole Ratio
Steps involved in working out the required calculation:
1. Write the balanced equation if not given in question.
2. Get mole ratio from equation
3. Convert information given (whether in mass, volume or concentration of
substance) in the question into moles
4. Use the mole ratio and work out by proportion the no. of moles of unknown
5. Convert no of moles to mass or volume or calculate concentration of
solution as required by question
Eg 1. What mass of magnesium oxide can be obtained from the combustion of 2.4 g
of magnesium?
Chemical Equation: 2Mg + O2 → 2MgO
No. of moles in 2.4g Mg = mass
Ar
(write it if not given)
= 2.4 = 0.1 ( Ar used, Mg made up of atoms)
24
From equation, 2 mol Mg produces 2 mol MgO
⇒ mole ratio is 2 mol Mg : 2 mol MgO
Hence, 0.1 mol Mg produces 0.1 x 2
2
= 0.1mol MgO
(Qn requires calculation of these)
(Using proportion from mole ratio)*
Mr of MgO = 24 + 16 = 40 ( Always show calculation of Mr value)
∴ Mass of 0.1 mol MgO produced = No. of moles x Mr g
= 0.1 x 40 g
= 4g
================================================================
*Some students may find it easier writing it as No.of moles of Mg
=
2
No. of moles of MgO
2
Substitute no. of moles of Mg , we have
0.1
= 2
No. of moles of MgO
2
Then, cross multiply to get no. of moles of MgO = 0.1 x 2 = 0.1
2
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Eg 2. Copper(II) oxide reacts with carbon to form copper and carbon dioxide.
Calculate the mass of copper(II) oxide used if 12.8g of copper was obtained
in the reaction.
Chemical Equation:
2CuO + C → 2Cu + CO2 (write it if not given)
No. of moles of 12.8g Cu = mass = 12.8 = 0.2 (mass is given in qn)
Ar
64
From equation , 2 mol Cu is obtained from 2 mol CuO
⇒ Mole ratio is 2 mol Cu : 2 mol CuO
* Hence, 0.2 mol Cu is obtained from 0.2 x 2
2
= 0.2 mol CuO
Mr CuO = 64 + 16 = 80 (essential working)
mass of 0.2mol CuO used = No. of moles x Mr g
= 0.2 x 80g
= 16g
* Alternative presentation :
No of moles of Cu = 2
No of moles of CuO
2
Substitute known mole
0.2
= 2
No. of CuO
2
Cross multiply to get No of moles of CuO = 0.2 x2 = 0.2
2
Eg 3. Calcium carbonate decomposes to form carbon dioxide and calcium oxide.
Calculate the volume of carbon dioxide (at r.t.p) produced from the
decomposition of 20g of calcium carbonate.
Chemical Equation :
CaCO3 (s) → CaO (s) + CO2 (g)
Mr CaCO3 = 40 +12 +(16x3) = 100 (essential working)
No. of moles in 20g CaCO3 =
From equation: _________ CaCO3 produces ____________ CO2 gas
⇒ Mole ratio is
Hence, 0.2mol CaCO3 produces
∴
volume of 0.2 mol CO2 =
Hence, 20g calcium carbonate decomposes to produce 4.8dm3 carbon dioxide.
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Questions
Mole Concept & Chemical Calculation
[Calculation from equation using mole ratio method]
(Note : All working must be clearly presented, final answers must be in 3 sig fig)
1. Copper(II) oxide reacts with hydrogen to form copper and water. What mass of
copper would be obtained from 16g of copper(II) oxide?
Chemical Equation :
2. 4.8kg of magnesium reduce copper(II) oxide to copper. What mass of copper is
obtained?
Chemical Equation :
3. In a reaction, silver nitrate decomposed on heating producing silver metal, nitrogen
dioxide and oxygen. Given that 42.5g of silver nitrate completely decomposed when
heated in an experiment.
Chemical Equation :
2AgNO3 →
2Ag
+
2NO2
+
O2
Calculate
(a) the mass in grams of the silver metal formed;
(b) the volume of nitrogen dioxide gas formed at r.t.p.
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4. 50cm3 of a 2M hydrochloric acid was added to excess zinc, producing zinc
chloride and hydrogen gas. Calculate the volume of hydrogen gas liberated at
r.t.p.
5. A solution of sulfuric acid contained 4.9g of H2SO4 per dm3.
20 cm3 of the acid reacted with 24 cm3 of a solution of NaOH. Calculate
concentration of NaOH in (a) mol/dm3 and (b) g/dm3
6. 25 cm3 of aqueous 0.1 mol/dm3 hydrochloric acid exactly neutralizes 20 cm3 of
aqueous sodium hydroxide.
The equation for this reaction is shown.
NaOH + HCl → NaCl +H2O
What is the concentration of the sodium hydroxide solution?
(GCE’O’Nov2009P1Q7)
(Hint : This question involves use of given concentration and volume of solution
to work out the no. of moles, using mole ratio to calculate unknown no. of
moles and then using calculated no. of mole and given volume of
solution to calculate the unknown concentration)
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Mole Concept & Chemical Calculation
Method 2 ( Using Mass Ratio )
Although the gram is the usual unit of mass used in calculations, other units such as the
kilogram or the tonne can be used, especially in industry. In cases where the kg or
tonne is used, it is simpler to use the Mass-Mass ratio method.
Steps:
1. Write the balanced equation if not given in question
2. Get mole ratio from equation
3. Convert mole ratio to mass ratio (use Ar or Mr, applying mass =mole x Ar or
mass = mole x Mr)
4. Use proportion to find unknown mass
E.g. 1. 4.8kg of magnesium reduced copper(II) oxide to copper. What mass of
copper is obtained?
Chemical Equation :
Mg + CuO → Cu + MgO
From equation, 1 mol Mg produces 1 mol Cu
Mole ratio : 1 mol Mg : 1 mol Cu
⇒ Mass ratio : 24 g Mg : 64 g Cu
⇒ 24 kg Mg : 64kg Cu
Explanation:
( Ar of Mg = 24; Ar of Cu=64,
mass of 1 mol Mg = 1 mol x 24 g = 24 g
mass of 1 mol Cu = 1 mol x 64g = 64 g)
Hence, 4.8 kg Mg produces 4.8 x 64 kg Cu
24
= 12.8 kg Cu
E.g. 2. Ammonia is produced from the reaction between hydrogen and nitrogen gas.
Calculate the mass of nitrogen needed to produce 17 tonnes of ammonia.
[ 1 tonne = 1000 kg = 10000 g ]
Chemical Equation :
N2 + 3H2
→
2NH3
Mole ratio is 2 mol of NH3 : 1 mol N2
⇒Mass ratio : (2 x17)g NH3 : (1x 28) g N2
⇒ 34 g NH3 : 28 g N2
⇒34 tonnes : 28 tonnes N2
Explanation:
Mr of NH3 = 14 + (3 x 1) = 17
Mr of N2 = 2 x 14 =28
mass = no. of mole x Mr g
Hence ,17 tonnes NH3 requires (17 x 28 ) = 14 tonnes
34
∴14 tonnes of nitrogen is needed.
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Questions :
[Calculations form equation using the mass ratio method ( when the reacting
masses are not in grams , say , in kg or in tonnes )]
1. Iron is made in the blast furnace by the following reaction:
Fe2O3 + 3CO → 2Fe + 3CO2
Starting from 28 tonnes of iron(III) oxide , how much iron can be made?
2. In industry, wolframite, (FeMn)WO4 ,is changed into tungsten(VI) oxide, WO3.
Metallic tungsten is formed by heating this oxide in hydrogen gas.
WO3 + 3H2 → W + 3H2O
Calculate the mass of tungsten in tonnes that could be formed from 20 tonnes
of tungsten(VI) oxide. [ Ar : H,1; O,16; W,184]
[Nov 2007 P3 5bii]
3. In the manufacture of metal uranium, uranium dioxide is first converted into a
fluoride. Balance this equation for the reaction.
[Nov 2008P3A7]
UO2
+
HF
→
UF4
+
H2 O
Uranium tetrafluoride is then reduced to metal uranium by heating with n
magnesium, according to this balanced chemical equation.
UF4 + 2Mg → 2MgF2 + U
Calculate the mass of magnesium that must be used to manufacture 10 tonnes
of uranium.
[Relative atomic masses: Ar: Mg, 24; U, 238 ]
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Mole Concept & Chemical Calculation
Method 3 – Using volume ratio (For calculations involving volumes of reacting gases,
it is simpler to use the volume-volume ratio method)
Steps:
1. Write the balanced equation if not given in question.
2. Get mole ratio from equation
3. Write down the volume ratio which is the same as the mole
ratio.
4. Use proportion to find volume of gas required
E.g. 1
Under suitable conditions, nitrogen and hydrogen combine to form ammonia.
Calculate the volume of ammonia formed if 100cm3 of nitrogen is reacted with
excess hydrogen. All gas volumes are measured at room temperature and
pressure.
Chemical Equation :
N2 + 3H2 → 2NH3
From equation, mole ratio is 1 mol of N2 : 2 mol NH3
⇒ volume ratio is 1 cm3 of N2 : 2 cm3 of NH3
Explanation:
( cm in ratio is used
as in the question)
3
Hence,100cm3 N2 produces (100 x 2 ) = 200 cm3 NH3
1
E.g. 2
What volume of CO is produced by completely reacting 15 cm3 CO2 with
charcoal?
Chemical Equation : C + CO2 → 2CO
From equation, Mole ratio is
1 mol of CO2 : 2 mol CO
⇒ Volume ratio is 1 cm3 of CO2 : 2 cm3 of CO
∴15 cm3 CO2 produces (15 x 2 ) = 30 cm3 CO
1
E.g. 3
A volume of 1000 dm3 methane is burnt completely in oxygen.
Calculate the volumes of gaseous products when measured
(i) above 100 0C,
(ii) below 100 0 C
[Nov 2003 P3B11b]
Chemical Equation :
CH4 + 2O2 → CO2 + 2H2O
Mole ratio is
1 mol CH4 : 1 mol CO2 : 2 mol H2O
⇒ volume ratio is 1 dm3 CH4 : 1 dm3 CO2 : 2 dm3 H2O
Hence, 1000 dm3 CH4 : 1000 dm3 CO2 : 2000 dm3 H2O
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(i)
Mole Concept & Chemical Calculation
Above 100 0C , gaseous products are CO2 and H2O ,thus total volume of
gaseous product measured is 1000dm3 + 2000 dm3 = 3000 dm3
(ii) Below 1000 C, carbon dioxide is the only gaseous product, thus volume
volume of gaseous product measured is 1000dm3 .
Questions
[Calculation involving volume of gases using volume-volume ratio]
1. Phosphine , PH3, burns in air. The change is represented by the following
equation.
4PH3(g) + 8O2(g) → P4O10 (s) + 6H2O(g)
(a) Calculate the relative molecular mass of phosphorus (V) oxide, P4O10.
(b) What volume of oxygen will be needed to burn completely 48dm3 of phosphine?
(c) What is the mass of 48dm3 of phosphine at room temperature and pressure?
2. 25 cm3 of butane (C4H10) was mixed with an excess of oxygen and exploded.
(All measurements were made at room temperature and pressure.)
Equation : 2C4H10 (g)
Calculate
+ 13O2 (g) →
8CO2 (g) +
10 H2O(g)
(a) the volume of carbon dioxide produced and
(b) the volume of oxygen required for the reaction
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Mole Concept & Chemical Calculation
8. Chemical Calculation involving Limiting and Excess Reactants
E.g.1
Nitrogen,N2 and hydrogen, H2 will react according to the chemical equation :
N2 + 3H2 → 2NH3
If I mole of nitrogen is reacted with 4 moles of hydrogen in the reaction vessel,
(a)
(b)
Which reactant will be in excess?
What is the maximum number of moles of ammonia produced?
Solution:
(a) From equation, Mole Ratio is 1 mol N2 : 3 mol H2
Thus, 1 mol of N2 will react only with 3 mol of H2
Since there is 4 mol of H2 in the reaction vessel, H2 is the reactant in
excess and N2 will be totally used up, it is the limiting reactant.
(b)
The amount of product formed depends on the amount of limiting
reactant used (N2)
From equation, Mole Ratio is 1 mol N2 : 2 mol NH3
Thus, maximum no. of moles of NH3 produced is 2 moles.
E.g. 2
A mixture of 125cm3 oxygen and 50cm3 hydrogen at room temperature is exploded in a
suitable apparatus. After reaction, the apparatus is allowed to cool to room temperature
again. Give the names of the remaining gases and also their volumes.
Chemical Equation : 2H2 + O2 → 2H2O
From equation,
Mole Ratio is 2 mol H2 : 1 mol O2
Volume Ratio: 2 cm3 H2 : 1 cm3 O2
⇒ 50cm3 H2 will react with ( 50 x 1 ) = 25 cm3 O2
2
or 125 cm3 O2 will need (125 x 2)cm3 = 250 cm3 H2 [note : only 50cm3 H2 is given]
so, not possible to totally
react 125cm3 O2]
Thus, hydrogen is the limiting reactant as it will be totally used up in the reaction and
oxygen is the excess reactant . So, the amount of water produced depends on the
amount of limiting reactant, H2.
From equation , Mole ratio is 2 mol H2 : 2 mol H2O
Volume ratio is 2 cm3 H2 : 2 cm3 H2O
∴
50cm3 H2 produces 50cm3 of H2O
3
∴ Thus, 50 cm of water vapour is produced and (125 – 25) cm3 = 100 cm3 of
oxygen remains in excess.
Questions:
Secondary 3 Science Chemistry
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Maris Stella High School
Mole Concept & Chemical Calculation
1. When 7g of iron reacts with 4g of sulfur, 11g of iron(II) sulfide is produced. What
will be produced if 7g of iron reacted with 7g of sulfur?
Chemical equation : Fe + S → FeS
A.
B.
C.
D.
11 g of iron(II) sulfide and 3 g of unchanged iron
11 g of iron(II) sulfide and 3 g of unchanged sulfur
11 g of iron(II) sulfide only
14 g of iron(II) sulfide only
(
)
2. 20 cm3 of oxygen are reacted with 20 cm3 of carbon monoxide. What are the
volumes of the gases remaining, at original temperature and pressure?
Chemical Equation :
2CO + O2 → 2CO2
A
B
C
D
Oxygen/cm3
0
0
10
10
Carbon monoxide/cm3
0
0
0
10
Carbon dioxide/cm3
20
40
20
20
(
)
3. Fe combines with Cl2 according to the equation : 2Fe + 3Cl2→ 2 FeCl3
How many grams of FeCl3 can be obtained by reacting 3 mol of Fe and 5 mol of
Cl2?
4. 200 cm3 of a 0.1 mol/dm3 solution of lead(II) nitrate was mixed with 100 cm3 of a
0.1 mol/dm3 solution of sodium sulfate. The insoluble lead(II) sulfate is precipitated
according to the chemical equation :
Pb(NO3)2 (aq)
+
Na2SO4 (aq)
→ PbSO4(s)
+
2NaNO3(aq)
[precipitate is an insoluble solid formed in a solution]
(i) What is the limiting reactant?
(ii) Calculate the number of moles of lead (II) sulfate precipitated.
(iii) Calculate the mass of lead (II) sulfate precipitated.
Summary of Key Points and Formulae in Mole Concept and Chemical Calculation
Secondary 3 Science Chemistry
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Maris Stella High School
Mole Concept & Chemical Calculation
- Define Relative Atomic Mass (Ar) and Relative Molecular Mass (Mr)
- One mole of a substance has 6 x 1023 particles ( be it ions, atoms , molecules or
electrons)
- Percentage composition
by mass of element
in a compound
- Mass of an element
in a sample of a
compound
= No. of the atoms of the element x Ar x 100 %
Mr of compound
= No. of the atoms of the element x Ar x mass of
Mr of compound
sample
- Molar mass = mass of one mole = Ar g ( for atoms)
- Molar mass = mass of one mole = Mr g ( for molecules )
- Molar volume of a gas is 24 dm3 at r.t.p
Mass
(in g)
No. of
particles
No. of
moles
6 x 1023
Ar
No. of
moles
Volume
(in cm3)
No. of
moles
Volume
(in dm3)
No. of
moles
24000 cm3
Mass
(in g)
24 dm3
No. of
moles
Mr
No. of
moles
Concentration
Mol/dm3
Secondary 3 Science Chemistry
Volume of
solution in dm3
22