AMOUNT OF SUBSTANCE The mole Reacting masses and atom economy Solutions and titrations The ideal gas equation Empirical and molecular formulae Ionic equations
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Topic 1.2
AMOUNT OF SUBSTANCE
The mole
Reacting masses and atom economy
Solutions and titrations
The ideal gas equation
Empirical and molecular formulae
Ionic equations
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THE
MOLE
Since atoms are so small, any sensible laboratory quantity of substance
must contain a huge number of atoms:
1 litre of water contains 3.3 x 1025 molecules.
1 gram of magnesium contains 2.5 x 1022 atoms.
100 cm3 of oxygen contains 2.5 x 1021molecules.
Such numbers are not convenient to work with, so it is necessary to
find a unit of "amount" which corresponds better to the sort of
quantities of substance normally being measured. The unit chosen for this
purpose is the mole. The number is
chosen so that 1 mole of a substance corresponds to its relative
atomic/molecular/formula mass measured in grams. A mole is thus defined as
follows:
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A mole of a substance is the amount of that substance that contains
the same number of elementary particles as there are carbon atoms in 12.00000
grams of carbon-12.
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One mole of carbon-12 has a mass of 12.0g.
One mole of hydrogen atoms has a mass of 1.0g.
One mole of hydrogen molecules has a mass of 2.0g.
One mole of sodium chloride has a mass of 58.5g.
The number of particles in one mole of a substance is 6.02 x 1023.
This is known as Avogadro's number, L.
Thus when we need to know the number of particles of a substance, we
usually count the number of moles. It is much easier than counting the number
of particles.
The number of particles can be calculated by multiplying the number of
moles by Avogadro’s number. The number of moles can be calculated by dividing
the number of particles by Avogadro’s number.
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(Number
of particles) = (number of moles) x L
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The mass of one mole of a substance is known as its molar mass, and has units of gmol-1.
It must be distinguished from relative atomic/molecular/formula mass, which is
a ratio and hence has no units, although both have the same numerical value.
The symbol for molar mass of compounds or molecular elements is mr.
The symbol for molar mass of atoms is ar.
Mass (m), molar mass (mr or ar) and number of
moles (n) are thus related by the following equation:
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MASS
= MOLAR MASS X NUMBER OF MOLES
or
m = mr x n
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Mass must be measured in grams and molar mass in gmol-1.
REACTING MASSES
It is possible to use the relationship moles = mass/mr to deduce the masses of
reactants and products that will react with each other.
When performing calculations involving reacting masses, there are two
main points which must be taken into account:
The total combined mass of
the reactants must be the same as the total combined mass of the products. This is known
as the law of conservation of mass.
The ratio in which species
react corresponds to the number of moles, and not their mass. Masses must
therefore all be converted into moles, then compared to each other, then
converted back.
i) reactions which go to completion
Eg What mass of aluminium will be needed to react with 10 g of CuO, and
what mass of Al2O3 will be produced?
3CuO(s) + 2Al(s) à Al2O3(s) + 3Cu(s)
10 g
= 10/79.5
= 0.126 moles of CuO
3:2 ratio with Al
so 2/3 x 0.126 = 0.0839 moles of Al, so mass of Al = 0.0839 x 27 = 2.3
g
3:1 ratio with Al2O3
so 1/3 x 0.126 = 0.0419 moles of Al2O3, so mass
of Al2O3 = 0.0419 x 102 = 4.3 g
ii) reactions which do not
go to completion
Many inorganic reactions go to completion. Reactions which go to
completion are said to be quantitative.
It is because the reactions go to completion that the substances can be
analysed in this way.
Some reactions, however, particularly organic reactions, do not go to
completion. It is possible to calculate the percentage yield of product by using the following equation:
% yield = amount of
product formed
x 100
maximum amount of
product possible
Eg 2.0 g of ethanol (C2H5OH) is oxidised to
ethanoic acid (CH3COOH). 1.9 g of ethanoic acid is produced. What is
the percentage yield? (assume 1:1 ratio)
Moles of ethanol = 2/46 = 0.0435
Max moles of ethanoic acid = 0.0435
so max mass of ethanoic acid = 0.0435 x 60 = 2.61 g
percentage yield = 1.9/2.61 x 100 = 73%
Eg When propanone (CH3COCH3) is reduced to
propan-2-ol (CH3CH2CH2OH), a 76% yield is
obtained. How much propan-2-ol can be obtained from1.4 g of propanone? (assume
1:1 ratio)
Moles of propanone = 1.4/58 = 0.0241 moles
So max moles of propan-2-ol produced = 0.0241 moles
So actual amount produced = 0.0241 x 76/100 = 0.0183 moles
So mass of propan-2-ol = 0.0183 x 60 = 1.1 g
When we carry out
a chemical reaction in order to make a product, we often make other products,
called by-products, as well.
Eg In the
production of NaOH from NaCl the following reaction takes place:
2NaCl + 2H2O
à 2NaOH + H2 + Cl2
The atom economy
of a reaction is the percentage of the total mass of reactants that can, in
theory, be converted into the desired product. It can be calculated as follows:
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% atom
economy = mass of desired product
x 100
total mass of products
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Assuming we start
with 2 moles of NaCl and 2 moles of H2O, we will make 2 moles of
NaOH, and 1 mole of H2 and Cl2.
So % atom economy
= (2 x 40)
x 100 = 52.3 %
(2 x 40) + (1 x 2) + (1 x 71)
The remaining
47.7% of the mass is converted into less useful products and is hence wasted.
So the higher the
atom economy, the less waste and the more efficient the product process
(assuming the reaction does actually go to completion).
All reactions
which have only one product have an atom economy of 100%
Atom economy is an
important consideration when considering how to make a particular useful
product.
SOLUTIONS
A solution is a homogeneous mixture of two or more substances in
which the proportions of the substances are identical throughout the mixture.
The major component of a solution
is called the solvent and the minor
components are called the solutes. In most
cases water is the solvent.
The amount of solute present in a
fixed quantity of solvent or solution is called the concentration of the solution.
It is usually measured in grams of solute per dm3 of solution or in
moles of solute per dm3 of solution. In the latter case (moldm-3)
it is also known as the molarity of the
solution.
The number of moles of solute,
molarity of the solution and volume of solution can thus be related by the
equation:
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Number
of moles = volume x molarity
n = C x V
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The volume of solution in this
case must always be measured in dm3 (or litres). If the volumes are
given in cm3 then V/1000 must be used instead.
If concentration is given in gdm-3,
it must be converted to molarity before it can be used in the above equation.
This can be done easily by dividing by the molar mass of the solute.
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Concentration
(gdm-3) = Molarity x molar mass
Or
Cg = Cm x mr
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The volume of one solution
required to react with a known volume of another can be deduced from the above
relationships and knowledge of the relevant chemical equation. Remember it is
moles which react in the ratio shown, so all quantities must be converted to
moles before the comparison can be made.
The quantitative investigation of
chemical reactions by comparing reacting volumes is known as volumetric analysis. The
procedure by which reacting volumes are determined is known as a titration.
In titrations, a solution whose
concentration is unknown is titrated against a solution whose concentration is
known. The solution of known concentration is always placed in the burette, and
the solution of unknown concentration is always placed in the conical flask.
Eg 28.3 cm3 of a 0.10
moldm-3 solution of NaOH was required to react with 25 cm3
of a solution of H2SO4. What was the concentration of the H2SO4 solution?
Equation: H2SO4 + 2NaOH à Na2SO4 + 2H2O
Moles of NaOH = 28.3/1000 x 0.1 =
2.8 x 10-3
2:1 ratio so moles of H2SO4 = 2.8 x 10-3/2
= 1.4 x 10-3
so concentration of H2SO4 = 1.4 x 10-3/25
x 1000 = 0.056 moldm-3.
Eg Calculate the volume of 0.50
moldm-3 nitric acid required to react completely with 5 g of lead
(II) carbonate.
Equation: PbCO3 + 2HNO3 à Pb(NO3)2 + CO2 + H2O
Moles of PbCO3 = 5/267 = 0.0187
1:2 ratio so moles
of HNO3 = 0.0187 x 2 = 0.0375
Volume of HNO3 = 0.0375/0.5 x 1000 = 74.9 cm3.
GASES
The volume
occupied by a gas depends on a number of factors:
--the temperature: the hotter the gas,
the faster the particles are moving and the more space they will occupy
--the pressure: the higher the pressure, the more compressed the gas
will be and the less space it will occupy
--the amount of gas: the more gas particles there are, the more space
they will occupy
The volume
occupied by a gas does not depend on what gas it is, however: one mole of any
gas, at the same temperature and pressure, will have the same volume as one
mole of any other gas.
The pressure,
temperature, volume and amount of gas can be related by a simple equation known
as the ideal gas equation:
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PV
= nRT
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P is the pressure
measured in pascals (Pa) or Nm-2. One atmosphere, which is normal
atmospheric pressure, is 101325 Pa.
V is the volume in
m3. Remember; 1 m3 = 1000 dm3 = 106
cm3.
T is the absolute
temperature, measured in Kelvin (K). Remember; 0 oC = 273 K.
R is the molar gas
constant and has a value of 8.31 Jmol-1K-1.
This equation can
be rearranged to find the density of gases and the RMM of gases, using the
relationship m = n x mr.
PV = mRT/mr,
so the mass of one mole is given by mr = mRT/PV, where m is the mass
in kg. The answer m will also be in kg so it must be converted into grams.
The density of a
gas, or mass/volume, is given by (m/V) = mrP/RT.
SUMMARY – USING MOLES
Using the four
relationships described, it is possible to calculate the amount of any
substance in a chemical reaction provided that the chemical equation is known
and the amount of one of the reacting species is also known. The procedure is
summarised in the table below:
These
relationships are frequently used in practical chemistry. Typical calculations
in AS Practical Chemistry involve:
i) Determining the concentration of a
solution
ii) Determining the relative molecular mass
of a solid
iii) Determining the percentage purity of a
solid
The percentage
purity of a substance can be calculated as follows:
Percentage purity
= mass substance would have if it was pure x 100
mass of
impure substance
EMPIRICAL AND MOLECULAR
FORMULAE
The empirical formula of a
compound is the formula which shows the simplest whole-number ratio in which
the atoms in that compound exist.
It can be calculated if the composition by mass of the compound is
known.
The molecular formula of a
substance is the formula which shows the number of each type of atom in the one
molecule of that substance.
It applies only to molecular substances, and can be deduced if the empirical
formula and molar mass of the compound are known.
The molecular formula is always a simple whole number multiple of the
empirical formula.
Eg a substance contains 85.8% carbon and 14.2% hydrogen, what is its
empirical formula? If its relative molecular mass is 56, what is its molecular
formula?
Mole ratio = 85.8 : 14.2
12 1
= 7.15 : 14.2
7.15 : 7.15
= 1 : 2 so
empirical formula = CH2
RMM = 56 = (CH2) so 14n = 56 and n = 56/14 = 4
Molecular
formula = C4H8
It is also possible to calculate the percentage composition by mass of
a substance, if its empirical or molecular formula is known.
Eg What is the percentage composition by mass of ethanoic acid, C2H4O2?
RMM = 60
% C = (12 x 2)/60 x 100 = 40.0%
% H = (1 x 4)/60 x 100 = 6.67%
%O = (16 x 2)/60 x 100 = 53.3%
FORMULAE OF IONIC COMPOUNDS
An ion is a species in which the number of electrons is not equal to
the number of protons. An ion thus has an overall charge, characteristic of the
difference in the number of protons and electrons. Ions with a positive charge
are known as cations and ions with a
negative charge are known as anions.
Compounds made up of ions are known as salts. They are all electrically neutral, so must all contain at
least one anion and at least one cation.
Salts do not have molecular formulae, as they do not form molecules.
They are written as unit formulae.
The unit formula of an ionic
compound is the formula which shows the simplest whole number ratio in which
the ions in the compound exist. This depends on the charges of the ions
involved. Some important ions and their charges are shown below:
i)
cations
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Charge
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Formula
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Name
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+1
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Na+
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Sodium
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+1
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K+
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Potassium
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+1
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Ag+
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Silver
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+1
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H+
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Hydrogen
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+1
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NH4+
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Ammonium
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+1
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Cu+
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Copper(I)
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+2
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Mg2+
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Magnesium
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+2
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Ca2+
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Calcium
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+2
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Fe2+
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Iron(II)
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+2
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Zn2+
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Zinc
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+2
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Pb2+
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Lead(II)
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+2
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Cu2+
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Copper(II)
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+2
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Ni2+
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Nickel(II)
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+3
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Al3+
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Aluminium
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+3
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Cr3+
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Chromium(III)
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+3
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Fe3+
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Iron(III)
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Note that some atoms can form more than one stable cation. In such
cases it is necessary to specify the charge that is on the cation by writing
the charge in brackets after the name of the metal.
ii) anions
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Charge
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Formula
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Name
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-1
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Hydroxide
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-2
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SO42-
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Sulphate
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-2
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CO32-
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Carbonate
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-1
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NO3-
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Nitrate
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-1
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HCO3-
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Hydrogencarbonate
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CHEMICAL EQUATIONS
The purpose of
chemistry is essentially to study chemical reactions. In chemical reactions,
elements or compounds react with each other to form other elements and/or other
compounds.
Chemical reactions
involve the movement of electrons between different species. The nuclei always
remain intact.
Every chemical
reaction can be represented by a chemical equation. A chemical equation
indicates the species involved in the reaction and shows the way in which they
react. Every chemical equation must contain three pieces of information:
--the identities of all the reactants and
products
The chemical
formulae of all the species involved in the reaction should be shown. Any
species left unchanged should be left out. Reactants must be written on the
left of the arrow and products on the right.
Remember that in
chemical reactions all the nuclei remain unchanged. Therefore the total number
of atoms of each type must be the same on each side of the equation. Atoms
themselves cannot be created or destroyed in chemical reactions; only
transferred from species to species.
--the reaction coefficients
Atoms, elements
and compounds combine with each other in simple whole number ratios, eg 1:1,
1:2, 1:3. The ratio in which the species react and in which products are formed
are shown in the reaction coefficients. These are the numbers which precede the
chemical formula of each species in the equation. If no coefficient is shown it
is assumed to be 1.
Deducing the
reaction coefficients for a reaction is known as balancing the equation. The
total number of atoms of each element must be the same on both sides of the
equation.
When balancing
chemical equations, always balance compounds first and elements second. It's
easier that way.
N.B. Reaction
coefficients in no way show the actual amount of a substance which is reacting.
They provide information only on the way in which they react.
--The state symbols
The state symbol
shows the physical state of each reacting species and must be included in every
chemical equation. There are four state symbols required for A-level chemistry:
(s) - solid
(l) - liquid
(g) - gas
(aq) - aqueous, or
dissolved in water
IONIC EQUATIONS
Many reactions that take place in aqueous
solution do not involve all of the ions that are written in the equation. Some
species remain in aqueous solution before and after the reaction. They
therefore play no part in the reaction and are known as spectator ions.
In ionic
equations, spectator ions are left out.
Eg BaCl2(aq) + Na2SO4(aq) à BaSO4(s) + 2NaCl(aq)
This reaction
involves the precipitation of barium sulphate.
Notice that the Cl-
ions and the Na+ ions remain in the aqueous state before and after
the reaction. They are therefore spectator ions.
The above reaction
can then be rewritten as follows:
Ba2+(aq)
+ SO42-(aq) à BaSO4(s)
Eg Al2(SO4)3(aq)
+ 6NaOH(aq) à 2Al(OH)3(s) + 3Na2SO4(aq)
This reaction
involves the precipitation of aluminium hydroxide.
The Na+
and SO42- ions are spectator ions and can be left out
The ionic equation
for the reaction is:
Al3+(aq)
+ 3OH-(aq) à Al(OH)3(s)
Ionic equations
are very useful for simplifying precipitation reactions.
They can also simplify acid-base
reactions:
Eg NaOH(aq) + HCl(aq) à NaCl(aq) + H2O(l)
The Na+
and Cl- ions are spectator ions, so the ionic equation for the
reaction is:
H+(aq) + OH-(aq)
à H2O(l)
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