Section 2 - Chemical Calculations student notes

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AS Chemistry
Unit 1: Theoretical Chemistry
Part 7: Relative mass
Imagine that you run a chemical factory producing iron sulphide. (Don’t get
excited, this will not prove profitable.) One way to make it is shown below:
Fe(s)
+
S(s)

FeS(s)
This tells you that 1 atom of iron will react with 1 atom of sulphur to form 1
particle of iron sulphide, i.e. they react in a 1:1 ratio. If you add extra atoms of
iron or sulphur they will be left over. This is an expensive mistake! Not only are
you wasting materials but it may be difficult (and costly) to separate out your
product.
To avoid this happening chemists need a way of counting atoms!! The easiest way
to do this is by ‘weighing’. Not such a silly idea. Think about bank tellers, they
often count large amounts of small change by weighing the coins.
Comparing the masses of atoms
Scientists have found a method for comparing the masses of atoms. This
includes the invention in 1919 of the Mass Spectrometer which we will look at
later. Chemists use a relative atomic mass scale to compare the masses of
different atoms. At first the element hydrogen was chosen as the standard
against which the masses of other atoms were compared. This makes sense; it
has the smallest atoms which could conveniently be assigned a relative atomic
mass (Ar) of 1. At a later date, when Ar values could be obtained more
accurately and scientists discovered the existence of isotopes, it became
necessary to choose a single isotope as the reference standard for relative
atomic masses.
In 1961, carbon-12 (12C) was chosen as the new standard. This was chosen
because carbon is a very common element. Being a solid, it is also easier to store
and transport than hydrogen, which is a gas.
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Unit 1: Theoretical Chemistry
The relative atomic masses of some elements
Element
Symbol
Relative
atomic
mass (Ar)
Carbon-12
12
Carbon
C
12.011
Chlorine
Cl
35.453
Copper
Cu
63.540
Hydrogen
H
1.008
Iron
Fe
55.847
Magnesium
Mg
24.312
Sulphur
S
32.064
C
12.000
Note: Ar values have no units. Copper atoms do not have a mass of 64g or 64
‘anythings’. They are just roughly 64 times heavier than hydrogen atoms.
Task 1
Q.1.
Why is the Ar of carbon given as 12.011 not 12.000?
_______________________________________________________
_______________________________________________________
_______________________________________________________
Q.2.
Roughly how many times are magnesium atoms heavier than
12
C atoms?
_______________________________________________________
Q.3.
Approximately how many times are carbon atoms heavier than hydrogen
atoms?
_______________________________________________________
AS Chemistry
Unit 1: Theoretical Chemistry
Q.4.
Which element has atoms approximately twice as heavy as sulphur atoms?
_______________________________________________________
Relative Formula Mass
Just as there are relative atomic masses for atoms, so chemists use relative
formula masses to compare other substances. The relative formula mass of a
substance can be worked out by adding together the relative atomic masses of
each of the atoms in the formula. In substances such as methane, where the
formula represents a discrete molecule, the relative formula mass is often
called the relative molecular mass. Both relative formula mass and relative
molecular mass are given the symbol, Mr.
Key learning points
Task 2
Can you complete the key equations below?
12

The isotope
C is the standard for relative mass.

Relative atomic mass (Ar) =

Relative molecular mass (Mr) =

Relative isotopic mass =
Using relative atomic masses to count atoms – the mole
Since one atom of carbon is 12 times as heavy as one atom of hydrogen, it
follows that 12g of carbon and one gram of hydrogen contain the same number
of atoms because the masses are in the same ratio. In fact, the relative atomic
AS Chemistry
Unit 1: Theoretical Chemistry
mass in grams of all elements will contain the same number of atoms.
Experiments show that this number is 6.023 x 1023. It is called the Avogadro
constant (L).
Chemists refer to this quantity of particles as one mole (mol). (You may like to
compare this with the number 12 and the term one dozen!!!) The particles
(atoms, molecules, ions, etc.) need to be carefully specified. E.g. 1 mol C contains
one mole of carbon atoms; 1 mol H20 contains one mole of water molecules; 1 mol
NaCl contains one mole of sodium chloride ion pairs (i.e. one mole of sodium ions
and one mole chloride ions). One mole of oxygen is ambiguous!!!
Number of moles =
A chemical equation usually implies quantities in moles. For example:
CH4
+
2O2

CO2
+
2H2O
1mol of methane molecules reacts with 2mol of oxygen molecules to from 1mol
of carbon dioxide molecules and 2mol of water molecules.
Key learning points

The Avogadro constant is defined as the number of atoms in exactly 12g
of 12C. It is a quantity and is given the symbol L.

L = 6.023 x 1023 particles mol-1

One mole is the amount of any substance that contains the same number
of particles as there are atoms in exactly 12g of carbon-12.
Learning Objectives:
Candidates should be able to:

define the terms relative atomic, isotopic, molecular and formula masses,
based on the 12C scale.

define the term mole in terms of the Avogadro constant.
Reference:
A-level Chemistry: pages 16-19.
Chemistry in Context: pages 4-6.
AS Chemistry
Unit 1: Theoretical Chemistry
AS Chemistry
Unit 1: Theoretical Chemistry
Part 8: The Mass Spectrometer
The relative atomic mass (Ar) of an isotopic mixture can be determined by
calculating the weighted mean of the individual relative atomic masses of the
isotopes. The abundance of the different isotopes is found using a mass
spectrometer.
Traditionally, there are four main stages:

Ionisation (of vaporised sample) – high energy electrons from an electron gun
knock out an electron from the gaseous particles forming positive ions.
M(g)
+
e-
M(g)
→
M+(g) +
→
This may be simplified to:
e-
Some doubly charged ions may also be produced, but in much smaller amounts
because this takes more energy. (Molecules may also be broken into many
different fragments by the high-energy electrons from the gun.)

Acceleration – Positive ions are attracted by (the negative plate of) an
electric field and focussed into a beam by passing them through a series of
slits.

Deflection – The fast-moving beam of ions is deflected by a strong magnetic
field. The magnitude of the deflection depends on the mass-to-charge ratio
AS Chemistry
Unit 1: Theoretical Chemistry
(m/z) of the ion. The smaller the m/z ratio the larger the deflection. The
magnetic field can be increased in order to deflect heavier ions into the
detector.

Detection – A charged plate creates a current when hit by ions. This sends a
signal to the computer or chart recorder and a ‘spectrum’ is produced. This
gives the relative abundance of each species.
The mass spectrum for boron
m/e
The number of isotopes
The two peaks in the mass spectrum shows that there are 2 isotopes of boron with relative isotopic masses of 10 and 11 on the 12C scale.
The abundance of the isotopes
The relative size of the peaks gives you a direct measure of the relative
abundances of the isotopes. The tallest peak is often given an arbitrary height
of 100 - but you may find all sorts of other scales used. It doesn't matter in
the least.
You can find the relative abundances by measuring the lines on the stick
diagram.
In this case, the two isotopes (with their relative abundances) are:
boron-10
23
boron-11
100
Working out the relative atomic mass
AS Chemistry
Unit 1: Theoretical Chemistry
The relative atomic mass (RAM) of an element is given the symbol Ar and is
defined as:
The relative atomic mass of an element is the weighted average of
the masses of the isotopes relative to 1/12 of the mass of a
carbon-12 atom.
A "weighted average" allows for the fact that there won't be equal amounts of
the various isotopes. The example coming up should make that clear.
Suppose you had 123 typical atoms of boron. 23 of these would be
would be 11B.
10
B and 100
The total mass of these would be (23 x 10) + (100 x 11) = 1330
The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3
significant figures).
10.8 is the relative atomic mass of boron.
Notice the effect of the "weighted" average. A simple average of 10 and 11 is,
of course, 10.5. Our answer of 10.8 allows for the fact that there are a lot more
of the heavier isotope of boron - and so the "weighted" average ought to be
closer to that.
The mass spectrum of chlorine
Chlorine is taken as typical of elements with more than one atom per molecule.
We'll look at its mass spectrum to show the sort of problems involved.
Chlorine has two isotopes, 35Cl and 37Cl, in the approximate ratio of 3 atoms of
35
Cl to 1 atom of 37Cl. You might suppose that the mass spectrum would look like
this:
m/e
You would be wrong!
AS Chemistry
Unit 1: Theoretical Chemistry
The problem is that chlorine consists of molecules, not individual atoms. When
chlorine is passed into the ionisation chamber, an electron is knocked off the
molecule to give a molecular ion, Cl2+. These ions won't be particularly stable,
and some will fall apart to give a chlorine atom and a Cl+ ion. The term for this is
fragmentation.
If the Cl atom formed isn't then ionised in the ionisation chamber, it simply
gets lost in the machine - neither accelerated nor deflected.
The Cl+ ions will pass through the machine and will give lines at 35 and 37,
depending on the isotope and you would get exactly the pattern in the last
diagram. The problem is that you will also record lines for the unfragmented Cl2+
ions.
Think about the possible combinations of chlorine-35 and chlorine-37 atoms in a
Cl2+ ion.
Both atoms could be 35Cl, both atoms could be 37Cl, or you could have one of
each sort. That would give you total masses of the Cl2+ ion of:
35 + 35 = 70
35 + 37 = 72
37 + 37 = 74
That means that you would get a set of lines in the m/z = 70 region looking like
this:
m/e
These lines would be in addition to the lines at 35 and 37.
The relative heights of the 70, 72 and 74 lines are in the ratio 9:6:1. If you
know the right bit of maths, it's very easy to show this. If not, don't worry.
Just remember that the ratio is 9:6:1.
What you can't do is make any predictions about the relative heights of the
lines at 35/37 compared with those at 70/72/74. That depends on what
AS Chemistry
Unit 1: Theoretical Chemistry
proportion of the molecular ions break up into fragments. That's why you've got
the chlorine mass spectrum in two separate bits so far. You must realise that
the vertical scale in the diagrams of the two parts of the spectrum isn't the
same.
The overall mass spectrum looks like this:
m/e
Learning Objectives:
Candidates should be able to:

Analyse mass spectra in terms of isotopic abundances.

Calculate the relative atomic mass of an element given the relative
abundances of its isotopes, or its mass spectrum.
Reference:
A-level Chemistry: page 17.
Chemistry in Context: pages 2-4.
AS Chemistry
Unit 1: Theoretical Chemistry
Part 9: Molarity
Task 1
Can you solve the anagrams given in bold below to make sense of the notes?
A ouinslto consists of a substance dissolved in a lnotesv. The dissolved
substance is called the euostl and may be a solid, a liquid or a gas. The solvent is
almost always a liquid, although dsoil solutions can occur. The oonnccttneari of a
solution measures how much of a sdsdilove substance is present per unit umvole
of a solution. A solution is described as ‘econdarecntt’ if it consists of a large
aquintty of solute in a llams quantity of solvent. A small quantity of solute in a
large quantity of solvent is described as ‘tidelu’. Do not use the words strong or
awke, as these have specific meanings in chemistry regarding dicas and abses.
Chemists usually express the concentration of a solution in terms of its molar
concentration (or molarity). By definition, the molar concentration (c) of a
solution is equal to the amount in moles of a solute (n) divided by the volume (V)
of the solution, i.e.
Molar concentration = no. of moles / volume
or
c=n/V
The units of molar concentration are moles per cubic decimetre (mol dm -3).
Often a 1.0 mol dm-3 solution is referred to as a 1.0 M (molar) solution.
Task 2
Worked example 1
250cm3 of a solution contains 5.85g of sodium chloride. Calculate the molar
concentration of sodium chloride in mol dm-3.
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Unit 1: Theoretical Chemistry
Worked example 2
The concentration of calcium ions in a sample of spring water is 100mg dm -3.
Calculate the molar concentration of the calcium ions in mol dm-3.
Learning Objectives:
Candidates should be able to perform calculations, including use of the mole
concept, involving volumes and concentrations of solutions.
Reference:
A-level Chemistry: pages 24-28.
Chemistry in Context: page 7.
AS Chemistry
Unit 1: Theoretical Chemistry
AS Chemistry
Unit 1: Theoretical Chemistry
Part 10: Empirical and molecular formulae
The empirical formula is the formula which represents the simplest ratio of
atoms of each element in a compound.
The molecular formula gives the actual number of atoms of each element in a
molecule (or the number of moles of each type of atom in 1mol of the
compound.)
E.g. Ethane has the molecular formula C2H6 and the empirical formula CH3.
Task 1
Complete the table below:
Substance
Molecular formula
Ethane
Benzene
Butane
Phosphorus (V) oxide
P4O10
Empirical formula
Task 2
Find the molecular formula for each of the following compounds from the
empirical formula and the relative molecular mass:
Empirical
formula
CF2
C2H4O
CH3
CH
Mr
Molecular
formula
100
88
30
78
Empirical
formula
CH2
CH3O
CH2Cl
C2HNO2
Mr
Molecular
formula
42
62
99
213
Calculation of empirical formulae
The empirical formula of a compound can be calculated from data which give the
percentage composition, by mass, of each element in the compound. This data is
determined by elemental analysis. You are not expected to know any details of
this method but a brief outline is given on page 381 of ‘Chemistry in Context’
and on page 21 of ‘A-level Chemistry’.
Example
AS Chemistry
Unit 1: Theoretical Chemistry
A compound containing carbon, hydrogen and oxygen gave, after elemental
analysis, the following percentages by mass: C 40%; H 6.7%. Find the empirical
formula.
Strategy
 Find the percentage of oxygen: 100 – (40 + 6.7) = 53.3 %

Assume there are 100g present. This gives the masses of the elements
as: C 40g; H 6.7g and O 53.3g.

Use Ar to calculate mole ratios: C 40/12.0 = 3.3; H 6.7/1.0 = 6.7; O
53.3/16.0 = 3.3.

Find the simplest mole ration by dividing through by the smallest number:
C 3.3/3.3 = 1; H 6.7/3.3 = 2 O 3.3/3.3 = 1

Give empirical formula: CH2O.
Calculation of molecular formulae
The molecular formula can be deduced from the empirical formula if the
relative molecular mass is known. A value for Mr can be determined from the
Ideal Gas Equation or from a mass spectrum. If the compound analysed above
has Mr = 180, the molecular formula can be calculated as follows.

Empirical formula mass of CH2O is 12.0 + 2.0 + 16.0 = 30.0

The ratio of Mr : empirical formula mass = 180:30 = 6:1

Therefore, in comparison with the empirical formula, the molecular
formula must contain 6 times the number of atoms.

Therefore the molecular formula is 6 x CH2O = C6H12O6.
AS Chemistry
Unit 1: Theoretical Chemistry
Worked example 1
An oxide of copper has the following composition by mass: Cu, 0.635g; O, 0.08g.
Calculate the empirical formula of the oxide.
Worked example 2
Compound X contains only boron and hydrogen. The percentage by mass of boron
in X is 81.2%. In the mass spectrum of X the peak at the largest value of m/z
occurs at 54.
(i) Use the percentage by mass data to calculate the empirical formula of X.
(ii) Deduce the formula of X.
Worked example 3
On complete combustion of 0.400g of a hydrocarbon, 1.257g of carbon dioxide
and 0.514g of water were produced.
(i) Calculate the empirical formula of the hydrocarbon.
(ii) If the Mr of the hydrocarbon is 84, what is its molecular formula?
AS Chemistry
Unit 1: Theoretical Chemistry
Worked example 4
2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g
of CO2 and 1.44g of H2O. The relative molecular mass of the compound was
found to be 60.
(i) What are the masses of carbon, hydrogen and oxygen in 2.4g of the
compound?
(ii) What are the empirical and molecular formulas of the compound?
Learning Objectives:
Candidates should be able to:

Define the terms empirical and molecular formulae.

Calculate empirical and molecular formulae, using combustion data or
composition by mass.
Reference:
A-level Chemistry: pages 21-22.
Chemistry in Context: page 6.
AS Chemistry
Unit 1: Theoretical Chemistry
Problems on empirical formulae
Please note: Q.1 should read 0.100 mol of nitrogen oxide
AS Chemistry
Unit 1: Theoretical Chemistry
Part 11: The Ideal Gas Equation
Chemists use chemical equations to work out the masses of reactants and
products involved in a reaction. If one or more of these is a gas, it is sometimes
more useful to know its volume rather than its mass.
This can be achieved using the Ideal Gas Equation:
pV = nRT
where:
p is the pressure of the gas in pascals, Pa;
V is the volume of the gas in cubic metres, m3 (1 m3 = 1000 dm3);
n is the amount of gas in moles;
R is a constant for all gases called the gas constant. Its value in SI
units is 8.31 JK-1mol-1;
T is temperature in Kelvin, K. (To convert from oC to K add 273.)
N.B. SI units should be used in calculations involving the gas equation.
Example
What is the volume, given in dm3, of 1 mol of an ideal gas at 20oC and 100kPa?
(This combination of temperature and pressure is often called ‘room
temperature and pressure’.)
pV = nRT
V = nRT/p
V = 1 x 8.31 x (20 + 273) / 1 x 105
V = 0.0243 m3 = 24.3 dm3 (usually approximated to 24 dm3)
Additional equations:

Moles = gas volume (cm3)/24000

Mr = mRT/pV
AS Chemistry
Unit 1: Theoretical Chemistry
Worked example 1
Calculate the volume of 0.5 mol of gas at 1x105 Pa and 20oC.
Worked example 2
What will be the volume of the gas in question 1 when the temperature is raised
to 100oC?
Worked example 3
Calculate the pressure inside the cathode-ray tube in a television set, assuming
that it contains 3.6 x 10-7 mol of gas in a volume of 5.0 dm3 at 20oC.
AS Chemistry
Unit 1: Theoretical Chemistry
Ideal gas
Real gases are complicated things. Scientists often use a simplified model of a
gas – an ideal gas.
Key assumptions:
•
the molecules have mass, but negligible size;
•
there are no intermolecular forces;
•
collisions between particles are perfectly elastic.
Limitations:
Gases deviate from ideal behaviour at:
•
Low temperatures – intermolecular forces become significant;
and
•
High pressures – volume is no longer negligible.
(In both these cases they are less like gases and more like liquids. We will
return to this topic in more detail later.)
Problems on the ideal gas equation (use R=8.31 JK-1mol-1)
1. A gas has a density of 7.149 g dm-3 at 50oC and 3.00 x105 Pa. What is its
molar mass?
2. A volume 485cm3 of a gas measured at 40oC and 200kPa has a mass of
1.715g. Calculate the molar mass of the gas.
3. 0.130g of a liquid vaporised at 100oC and 101kPa occupied a volume of 85.0
cm3. Calculate its relative molecular mass.
If the percentage composition of the liquid is C 52.2%, H 13.0%, O 34.8%,
calculate the empirical formula of the compound. Can you suggest its
identity?
AS Chemistry
Unit 1: Theoretical Chemistry
4. A compound of phosphorus and fluorine contains 24.6% by mass of
phosphorus. 1.000g of this compound has a volume of 196cm3 at 300K and
101kPa. Deduce the molecular formula of the compound.
Learning Objectives:
Candidates should be able to state and use the general gas equation pV=nRT in
calculations, including the determination of Mr.
Reference:
A-level Chemistry: page 17.
Chemistry in Context: pages 2-4.
AS Chemistry
Unit 1: Theoretical Chemistry
Part 12: Calculations based on chemical equations
Many of the calculations you will come across at AS level can be solved using the
same basic strategy:
Write a balanced equation for the reaction.
1. You will have enough information about one substance to calculate its
amount in moles.
You can then use the balanced chemical equation to state the mole ratios of
the substances in the equation. These are known as the stoichiometric
amounts.
2. You can then find the amount in moles of the substance in which you are
interested (the unknown).
3. You can then use the amount in moles to find the missing information
about that substance.
Steps 1 and 3 will generally involve one of the following equations:
n
m
Mr
or
n = c.V
or
n = gas volume (cm3)/24000
Task 1 - An easy example
What mass of magnesium oxide is formed when 6g of magnesium are burned in
excess oxygen?
Answer:
AS Chemistry
Unit 1: Theoretical Chemistry
Three-step strategy
Step 1
Step 2
Step 3
Masses
Masses
Balanced
equation
Volume of a gas
Moles of
Moles of
‘known’
‘unknown’
Volume of gases
Volumes or
Volumes or
molarities of
molarities of
solutions
solutions
Key equations:
n
m
Mr
or
n = c.V
or
n = gas volume (cm3)/24000
AS Chemistry
Unit 1: Theoretical Chemistry
Learning Objectives:
Candidates should be able to:

write and/or construct balanced equations

perform calculations, including use of the mole concept, involving:
o reacting masses (from formulae and equations)
o volumes of gases (e.g. in the burning of hydrocarbons)
o volumes and concentrations of solutions

deduce stoichiometric relationships from calculations such as those
above.
References:
A-level Chemistry: page 23-27.
Chemistry in Context: pages 10-18.
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