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FORMULAE, EQUATIONS AND MOLES
Important terms
In order to use formulae and equations it is necessary to understand the meaning of important
terms
...

Ion
An atom or group of atoms possessing a negative charge
Element
A substance containing just one type of atom
...

Molecular
The actual numbers of each type of atom in a molecule of the
formulae
substance
...
This is done using the mole
...

If we take a mole of atoms of a particular element, the mass will be the Relative Atomic Mass
for that element
...
023 x 1023)
The mole is the amount of a substance that contains the same number of particles as there
are atoms in 12
...
This number of atoms is 6
...

Amount of substance = number of particles
Avogadro constant

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Moles from mass
To find the number of moles of atoms we use the formula:
Moles =

mass of substance
relative atomic mass

If we take a mole of ionic lattice or molecules of a material, the mass will be the molar mass
To find the number of moles of molecules or compounds we can use the formula:
Moles =

mass
molar mass (RFM, M r )

Examples
1
...
)
Molar mass of MgSO 4 = 24 + 32 + (4x 16) = 120
Moles = 10 /120 = 0
...
Calculate the mass of 0
...
5 + (2 x 14) + (6x 16) = 178
...
04 x 178
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5g

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Formulae determination
An empirical formula is the simplest formula which shows the ratio of each type of atom
present in a compound
...

The formula of a compound can be calculated from the mass of elements in it
...

Step 2:
From the number of moles find the simplest ratio
...

Example
A hydrocarbon consists of 14
...
2g hydrogen
...

C
14
...
2 mol

H
1
...
2 mol

Ratio = 1
:
1
So the empirical formula is CH
If we try to work out the structure of this we can see that this is not the formula of the
molecule
...

If the molar mass of this compound is 78gmol-1, what is the molecular formula?
CH has a mass of 13
the number of these units is 78/13 = 6
so the formula is C 6 H 6
A similar process is used to find formulae from percentage composition
...
00% Carbon, 6
...
33% Oxygen
...
Calculate the molecular formula
...
00 / 12
= 3
...
67 / 1
= 6
...
33 / 16
= 3
...
(This is Avogadro’s Law)
...
5 moles of oxygen
...
5 volumes oxygen
...
5 = 97
...

At room temperature and 1Atm this is 24dm3 (24000cm3)
...
4dm3)
...

Moles = Volume (in dm3)
24
Example
...

Moles = 480 / 24,000 = 0
...

Concentration =

Moles
Volume (in dm3)

The concentration of a solution can be stated as the mass of solute per cubic decimeter of
solution (g/dm3) or the amount in moles of a solute present in 1dm3 of solution (mol/dm3)
...

Examples
1
...
4g MgSO 4 in 500cm3 of solution
...
4 / 120 = 0
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02mol / 0
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04moldm-3
2
...
2 mol dm-3 solution
...
2 x 0
...
02mol
Mass = Moles x Formula mass
= 0
...
8g

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Parts per million
This is a way of expressing very dilute concentrations of substances (usually pollutants)
...

Usually describes the concentration of something in water or soil
...

or the volume of a gas pollutant in air
...

Calculate the mass of gold in the 250cm3 sample of river water
...
056 mg of gold
25 cm3 of water contains 0
...
4 mg = 1
...

Would you be at serious risk if you were exposed 380 cm3 of carbon monoxide in a 3500 dm3
room?
380 cm3 in 3500000 cm3 = 380 cm3 in 3
...
5 = 108
...

Whether using masses, volumes of solutions (of known concentration), or volumes of gases
the same general method can be used
...

Step 3: From the number of moles of this substance calculate the mass / volume /
concentration of the desired substance
...
84g of H 2 SO 4
...
84 / 98 = 0
...
08
Step 3 – Mass of Iron(II)sulphate = moles x RFM = 0
...
2g (3 sig
...
)
Example 2:
What mass of aluminium is needed to react with 7
...
5 = 0
...

So 1 mol of CuSO 4 reacts with ⅔ mol of Al
...
0439 mol CuSO 4 reacts with 0
...
0293 mol Al
Step 3 – Mass of Al = moles x RFM = 0
...
790 g (3 sig
...
)
Example 3:
Calculate the volume carbon dioxide gas which will be produced from the combustion of
2
...

C 3 H 8 + 5O 2
3CO 2 + 4H 2 O
Step 1 - Moles of propane used = mass / RFM = 2
...
05 mol
Step 2 - From equation 1mol propane forms 3mol carbon dioxide
So mole of carbon dioxide = 0
...
15 x 24,000 = 3,600cm3

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Titrations
A titration is a method of finding out how much of one material will react with how much of
another
...
A pipette-filler is added to the volumetric pipette
...
Some of the solution is drawn into the pipette
...

3
...

4
...

5
...

6
...
Some of the liquid will remain in the tip and this should be left
as the pipette is calibrated to allow for this
...
A suitable indicator should be added to the conical flask
8
...

9
...

10
...
Add the solution from the burette until the indicator
changes colour
...
This is the rough reading
...
Discard the contents of the flask and rinse it with tap water and then distilled water
...
Repeat the process, adding the solution from the burette fairly slowly with continual
stirring
...
When one drop changes the colour of the indicator, allow the
solution to drain down the sides of the burette before taking the reading
...
Accurate burette readings should be recorded to two decimal places, the second
decimal place being 0 or 5
...
Readings should continue to be taken in this way until one rough and two accurate
readings that are within 0
...

15
...


Titration calculations
Use the same three step method as before
...
45 cm3 of 0
...
0cm3 of sulphuric acid
...

2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O
Step 1 - Moles of NaOH = 23
...
2 = 0
...
00469 / 2 = 0
...
002345 / 0
...
0938 mol dm-3 (3 sig
...
)

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Percentage Yield
Chemical reactions are carried out in the laboratory and by industry to make new materials
...

The efficiency of the conversion process is measured using percentage yield or atom
economy
...
00g of the pure
acid, is reacted with an excess of copper oxide
...
37g of the hydrated copper sulphate, CuSO 4
...
Calculate the % yield
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00 / 98 = 0
...
0204mol
From equation 1mol sulphuric acid forms 1mol copper sulphate
So moles of copper sulphate formed = 0
...
5 + 32 + 64 + 90 = 249
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5 x 0
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09g
% Yield = Actual yield / theoretical x 100 = 4
...
09 x 100 = 85
...

Molar mass of useful product
% atom economy=

x 100
Total molar mass of starting materials

Taking the laboratory preparation of copper(II) sulphate from sulphuric acid and copper(II)
oxide as an example
...
5 + 16 = 79
...
5 + 32 + 64 = 159
...
5
159
...
9%
177
...

C 2 H 4 + H 2 O  C 2 H 5 OH
% atom economy = 46 / 46 x 100 = 100% (there are no unwanted products)
...

Equations can be written to explain simple test tube reactions such as displacement and
precipitation
...
74g of manganese(IV) oxide is added to an excess of concentrated hydrochloric acid and
warmed
...
The mass of chlorine given off
was 1
...
Another product of the reaction was purified and analysed
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10g of manganese and 1
...

(a) Determine the formula of the salt
...

(c) Calculate the number of moles of chlorine and salt produced in the reaction
...

(a)

Mn
1
...
02 mol
Ratio = 1

Cl
1
...
5
=0
...
74 / 87 = 0
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42 / 71 = 0
...
52 / 126 = 0
...
02
1



MnCl 2 + Cl 2
0
...
02
1
1

+

?

Balancing for Cl:
MnO 2 + 4HCl  MnCl 2 + Cl 2 + ?
Balancing for H & O: MnO 2 + 4HCl  MnCl 2 + Cl 2 + 2H 2 O
Example 2
For the following reactions, write chemical and ionic equations using state signs
...

Na 2 CO 3(aq) + 2HNO 3(aq)  2NaNO 3(aq) + H 2 O (l) + CO 2(g)
CO 3 2- (aq) + 2H+ (aq)

H 2 O (l) + CO 2(g)
(b) When iron filings are added to copper sulphate solution, the blue solution turns green
and the grey solid is replaced by a red-brown solid
...

Na 2 CrO 4(aq) + 2AgNO 3(aq)  2NaNO 3(aq) + Ag 2 CrO 4(s)
CrO 4 2- (aq) + 2Ag+

Ag 2 CrO 4(s)

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