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Moles and Equations
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Moles and Equations
ATOMIC SIZE AND ATOMIC MASS
Atomic size
The atom is extremely small and chemists use a unit called the nanometre to measure
such tiny objects.
1 nanometre, nm, is one-thousand-millionth of a metre (0.00000001 m.).
Measuring atomic masses
The mass of an atom is too small to be measured on a balance.
The carbon–12 isotope is chosen as the international standard for the measurement
of atomic mass.
 One atom of carbon–12 is defined as having a mass of exactly 12 u
‘u’ is the ‘unified mass unit’.
1 u has a mass of one-twelfth the mass of one atom of carbon–12.
1 u = 1.661 x 10–24 g
Problem:
What is the mass of one atom of carbon-12?
 A typical sample of carbon contains atoms of the carbon–12 isotope (98.89%), with
small proportions of the heavier isotopes, carbon–13 and carbon–14.
Chemists measure atomic masses as relative masses This is done by comparing the
masses of atoms with the mass of the carbon–12 isotope.
Relative isotopic mass
The relative isotopic mass is the mass of an atom
of an isotope of an element compared to onetwelfth of the mass of the carbon–12 isotope.
The relative isotopic masses of the three carbon isotopes are shown below.
isotope
relative mass
12
exactly 12
13
13.00
14
14.00
C
C
C
The average mass of a carbon atom after taking into account the relative abundance
of each isotope is 12.0111.
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Moles and Equations
Relative atomic mass
Chemists uses relative masses of atoms to compare the masses of the atoms of
different elements.
The relative atomic mass, Ar, of an element is the weighted
mean mass of an atom of the element compared to onetwelfth of the mass of the carbon–12 isotope.
Weighted mean mass
The weighted mean mass takes into account
 the mass of each isotope
 the percentage abundance of each isotope.
Bromine contains 53.0% of bromine-79 and 47.0% of bromine-81.
………
mass from 79Br = …….. x ………
………
mean from 81Br = …….. x ………
relative atomic mass (weighted mean mass) of bromine = ………….
Problems
Calculate the relative atomic mass, Ar, of the following to 4 significant figures.
(a)
Chlorine contains 75.53% 35Cl and 24.47% 37Cl.
(b)
Iridium contains 38.5% 191Ir and 61.5% 193Ir.
(c)
Silicon contains 92.18% 28Si, 4.71% 29Si and 3.12% 30Si.
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Moles and Equations
MORE RELATIVE MASSES
Chemists use relative masses for all chemicals.
Relative molecular mass
This is the relative mass of simple molecular substances, such as chlorine, Cl2, and
water, H2O.
The relative molecular mass, Mr, is the weighted
mean mass of a molecule of a compound compared
to one-twelfth of the mass of the carbon–12 isotope.
The relative molecular mass can easily be calculated by adding together the relative
atomic masses of the atoms within the molecule:
molecule
Mr
Cl2:
35.5 x 2
H2O:
1.0 x 2 + 16.0
= 71.0
= 18.0
Relative formula mass
This is the relative mass of any formula unit. It is especially useful for ionic
compounds, such as sodium chloride, NaCl, compounds that do not exist as simple
molecules.
The relative formula mass is the weighted mean mass
of the formula unit of a compound compared to onetwelfth of the mass of the carbon–12 isotope.
Find a relative formula mass by simply add together the relative atomic masses of
each element in the formula:
compound
relative formula mass
NaCl:
23.0 + 35.5
= 58.5
CaBr2:
40.1 + 79.9 x 2
= 199.9
Some relative questions
Use the relative atomic masses from the periodic table to calculate the relative
formula mass of:
(a) HCl;
(b) CO2;
(c) H2S;
(d) NH3;
(e) Fe2O3;
(e) H2SO4:
(f) HNO3;
(g) NH4Br;
(h) C6H12O6;
(i) Pb(NO3)2.
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Moles and Equations
THE MOLE
Chemists are able to count atoms using the mole concept.
The mole is the S.I. unit of amount of substance and the abbreviation for the mole is
mol.
1 mole is defined as the amount of substance which
contains as many particles as there are carbon atoms
in exactly12 g of the carbon–12 isotope.
The number of atoms in 1 mole of carbon–12 is a number called the Avogadro
constant, NA.
The Avogadro constant is
or
602,000,000,000,000,000,000,000 per mole
6.02 x 1023 mol1.
My name is:
‘Count Lorenzo
Romano
Amedeo Carlo
Avogadro di
Quaregna e
Cerreto’
To the nearest whole number,
12 g of carbon contains 1 mole of atoms (6.02 x 1023 atoms).
1 g of hydrogen contains 1 mole of atoms (6.02 x 1023 atoms).
207 g of lead contains 1 mole of atoms (6.02 x 1023 atoms).
Thus, for carbon,
12 g of carbon contains 1 mole of atoms (6.02 x 1023 atoms).
6 g of carbon contains 0.5 mol. of atoms (3.02 x 1023 atoms).
24 g of carbon contains 2 mol. of atoms (12.04 x 1023 atoms).
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Moles and Equations
Just how big is 1 mole?
The table below shows how the numerical value for Avogadro's Number has been
modified through the years in the quest for more digits and accuracy for the number.
It is a miniature version of the quest for digits for PI: .
Year
Value
1931
6.061 x 1023
1958
6.02 x 1023
1981
6.022045(31) x 1023
1993
6.0221367(36) x 1023
1 mole (6.02 x 1023) standard soft drink cans would cover the surface of the earth to
a depth of over 200 miles.
If you had 1 mole (6.02 x 1023) of popcorn kernels, and spread them across the
United States of America, the country would be covered in popcorn to a depth of over
9 miles.
If we were able to count atoms at the rate of 10 million per second, it would take
about 2 billion years to count the atoms in one mole.
1 mole of marbles would cover
the whole surface of the U.K. to
a depth of 1,500 km
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Moles and Equations
MOLES OF ATOMS
Refer to the periodic table to complete the table below.
Element
Hydrogen
Symbol
Relative
atomic
mass, Ar
1
Number
of
moles
Mass of
element /g
2
2g
Carbon
1
Sulfur
2
Oxygen
3
Calcium
0.5
Sodium
Number of
atoms
12.04 x 1023
46 g
Potassium
6.02 x 1023
Magnesium
3.01 x 1023
Gallium
34.85 g
Strontium
21.9 g
Bromine
18.06 x 1023
Arsenic
1.204 x 1023
Molybdenum
2.408 x 1023
Lithium
34.5 g
Rubidium
8.55 g
Silicon
5.62 g
Scandium
9g
36.12 x 1023
Cobalt
Chromium
0.2
Nitrogen
0.01
Uranium-238
8
Phosphorus
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0.31 g
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Moles and Equations
MOLAR MASS, M
The molar mass of a substance is the mass of
one mole of the substance.
The units of molar mass are g mol1. This is a very useful term because it can be
applied universally to elements and compounds.
The molar mass, M, is equal to the relative mass in grammes of the substance
For carbon,
M(C)
=
12.0 g mol1.
For water,
M(H2O)
=
18.0 g mol1.
Using molar masses,
mass
number of moles = molar mass
In 4 g of C, amount (in moles) of C
4
= 12 = 0.33 mol
In 11 g of CO2, amount (in moles) of CO2
11
= 44 = 0.25 mol
In 5.85 g of NaCl, amount (in moles) of NaCl
5.85
= 58.5 = 0.100 mol
Remember:
m
n = M
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Moles and Equations
Moles of molecules/formulae
Refer to the periodic table to complete the table below.
Compound
Formula
molar mass
/g mol1
Number of
moles
Mass of
compound
/g
Number of
molecules
Water
H2O
18
1
18 g
6.02 x 1023
Ammonia
NH3
1
Carbon dioxide
0.5
Carbon
disulfide
7.62 g
Phosphine
PH3
68 g
Methane
CH4
64 g
Methanol
CH3OH
Dichlorine
oxide
Cl2O
3.01 x 1023
3
12.04 x 1023
Hydrogen
fluoride
Acetone
C3H6O
29 g
18.06 x 1023
Hydrogen
bromide
Nitrogen
trichloride
Ethane
0.1
C2H6
0.25
Ethanol
1.5
Benzene
C6H6
Phosphorus
pentachloride
PCl5
Octane
C8H18
468 g
10
348 g
PCl3
Silane
Glucose
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0.2
SiH4
6.02 x 1024
C6H12O6
6.02 x 1022
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Moles and Equations
More Mole questions
1.
2.
Calculate the mass of each of the following
(a)
1 mole of hydrogen molecules, H2
(b)
1 mole of hydrogen atoms, H
(c)
1 mole of silicon dioxide, SiO2
(d)
0.5 mole of carbon dioxide, CO2
(e)
0.25 mole of hydrated sodium carbonate Na2CO3•10H2O.
How many moles are each of the following?
(a)
32 g of oxygen molecules, O2
(b)
32 g of oxygen atoms, O
(c)
31 g of phosphorus molecules, P4
(d)
32.1 g of sulfur molecules, S8
(e)
50.05 g of calcium carbonate, CaCO3.
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Moles and Equations
TYPES OF CHEMICAL FORMULA
Empirical formula
The empirical formula of a compound is the simplest,
whole-number ratio of elements in a compound.
An empirical formula is often calculated from experimental results using the mole
concept.
Worked Example 1
In an experiment, 10.025 g of calcium combined with 9.5 g of fluorine.
[Ar: Ca, 40.1; F, 19.0.]
molar ratio of atoms
Ca
:
F
=
………
……..
:
………
……..
……..
:
……..
……..
:
……..
divide by smallest number
Empirical formula:
……..
Worked Example 2
Analysis of a compound produced the percentage composition by mass of
potassium: 70.9%; sulfur: 29.1%. [Ar: K, 39.1; S, 32.1.]
100.0 g of the compound contains 70.9 g of potassium and 29.1 g of sulfur
molar ratio of atoms
=
divide by smallest number
K
………
……..
S
:
………
……..
……
:
…….
……
:
…….
Empirical formula:
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:
……
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Moles and Equations
Formula Determination
1.
2.
Find the formula of the following:
(a)
calcium oxide (4.01 g of calcium reacts to form 5.61 g of calcium oxide).
(b)
sodium oxide (2.3 g of sodium reacts to form 3.1 g of sodium oxide).
(c)
iron oxide (11.16 g of iron reacts to form 15.96 g of iron oxide).
Use the following percentage compositions by mass to find the empirical
formula of the compound.
(a)
A compound of iron and bromine [Fe, 18.9 %; Br, 81.1 %].
(b)
A compound of sulfur and oxygen [S, 40.0 %; O, 60.0 %]
(c)
A compound of iron, sulfur and oxygen [Fe, 36.8 %; S, 21.1 %; O, 42.1 %]
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Moles and Equations
Molecular formula
The molecular formula of a compound is the actual
number of atoms of each element in a molecule.
A molecular formula is determined by comparing the empirical formula with the
relative molecular mass, Mr, of the compound.
Worked Example
A compound has an empirical formula of CH2 and a relative molecular mass, Mr, of
56. What is its molecular formula?
empirical formula mass of CH2:
12 + 1 x 2 = 14
number of CH2 units in a molecule:
56
14 = 4
molecular formula:
C4H8
Questions
1.
An empirical formula of CH2 can apply to different molecules, depending on Mr.
Complete the table below to find the different molecular formulae from CH2.
empirical formula
empirical formula mass
Mr
28
multiple
x2
molecular formula
C2H4
42
CH2
70
14
112
154
322
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Moles and Equations
2.
Use the following information to find the empirical and molecular formula of
each compound.
(a)
A compound of carbon and hydrogen formed when 2.4 g of carbon
combines with 0.6 g of hydrogen, Mr 30.
(b)
A compound of carbon and hydrogen and oxygen formed when 0.3 g of
carbon combines with 0.1 g of hydrogen and 0.4 g of oxygen, Mr 32.
(c)
A compound of carbon and hydrogen and oxygen with the composition by
mass: C, 40.0 %; H, 6.7 %; O, 53.3 %; Mr 180.
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Moles and Equations
WATER OF CRYSTALLISATION
The familiar blue crystals of ‘copper sulfate’ owe their colour to the presence of water
of crystallisation. When copper sulfate crystals are heated they turn a paler blue until
all the water is removed. You are left with just a white powder.
Hydrated describes the crystalline compound which contains water molecules in its
structure.
Anhydrous describes the compound when all the water of crystallisation has been
removed.
When compounds are crystallised from water, they frequently contain water molecules
within their structure giving a crystal appearance to the compound.
This water is known as the water of crystallisation.
We can work out the amount, in moles, of water in a crystal structure using
percentage or mass composition data. The method is similar to that used in empirical
formula calculations.
Example 1:
A sample of 7.392 g of MgSO4 crystals lost 3.780 g of water on heating.
Calculate n in the formula MgSO4•nH2O.
molar ratio
=
divide by smallest number
Notice the dot in the
middle of the formula.
MgSO4
………
……..
:
……..
:
……..
……..
:
……..
Formula:
:
H2O
………
……..
……..……..
Example 2
A sample of 9.98 g of CuSO4•xH2O was heated until the blue crystals became
colourless once all of the water had been removed. The residue was re-weighed and
the mass was found to be 6.38 g.
Calculate x in the formula CuSO4•xH2O.
molar ratio
=
divide by smallest number
Formula:
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CuSO4
………
……..
:
……..
:
……..
……..
:
……..
:
H2O
………
……..
……..……..
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Moles and Equations
Questions
1.
From the experimental results below, work out the formula of the hydrated salt.
Mass of BaCl2•xH2O
Mass of BaCl2
2.
= 7.329 g
= 6.249 g
1.1895 g of CoCl2•xH2O was heated to drive off all of the water of
crystallisation. The mass of the anhydrous cobalt chloride was 0.6495 g.
Calculate the value of x to determine the formula of the hydrated salt.
3.
12.9375 g of ZnSO4•xH2O was heated to drive off all of the water of
crystallisation. The mass of the anhydrous zinc sulfate was 7.2675 g.
Calculate the value of x to determine the formula of the hydrated salt.
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Moles and Equations
EQUATIONS
Chemical reactions involve the rearrangement of atoms and ions.
Chemical equations provide two types of information :

qualitative
-
which atoms or ions are rearranging

quantitative
-
how many atoms or ions are rearranging.
Qualitative information
At its simplest qualitative level, a word equation can summarise a reaction:
e.g.
hydrogen + oxygen  water
The formula of each substance provides information about the chemicals that are
reacting together:
H2 + O2  H2O
State symbols can be added to provide information about the physical states of each
species under the conditions of the reaction :
gaseous state
liquid state
solid state
aqueous solution
(g)
(l)
(s)
(aq)
H2(g) + O2(g)  H2O(l)
Balancing an equation accounts for all of the species in the reaction.
In a correctly balanced equation, there will be the same number of each element on
each side of the equation:
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Moles and Equations
Balancing equations
Balance each of the following equations:
1.
Fe(s) + S(s)  FeS(s)
2.
Na(s) + Cl2(g)  NaCl(s)
Balancing is
done by trial
and error
3.
N2(g) + O2(g)  NO2(g)
4.
H2(g) + Cl2(g)  HCl(g)
5.
Mg(s) + N2(g)  Mg3N2(s)
6.
K(s) + O2(g)  K2O(s)
7.
Al(s) + O2(g)  Al2O3(s)
8.
S(s) + O2(g)  SO3(g)
9.
P4(s) + O2(g)  P2O5(s)
10.
Fe(s) + Cl2(g)  FeCl3(s)
11.
Ca(s) + HCl(aq)  CaCl2(aq) + H2(g)
12.
KI(aq) + Cl2(g)  KCl(aq) + I2(aq)
13.
NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l)
14.
Al(s) + HCl(aq)  AlCl3(aq) + H2(g)
15.
CH4(g) + O2(g)  CO2(g) + H2O(l)
16.
N2(g) + H2(g)  NH3(g)
17.
MgCO3(s) + HCl(aq)  MgCl2(aq) + H2O(l) + CO2(g)
18.
NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq)
19.
Fe(s) + H2SO4(aq)  Fe2(SO4)3(aq) + H2(g)
20.
Al2O3(s) + H2SO4(aq)  Al2(SO4)3(aq) + H2O(l)
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Moles and Equations
EQUATIONS AND THE MOLE
The balancing number is the amount of each substance in moles:
O2(g)

2 H2O(l)
… mol
... mol

… mol
………. g
………. g

………….. g
….… g
……..g

……..g
equation
2 H2(g)
moles
reacting masses
+
 ……. g of H2(g) reacts with …….. g O2(g) to form ……… g of H2O(l)
Now work out the masses of reactants and products from each of the balanced
equations below.
1.
2 SO2(g)
2.
Mg (s)
+
2 HCl(aq)
3.
4 Al(s)
+
3 O2(g)
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+
O2(g)



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2 SO3(g)
MgCl2(aq)
+
H2(g)
2 Al2O3(s)
© Rob Ritchie, 29/04/17
Moles and Equations
Working out reacting quantities
We can use this idea to work out quantities involved in any reaction. It is often better
to work in moles right from the start.
Example
Nitrogen and oxygen react to form nitrogen monoxide, NO.
N2(g) + O2(g)  2 NO(g)
Calculate the masses of N2 and O2 required to form 3 g of NO.
STAGE 1
Work out the number of moles of NO that is formed.
m ……….
n(NO) = M = ………. = ……… mol
n = number of moles;
m = mass; M = Molar mass
STAGE 2
Write down a BALANCED EQUATION to find the moles used in this reaction.
We do this by scaling the balancing numbers to match 0.1 mol NO.
N2(g)
+
O2(g)

2 NO(g)
moles
............. mol
+
............. mol

............. mol
moles used
............. mol
............. mol

............. mol
equation
STAGE 3
Find the masses of N2(g) and O2(g) from their moles.
m
n(NO) = M
m=nxM
Remember:
mass of N2 = …….. x ……….. = …………….g
mass of O2 = …….. x ……….. = …………….g
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m
n = M
© Rob Ritchie, 29/04/17
Moles and Equations
Some quantitative questions
1.
For the reaction:
H2(g) + F2(g)  2 HF(g)
Calculate the masses of hydrogen and fluorine required to form 5 g of hydrogen
fluoride, HF.
2.
For the reaction:
N2(g) + 3 H2(g)  2 NH3(g)
Calculate the mass of ammonia, NH3 that could be formed by reacting 30 g of
hydrogen with an excess of nitrogen.
3.
For the reaction:
4 Na(s) + O2(g)  2 Na2O(s)
Calculate the masses of sodium and oxygen required to form 5.25 g of sodium
oxide, Na2O. Give your answers to 3 significant figures.
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Moles and Equations
GASES AND THE MOLE
In 1811, Avogadro put forward a hypothesis:
In English:
Equal volumes of gases contain the same number of
molecules under the same conditions of temperature
and pressure.
or
1 mole of molecules of any gas occupies 24.0 dm3
(24,000 cm3) at room temperature and pressure (RTP).
Working out moles from gas volumes
We can work out the number of moles of gas molecules present by simply measuring
the volume:
For a gas at room temperature and pressure,
volume in dm3:
Number of moles =
V (in dm3)
24.0
volume in cm3:
Number of moles =
V (in cm3)
24 000
Worked Example
How many moles of gas molecules are in 480 cm3 of a gas at RTP?
Number of moles =
V (in cm3)
24 000
 Number of moles = ……..…. = ………. mol
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Moles and Equations
Questions
1.
How many moles of gas molecules are in the following gas volumes at r.t.p.
(a) 96 dm3
…………
(b) 3600 dm3
…………
(c) 2.0 dm3
…………
(d) 24 cm3
…………
(e) 2 cm3
…………
(f) 1.6 cm3
…………
2.
What is the volume of the following moles of gas at r.t.p.
(a) 10 mol
…………
(b) 75 mol
…………
(c) 0.4 mol
…………
(d) 0.006 mol
…………
(e) 0.0025 mol
…………
(f) 1.82 x 105 mol
…………
3.
What is the volume, at r.t.p., of
4.
What is the mass, at r.t.p., of
(a) 44 g CO2(g);
(a) 1.2 dm3 O2;
(b) 7 g N2(g);
(b) 720 cm3 CO(g);
(c) 5.1 g NH3(g)?
(c) 48 cm3 CH4(g)?
V (in dm3)
n =
24.0
V (in cm3)
n = 24 000
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Moles and Equations
Calculations involving gas volumes
As with masses, we can use the mole concept with gas volumes.
Example
Hydrogen peroxide, H2O2, decomposes into water and oxygen.
2 H2O2(l)  2 H2O(l) + O2(g)
What volume of oxygen, at RTP, is formed by the decomposition of 17 g of
hydrogen peroxide, H2O2(l)?
STAGE 1
m ……….
n(H2O2) = M = ………. = ……… mol
n = number of moles;
m = mass; M = Molar mass
STAGE 2
Write down a BALANCED EQUATION to find the moles used in this reaction.
We do this by scaling the balancing numbers to match 0.5 mol H2O2.
equation
moles
moles used
2 H2O2(l)

2 H2O(l)
+
O2(g)
2 mol H2O2(l)

2 mol H2O(l)
+
1 mol O2(g)
0.5 mol H2O2(l)

0.5 mol H2O(l)
+
0.25 mol O2(g)
STAGE 3
Find the volume of O2 from its moles.
V (in cm3)
n = 24 000
 V = n x 24 000
volume of O2 = …….. x ……….. = …………….cm3
V (in dm3) V (in cm3)
n =
= 24 000
24.0
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More quantitative questions
1.
For the reaction:
2NaNO3(s)  2NaNO2(s) + O2(g)
What volume of O2, in cm3, is formed at RTP from 4.25 g of NaNO3?
2.
For the reaction:
2CO(g) + O2(g)  2CO2(g)
What volumes, in dm3, of CO and O2 react together to form 13.2 g of CO2?
4.
Mg was reacted with O2 to form 1.612 g MgO
(a)
Write a balanced equation for the reaction
(b)
What mass of Mg was reacted?
(c)
What volume, in cm3, of O2, was reacted at RTP?
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2.
5.
Balance the equation:
Al(s) +
Cl2(g) 
AlCl3(s)
(a)
What mass of AlCl3 forms by reacting 5.4 g of Al with an excess of Cl2?
(b)
9.0 g of Al reacted with 9 dm3 of Cl2 at RTP.
Which reactant was in excess?
A compound of nitrogen and oxygen, A, was formed by reacting 20 cm3 of
nitrogen, gas with 10 cm3 of oxygen.
(a)
Determine the formula of compound A.
(b)
Write an equation for this reaction.
(c)
What volume of compound out formed at r.t.p.?
(d)
What volumes of nitrogen and oxygen would have reacted to form 30 cm3
of a gaseous oxide of nitrogen with the formula
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(i)
NO2;
(ii)
NO?
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Moles and Equations
SOLUTIONS AND THE MOLE
The concentration of a solution is usually expressed in the moles of dissolved
substance per cubic decimetre, (1000 cm3) of solution.
1 dm3 is equivalent to 1 litre, l.
1 cm3 is equivalent to the millilitre, ml.
A concentration of 1 mol dm–3 contains 1 mole of dissolved solute in 1 dm3 of
solution.
A concentration of 2 mol dm–3 contains 2 moles of dissolved solute in 1 dm3 of
solution.
Solutions of sodium chloride, NaCl(aq)
1 mole of NaCl has a mass of 23.0 + 35.5 = 58.5 g.
A 1 mol dm–3 aqueous solution of NaCl contains 58.5 g of NaCl dissolved in sufficient
water to make 1 dm3 of an aqueous solution.
A 2 mol dm–3 aqueous solution of NaCl contains 2 x 58.5 g = 117.0 g per cubic
decimetre of aqueous solution.
Working out moles from volumes of solutions
Provided we know the molar concentration of a solution, we can work out the number
of moles by measuring the volume:
For a solution of concentration c mol dm3,
volume in dm3:
Number of moles = c x V (in dm3)
volume in cm3:
V (in cm3)
Number of moles = c x 1000
n = c x V (in dm3)
V (in cm3)
n = c x 1000
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Concentration questions
1.
Find the number of moles dissolved in:
(a) 1 dm3 of a 1 mol dm–3 solution,
(b) 2 dm3 of a 1 mol dm–3 solution,
(c) 4 dm3 of a 5 mol dm–3 solution,
(d) 250 cm3 of a 1.00 mol dm–3 solution,
(e) 10 cm3 of a 2.0 mol dm–3 solution,
(f)
20cm3 of a 0.10 mol dm–3 solution
(g) 26.2 cm3 of a 0.500 mol dm–3 solution,
(h) 29.6 cm3 of a 0.123 mol dm–3 solution,
2.
(i)
13.2 cm3 of a 0.0260 mol dm–3 solution,
(j)
7.2 cm3 of a 0.0032 mol dm–3 solution.
Find the concentration in mol per cubic decimetre of solution for:
(a) 2 moles dissolved in 1 dm3 of solution,
(b) 4 moles dissolved in 2 dm3 of solution,
(c) 0.500 moles dissolved in 250 cm3 of solution,
(d) 2.00 moles dissolved in 200 cm3 of solution,
(e) 0.0100 moles dissolved in 100 cm3 of solution.
3.
Find the concentration in grams per cubic decimetre of solution for:
(a) 2 moles of NaOH dissolved in 1 dm3 of solution,
(b) 4 moles of HCl dissolved in 2 dm3 of solution,
(c) 0.500 moles of H2SO4 dissolved in 250 cm3 of solution,
(d) 2.00 moles of HNO3 dissolved in 200 cm3 of solution,
(e) 0.0100 moles of Ca(OH)2 dissolved in 100 cm3 of solution.
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CALCULATIONS FROM TITRATIONS
The manipulation of titration results follow a set pattern.
There are 3 stages involved and these MUST be learnt.
Worked example
In a titration 25.0 cm3 of 0.100 mol dm–3 aqueous sodium hydroxide were found to
react with 10.4 cm3 of sulfuric acid.
Find the concentration of the sulfuric acid.
STAGE 1
Work out the number of moles of WHAT YOU CAN.
Number of moles of NaOH
= cx
V (in cm3)
mol
1000
= …………….. mol
= ……….. mol NaOH
STAGE 2
Write down a BALANCED EQUATION to find out the number of moles of the other
reactant.
2 NaOH(aq)
+
H2SO4(aq)

Na2SO4(aq)
…………mol
………. mol (balancing numbers)
……….. mol
……… mol
+
2 H2O(l)
STAGE 3
Find the concentration of H2SO4 from its moles and volume.
V (in cm3)
n = c x 1000
1000
 c = n x V (in cm3)
..........
concentration of H2SO4 = …….. x .......... = ……………. mol dm–3
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Problems to concentrate the mind
1.
In a titration 25.0 cm3 of 0.250 mol dm–3 aqueous sodium hydroxide were found
to react exactly with 27.8 cm3 of hydrochloric acid. Find the concentration of the
hydrochloric acid.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
2.
In a titration 25.0 cm3 of 0.200 mol dm–3 aqueous potassium hydroxide were
found to react exactly with 15.4 cm3 of hydrochloric acid. Find the concentration
of the hydrochloric acid.
HCl(aq) + KOH(aq)  KCl(aq) + H2O(l)
3.
In a titration 25.0 cm3 of 0.500 mol dm3 aqueous sodium hydroxide were found
to react exactly with 23.2 cm3 of nitric acid. Find the concentration of the nitric
acid.
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
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4.
In a titration 25.0 cm3 of 1.00 mol dm3 aqueous sodium hydroxide were found to
react exactly with 26.6 cm3 of sulfuric acid. Find the concentration of the sulfuric
acid.
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
5.
A standard aqueous solution of sodium hydroxide was prepared containing
1.00 g in 250 cm3 of solution. In a titration 25.0 cm3 of this solution were found to
react exactly with 20.5 cm3 of hydrochloric acid.
Find the concentration of the standard solution and hence the concentration of
the hydrochloric acid.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
6.
A standard aqueous solution of sodium carbonate was prepared containing
10.6 g in 100 cm3 of solution. In a titration 25.0 cm3 of this solution were found to
react exactly with 11.3 cm3 of hydrochloric acid.
Find the concentration of the standard solution and hence the concentration of
the hydrochloric acid.
2 HCl(aq) + Na2CO3(aq)  2 NaCl(aq) + H2O(l) + CO2(g)
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