### Calculations_in_Chemistry_2 124KB

```Calculations in Chemistry
•
You need to know how to carry out several calculations in Additional
and Triple Chemistry
•
This booklet gives you a step by step guide to carrying out these
equations and examples of each one. Follow the same rules for the
questions in your exam and chemistry calculations are actually quite
easy, honest!
•
These are:
• Relative formula mass
• Mole masses
• Percentage of element in a compound
• Formula of a compound from its percentage composition
• Masses of reactants and products
• Percentage yield
• Atom economy
– Triple:
• Titration calculations
• Bond energy calculations
Calculations
Relative atomic and Formula Masses
The mass of an atom is too small to deal with in real terms, so we use
‘relative’ masses – Carbon is given a mass of 12, and everything else is
compared with it and given a mass, e.g. Oxygen is ‘heavier’, so its
relative mass is 16.
The relative atomic mass can be found by looking at
the periodic table, It is always the larger of the two
numbers.
Relative formula mass can be found by adding up the relative atomic
masses of each element in a compound.
E.g. Carbon Dioxide (CO2)
Carbon has a relative atomic mass of 12
Oxygen has a relative atomic mass of 16
The relative formula mass of Carbon Dioxide is therefore:
12 + (16 x 2) = 44
x 2 because
it’s “O2”
Moles
Because saying ‘relative formula (or atomic!) mass in grams’ is a bit
clumsy, we simply say ‘moles’ instead. This means that 1 mole of Carbon
Dioxide is 44 grams, or 44g. Simple!
Percentage of an element in a compound
We can use the relative atomic mass (Ar) of elements and the relative
formula mass (Mr) of compounds to find out the percentage composition
of different elements.
E.g. What percentage mass of white Magnesium Oxide is actually
Magnesium, and how much is Oxygen?
1.
2.
3.
Work out the mass of MgO
24 + 16 = 40
Convert to grams
40g
1. Work out the formula mass of
the compound
2. Convert this into grams
3. Work out the percentage by
using this equation:
Mass of element
x 100%
Total mass of compound
Work out the percentage
24 X 100% = 60% is Magnesium, so 40% must be Oxygen!
40
Formula of a compound from its percentage composition
We can also do this backwards! If we know the percentage composition
of a compound we can work out the ratio of atoms. This is known as the
Empirical Formula. Sometimes this is the same as the molecular
formula, but not always (e.g. water has an empirical and molecular
formula of H2O. Hydrogen peroxide's empirical formula is HO, but it’s
molecular formula is H2O2.
E.g. If 9g of Aluminium react with 35.5g of Chlorine, what is the
empirical formula of the compound formed?
Aluminium
9
= 1/3 moles of Aluminium atoms
27
Chlorine
35.5
= 1 mole of Chlorine atoms
35.5
1. Divide the mass of each
element by its relative
atomic mass to find out the
number of moles reacted
2. Create a ratio and simplify if
necessary
3. Write a formula based on
the ratio
Al : Cl
1/3 : 1
1 : 3
AlCl3
Masses of reactants and products
This is an important calculation when we want to know how much of each
reactant to react together. For example, sodium hydroxide reacts with
chlorine gas to make bleach. If we have too much Chlorine, some will be
wasted. Too little and not all of the sodium hydroxide will react.
2NaOH + Cl2  NaOCl + NaCl + H2O
Sodium Chlorine Bleach
Hydroxide
Salt
Water
How much Chlorine gas should we bubble through 100g of Sodium
Hydroxide to make Bleach?
1. NaOH
23 + 16 + 1 = 40g is one mole of NaOH
2. We have 100g in our reaction so…
100 = 2.5 moles
40
1. Work out the mass of one
mole of Sodium Hydroxide
2. Calculate how many moles
3. Work out how many moles
of Chlorine you need
4. Convert this into a mass
for Chlorine
3. The chemical equation tells us that we need 2 moles of Sodium
Hydroxide (2NaOH) for every mole of Chlorine (Cl2).
So we need: 2.5 = 1.25 moles of Chlorine
2
4. 35.5 x 2 = 71g is one mole of Cl2
So we need 1.25 x 71 = 88.75g of Chlorine to react with 100g of
Sodium Hydroxide.
Percentage Yield
Rather than talk about the yield of a chemical reaction in terms of mass
(grams, tonnes etc.) we can talk about the percentage yield. This gives
us an idea of the amount of product that the reaction really makes,
compared to what it could possibly make under perfect conditions.
There are many reasons why we don’t make 100% every time, such as:
–
–
–
–
The reaction may be reversible
Some product could be left behind in the apparatus
The reactants may not be pure
It may be difficult to separate the products if more than one are made.
Using this reaction “A + B  C”, it was found that in perfect conditions,
scientists could make 2.5g of C. However, when they tried it out, they
only made 1.5. What is the percentage yield of this reaction?
Amount of product produced
x 100%
Maximum amount of product possible
1.5 x 100% = 60% percentage yield
2.5
The higher the
percentage yield and
atom economy, the
better the reactions are
for the Earth’s
resources, as there’s less
waste!
Percentage Atom Economy
Sometimes reactions will make more than one product. Working out the
atom economy is a way of finding out how much of the reactant ends up
in the useful product.
Relative formula mass of useful product x 100%
Ethanol (C2H5OH) can be converted into ethene (C2H4) to make
polyethene. However, the reaction also produces water.
C2H5OH  C2H4 + H2O
Mass of useful product = (12 x 2) + (1 x 2) = 28
Mass of all products = 28 + [(1x4) + 16] = 28 + 20 = 48
28
48
x 100% = 58% atom economy
Triple Chemistry Calculations
Titration Calculations
Titrations are used to find out the volumes of solutions that react
exactly. If you know the concentration of one of the solutions, and the
volumes that react together are known as well, you can use this to work
out the concentration of the other solution.
It is important that you know that mol/dm3 is a term for concentration of a
solution. It means that X moles of substance are dissolved in 1000cm3 (1dm3) of a
solvent.
1mol/dm3 of Sodium Hydroxide means that 40g (the mass of one mole) is dissolved
in 1000cm3 of the solvent. 20g dissolved in the same volume would be 0.5mol/dm3.
E.g. a student put 25.0cm3 of sodium hydroxide solution with an
unknown concentration into a conical flask using a pipette. The sodium
hydroxide reacted with exactly 20.0cm3 of 0.50mol/dm3 hydrochloric
acid added from a burette. What was the concentration of the sodium
hydroxide solution?
1.
The equation for this reaction is:
Calculate the number of moles
reacted in the known substance:
(Concentration/1000) x volume used
NaOH + HCl  NaCl + H2O
0.50 moles of Hydrochloric acid are dissolved in 1000cm3 of acid. We
used 20cm3 in this reaction. So…
0.50
X 20 = 0.10 moles of HCl in 20cm3 of acid.
1000
Titration Calculations continued…
We know from the equation that 1 mole of HCl reacts with 1 mole of
NaOH. So, if there is 0.10 moles of HCl in the reaction, there must be
0.10 moles of NaOH as well.
Now we can work out the concentration of NaOH.
If 0.10 moles of NaOH are in 25cm3
of solution, we can use the equation in the
box on the right to work out how many are
in 1000cm3, therefore the concentration!
2.
Use the volume of the
second unknown solution to
work out the
concentration in 1dm3
(number of moles/volume
of solution) x 1000
0.10 X 1000 = 0.40 moles of NaOH in 1000cm3 of solution = 0.4 mol/dm3
25
This is the concentration!
And Finally…
Bond Energy Calculations
The energy needed to break a bond between two atoms is called the
bond energy. It is measured in kJ/mol – kilojoules per mole. We can use
this to work out the energy change (ΔH) for chemical reactions.
However, we need a list of the bond energies to do this:
Bond
Bond energy
(kJ/mol)
Bond
Bond energy
(kJ/mol)
C-C
347
H-Cl
432
C-O
358
H-O
464
C-H
413
H-N
391
C-N
286
H-H
436
C-Cl
346
O=O
498
Cl-Cl
243
NΞN
945
You don’t need to
remember this!
To calculate bond energy reactions we need to know two things:
–
–
How much energy is needed to break the chemical bonds in the reactants
How much energy is released when the new bonds are formed in the products
e.g. Hydrogen and Nitrogen are used in the Haber process to make
ammonia. What is the overall energy change in this reaction?
3H2 + N2 = 2NH3
2. Energy needed to break one mole of
nitrogen, and three moles of hydrogen
945 + (3 x 436) = 2253 kJ/mol
3. Energy released when 3 N-H bonds are
formed in one mole of ammonia = 3 x -391 = 1173
Energy released in two moles of ammonia = 2
x -1173 = -2346 kJ
Overall energy change = +2253 – 2346 = -93kJ
1. Draw out the structure of each
atom so you can work out what
bonds are involved
2. Work out the energy needed to
break the reactants into atoms.
3. Work out the energy released
when the bonds in the products
are formed
4. Work out the energy change
The bond energies are:
NΞN = 945 kJ/mol
H-H = 391 kJ/mol
```