This is free sample of the OCR AS Chemistry revision guide and contains the first two topics of Module 2.To get access to the complete revision notes, you can sign up for a premium membership here:https://alevelchemistry.co.uk/membership

Module 1 – Development of practical skills in chemistry ( not covered in this booklet)

• Practical skills assessed in a written examination

Module 2 – Foundations in chemistry

• Atoms, compounds, molecules and equations

• Amount of substance

• Acid–base and redox reactions

• Electrons, bonding and structure

Module 3 – Periodic table and energy

• The periodic table and periodicity

• Group 2 and the halogens

• Qualitative analysis

• Enthalpy changes

• Reaction rates and equilibrium (qualitative)

Module 4 – Core organic chemistry

• Basic concepts

• Hydrocarbons

• Alcohols and haloalkanes

• Organic synthesis

• Analytical techniques (IR and MS)

Module 2 – Foundations in chemistry Atoms elements, compounds and formulae

The ratio of the average mass of one atom of an element to one twelfth of the mass of an atom of carbon-12.

Key formula and triangles you need to know:

Definitions Mole: The mole is the amount of substance in grams that has the same number of particles as there are atoms in 12 grams of carbon-12.

Relative atomic mass is the average mass of one atom compared to one twelfth of the mass of one atom of carbon-12

Molar Mass is the mass in grams of 1 mole of a substance and is given the unit of g mol-1

Used for pure substances For gases

amount = mass /Molar Mass Gas Volume (dm3 )= amount x 24

This equation give the volume of a gas at room pressure (1atm)

and room temperature 25oC.

For solutions

Concentration = amount/ volume

Unit of concentration: mol dm-3 or M

Unit of Volume: dm3

Converting volumes: cm3 dm3 ÷ 1000 cm3 m3 ÷ 1000 000 dm3 m3 ÷ 1000

Examples

1) How many moles are there in each of the following?

a) 72.0 g of Mg b) 4.00 kg of CuO c) 39.0 g of Al(OH)3 d) 1.00 tonne of NaCl

e) 20.0 mg of Cu(NO3)2

2) What is the mass of each of the following?

a) 5.00 moles of Cl2

b) 0.200 moles of Al2O3

c) 0.0100 moles of Ag

d) 0.00200 moles of (NH4)2SO4

e) 0.300 moles of Na2CO3.10H2O

3) a) Calculate the number of moles of CO2 molecules in 11.0 g of carbon dioxide.

b) Calculate the number of moles of C atoms in 11.0 g of carbon dioxide.

a) Calculate the number of moles of O atoms in 11.0 g of carbon dioxide.

4) a) Calculate the number of moles of Al2O3 in 5.10 g of Al2O3.

b) Calculate the number of moles of Al3+ ions in 5.10 g of Al2O3.

a) Calculate the number of moles of O2- ions in 5.10 g of Al2O3.

5) An experiment was carried out to find the Mr of vitamin C (ascorbic acid). It was found that 1.00 g contains 0.00568 moles of Vitamin C molecules.

Calculate the Mr of vitamin C.

Answers:

1 a 2.96 b 50.3 c 0.500 d 17100 e 0.000107

2 a 355 g b 20.4 g c 1.08 g d 0.264 g e 85.8g

3 a 0.250 b 0.250 c 0.500

4 a 0.0500 b 0.100 c 0.150

5. 176

Avogadros constant

Remember: Avogadro's Constant There are 6.02 x 1023 atoms in 12 grams of carbon-12. Therefore explained in simpler terms 'One mole of any specified entity contains 6.02 x 1023 of that entity'

No of particles = amount of substance (in mol) X Avogadro's constant

Example

How many chloride ions are there in a 25.0 cm3 of a solution of magnesium chloride of concentration 0.400 moldm-3 ?

Number ions of Cl- = amount x 6.02 x 1023 = 0.0200 x 6.02 x 1023 = 1.204 x1022

amount= concentration x Volume = 0.400 x 0.025 = 0.0100 mol

MgCl2 Amount of chloride ions = 0.0100 x2 = 0.0200

There are two moles of chloride ions for every one mole of MgCl2

Ideal gas equation

Used to calculate volume of gas given the pressure, temperature and number of moles.

Example: What is the mass of Cl2 gas that has a pressure of 100kPa, temperature 293K, volume 500cm3 . (R = 8.31 JK–1mol–1 )

1. moles = PV/RT = 100 000 x 0.0005 / (8.31 x 293) = 0.0205 mol 2. 100 kPa = 100 000 Pa 3. 500 cm3 = 0.0005 m3 4. Mass = moles x Mr = 0.0205 x (35.5 x2) = 1.46 g

Questions 1) Convert the following into SI units.

a) 200ºC

b) 98 kPa

c) 50 cm3

d) -50ºC

e) 0.1 MPa

f) 3.2 dm3

2) Calculate the volume that 0.400 moles of an ideal gas occupies at 100ºC (3sf) and a pressure of 1000 kPa (4sf).

3) How many moles of gas occupy 19400 cm3 at 27.0ºC and 1.00 atm pressure?

4) Calculate the pressure that 0.0500 moles of gas, which occupies a volume of 200 cm3 (3sf) exerts at a temperature of 50.0 K.

5) 0.140 moles of a gas has a volume of 2.00 dm3 at a pressure of 90.0 kPa. Calculate the temperature of the gas.

6) At 273 K and 101000 Pa, 6.319 g of a gas occupies 2.00 dm3 . Calculate the relative molecular mass of the gas.

7) Find the volume of ethyne (C2H2) that can be prepared from 10.0 g of calcium carbide at 20.0ºC and 100 kPa (3sf).

CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g)

8) What mass of potassium chlorate (V) must be heated to give 1.00 dm3 of oxygen at 20.0ºC and 0.100 MPa. 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

9) What volume of hydrogen gas, measured at 298 K and 100 kPa, is produced when 1.00 g of sodium is reacted with excess water?

2 Na + 2 H2O o 2 NaOH + H2(g)

10) What volume of carbon dioxide gas, measured at 800 K and 100 kPa, is formed when 1.00 kg of propane is burned in a good supply of oxygen? C3H8 + 5 O2 o 3 CO2 + 4 H2O

Answers: 1 (a) 473 K (b )98000 Pa (c) 50 x 10-6 m 3 (d) 223 K (e)100000 Pa (f)3.2 x 10-3 m 3 (2) 1.24 x 10-3 m 3 (3) 0.786 (4) 104000 Pa (5) 155 K (6) 71.0 (7) 0.00380 m3 (8) 3.36 g (9) 0.000538 m3 (10) 4.53 m3

Empirical formula Definition: Simplest ration of elements in a formula

Method:

Worked example:

Atomic mass could be given as a percentage at times.

Use the percentages to represent atomic mass numbers

Molecular formula (Mr)

Molecular formula can be worked out from percentage composition/empirical formulae.

work out the molecular formula for the compound with an empirical formula of C3H6O and a Mr of 116

C3H6O has a mass of 58

The empirical formula fits twice into Mr of 116

So molecular formula is C6H12O2

Hydrated salts and calculation of unknown A Hydrated salt contains water of crystallisation

Cu(NO3 )2 .6H2O

hydrated copper (II) nitrate(V).

Cu(NO3 )2

Anhydrous copper (II) nitrate(V)

Th amount of water in hydrated salts can be measured by measuring the mass of the salt and crucible then heating the salt and measuring the end mass (Mass reading that stays constant)

Key precautions have to be taken to increase reliability and accuracy of results:

x Using too much salt will lead to incomplete decomposition x Apparatus has to be dried before use x Covering the crucible improves accuracy of results x Small amount of salt should not be used to minimise weighing errors

Volume of gas reactions

Simple ratio can be used to calculated volumes of gasses. This is based on the ideal gas law therefore 1 mol of gas at the same pressure and temperature will occupy the same volume.

An important reaction which occurs in the catalytic converter of a car is

2CO(g) + 2NO(g) ¡ 2CO2 (g) + N2 (g)

In this reaction, when 500 cm3 of CO reacts with 500 cm3 of NO at 650 °C and at 1 atm. Calculate the total volume of gases produced at the same temperature and pressure ?

Ratio of NO to N2 IS 2:1 therefore volume of N2 produced is 250 cm3.

Ratio of CO to CO2 is 1:1 so 500 cm3 of CO2 is produced.

In total: 750cm3 of gas is generated from 1000cm3 of reactants.

Gas syringes in experiments If drawing a gas syringe make sure you draw it with some measurement markings on the barrel to show measurements can be made.

The volume of a gas depends on pressure and temperature so when recording volume it is important to note down the temperature and pressure of the room.

Make sure you don’t leave gaps in your appparatus where gas could escape

Gas syringes can be used for a variety of experiments where the volume of a gas is measured, possibly to work out moles of gas or to follow reaction rates. Moles of gas can be calculated from gas volume (and temperature and pressure) using ideal gas equation

Potential errors in using a gas syringe

•gas escapes before bung inserted

•syringe sticks

• some gases like carbon dioxide or sulphur dioxide are soluble in water so the true amount of gas is not measured.

Making volumetric solutions x A prior calculation is carried out to estimate the approximate mass of the primary standard

required to make up a known volume of standard solution.

x The primary standard is accurately weighed out on an electronic balance using a weighing bottle

or boat.

x The solid is transferred to the volumetric flask and the weighing bottle rinsed into the volumetric

flask.

x A volume of distilled water is added to the flask and the mixture swirled until all of the solute

has dissolved.

x Distilled or deionised water is added to the mark (an engraved line on a volumetric flask).

x The volumetric flask is inverted several times to thoroughly mix the contents.

x The flask is labelled and set to one side.

Example: Preparation a standard solution of sodium carbonate

Sodium carbonate has the formula Na2CO3. It has a relative formula mass of 106

If we wish to prepare 250 ml of a 0.1 mol dm-3 solution then we need a total of 0.25 x 0.1 = 0.025 moles = 0.025 x 106 g = 2.65 g

Approximately 2.65 g is accurately weighed on an electronic balance in a weighing bottle.

The sodium carbonate is transferred into a 250 ml volumetric flask and about 100 ml of deionised water is added, rinsing out the weighing bottle.

The mixture is shaken in the volumetric flask until the sodium carbonate dissolves.

Deionised water is added to the volumetric flask up to the mark.

% yield in a process can be lowered through incomplete reactions, side reactions, losses during transfers of substances, losses during purification stages

Do take into account balancing numbers when working out % atom economy.

High atom economy minimises waste in chemistry. Alternatively any waste can sold as bi products to maximise return.

In reactions that produce only one product, the atom economy is 100% making it ideal economically.

Experimental

25 ml of dilute base are pipetted into a conical flask. A few drops of indicator solution are added. Dilute acid is added from burette a few drops at a time with swirling until the end-point is reached (colour change).

The titration is repeated until concordant results are obtained.

Typical results - titration of dilute hydrochloric acid and sodium hydroxide

x Molarity of sodium hydroxide = 0.098 mol dm-3

x Molarity of hydrochloric acid = approximately 0.05 - 0.2 mol dm-3

Calculation

Moles of sodium hydroxide in 25.0 ml = 0.025 x 0.098 = 2.45 x 10-3

Equation for the reaction:

NaOH + HCl NaCl + H2O

Hence, moles of acid = moles of base at the end-point.

Moles of acid = 2.45 x 10-3

Volume of acid from titration = 0.01845 dm3

Hence, concentration of acid = moles/volume = 2.45 x 10-3/ 0.01845 = 0.133 mol dm-3

Evaluation

The procedure of titration is very accurate when the apparatus is used correctly, but even so each step has a corresponding inaccuracy due to the manufactured tolerance of the glassware.

For example:

Grade B 25 ml pipette has an accuracy of ± 0.04 ml

This corresponds to a percentage inaccuracy of 100 x 0.04/25 = ± 0.16 %

The burette readings have an inaccuracy of one half of the minimum reading = ± 0.05 ml

Two readings per titre gives an inaccuracy of ± 0.1, corresponding to a percentage inaccuracy of 100 x 0.1/18.45 = ± 0.54% (this titration)

All of the percentage inaccuracies may be added in a procedure and converted back to an absolute inaccuracy of the final 'answer'.

Total percentage = 0.16 + 0.54 = 0.70 %

Hence, the calculated molarity of hydrochloric acid = 0.133 ± 0.001 mol dm-3

Note that only the two steps have been taken into account in this example. In a real experiment you would also have to take into account inaccuracy in any solutions prepared, masses weighed out etc.

Examples of acid base curves

a) Strong acid – strong base

b) Strong acid- weak base

Weak acid – strong base

Uncertainity Key terms:

Measurements: the values taken as the difference between the judgements of two values (e.g. using a burette in a titration)

Readings: the values found from a single judgement when using a piece of equipment

The uncertainty of a reading (one judgement) is at least ±0.5 of the smallest scale reading. The uncertainty of a measurement (two judgements) is at least ±1 of the smallest scale reading.

Each type of apparatus has a specific uncertainity.e.g

Balance +/- 0.001g

Volumetric flask +/- 0.1cm3

Burette +/- 0.05cm3

25cm3 pipette +/- 0.1cm3

Decreasing the sensitivity by using higher resolution apparatus will reduce uncertainties.

In burette use, two readings are taken to decrease uncertainty e.g if a buretteused in a titration has an uncertainty of +/-0.05 then the uncertainty would be +/-0.10 after taking two readings.

Reducing uncertainties in titrations

x Decrease concentration of substance in burette OR x Increase volume and concentration of substance in the conical flask x Avoid use of measuring cylinders and stick to using burettes and pipettes

Reducing uncertainties in mass measurements

x Use a larger mass x More accurate balance x Calculating mass difference

If the %uncertainty due to the apparatus < percentage difference between the actual value and the calculated value then there is a discrepancy in the result due to other errors.

If the %uncertainty due to the apparatus > percentage difference between the actual value and the calculated value then there is no discrepancy and all errors in the results can be explained by the sensitivity of the equipment.