Spec2017

1:01 understand the three states of matter in terms of the arrangement, movement and energy of the particles

Solid

Arrangement: Particles are close together and regularly packed.

Movement: Particles vibrate around a fixed point.

Energy: Particles have less kinetic energy than both liquids and gasses.

Liquid

Arrangement: Particles are close together but irregular.

Movement: Particles are free to move.

Energy: Particles have less kinetic energy than gasses but more than solids.

Gas

Arrangement: Particles are far apart and there are no forces between them.

Movement: Particles are free to move.

Energy: Particles have more kinetic energy than liquids and solids.

 

1:02 understand the interconversions between the three states of matter in terms of: the names of the interconversions, how they are achieved and the changes in arrangement, movement and energy of the particles

Melting: When a solid is heated, the energy makes the particles vibrate fast enough so that the forces of attraction between the particles break. For example   H2O(s) –> H2O(l)

Freezing: When a liquid is cooled, the particles move slow enough so that the forces of attraction between them will hold them into a solid. For example   H2O(l) –> H2O(s)

Boiling: When a liquid is heated strongly, the energy makes the particles move fast enough so that all forces of attraction are broken. For example   H2O(l) –> H2O(g)

Condensing: When a gas is cooled, the particles move slow enough so that the forces of attraction between them will hold them as a liquid. For example   H2O(g) –> H2O(l)

Sublimation: A small number of substances have the ability to change directly from a solid to a gas when heated. For example   CO2(s) –> CO2(g)

1:03 understand how the results of experiments involving the dilution of coloured solutions and diffusion of gases can be explained

Diffusion is the spreading out of particles in a gas or liquid. There is a net movement of particles from areas of high concentration to areas of low concentration until a uniform concentration is achieved.

 

i) dilution of coloured solutions

Dissolving potassium manganate(VII) in water demonstrates that the diffusion in liquids is very slow because there are only small gaps between the liquid particles into which other particles diffuse.

The random motion of particles cause the purple colour to eventually be evenly spread out throughout the water.

Adding more water to the solution causes the potassium manganate(VII) particles to spread out further apart therefore the solutions becomes less purple. This is called dilution.

 

ii) diffusion experiments

When ammonia gas and hydrogen chloride gas mix, they react together to form a white solid called ammonium chloride.

ammonia                  +              hydrogen chloride                 –>            ammonium chloride

NH3(g)                     +              HCl(g)                                     –>            NH4Cl(s)

A cotton wool pad was soaked in ammonia solution and another was soaked in hydrogen chloride solution. The two pads were then put into opposite ends of a dry glass tube at the same time.

The white ring of ammonium chloride forms closer to the hydrochloric acid end because ammonia particles are lighter than hydrogen chloride particles and therefore travel faster.

Even though these particles travel at several hundred metres per second, it takes about 5 min for the ring to form. This is because the particles move in random directions and will collide with air particles in the tube.

1:06 (Triple only) understand how to plot and interpret solubility curves

The solubility of solids changes as temperature changes. This can be plotted on a solubility curve.

Image result for solubility curve

The salts shown on this graph are typical: the solubility increases as temperature increases.

For example, the graph above shows that in 100g of water at 50⁰C the maximum mass of potassium nitrate (KNO₃) which will dissolve is 80g.

However, if the temperature were 80⁰C a mass of 160g of potassium nitrate (KNO₃) would dissolve in 100g of water.

1:07 (Triple only) practical: investigate the solubility of a solid in water at a specific temperature

At a chosen temperature (e.g. 40⁰C) a saturated solution is created of potassium nitrate (KNO₃) for example.

Some of this solution (not any residual solid) is poured off and weighed. The water is then evaporated from this solution to leave a residue of potassium nitrate which is then weighed.

The difference between the two measured masses is the mass of evaporated water.

The solubility, in grams per 100g of water, is equal to 100 times the mass of potassium nitrate residue divided by the mass of evaporated water.

 solubility (g/100g) = \frac{mass Of Solute}{mass Of Solvent} \times 100

1:08 understand how to classify a substance as an element, a compound or a mixture

Element: The simplest type of substances made up of only one type of atom.

Compound: A substance that contains two or more elements chemically joined together.

Mixture: Different substances in the same space, but not chemically combined.

Note: elements such as oxygen (O2) are described as diatomic because they contain two atoms.

The full list of elements that are diatomic is:

  • Hydrogen (H2)
  • Nitrogen (N2)
  • Fluorine (F2)
  • Oxygen (O2)
  • Iodine (I2)
  • Chlorine (Cl2)
  • Bromine (Br2)

1:09 understand that a pure substance has a fixed melting and boiling point, but that a mixture may melt or boil over a range of temperatures

Pure substances, such as an element or a compound, melt and boil at fixed temperatures.

However, mixtures melt and boil over a range of temperatures.

Example: although pure water boils at 100⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will raise the boiling point to 100.2⁰C.

Example: although pure water melts at 0⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will lower the melting point to -0.6⁰C.

1:10 describe these experimental techniques for the separation of mixtures: simple distillation, fractional distillation, filtration, crystallisation, paper chromatography

Simple distillation

This method is used to separate a liquid from a solution. For example: separating water from salt water.

The salt water is boiled. The water vapour condenses back into a liquid when passed through the condenser. The salt is left behind in the flask.

Note: cold water is passed into the bottom of the condenser and out through the top so that the condenser completely fills up with water.

 

Fractional distillation

This method is used to separate a mixture of different liquids that have different boiling points. For example, separating alcohol from a mixture of alcohol and water.

Water boils at 100oC and alcohol boils at 78oC. By using the thermometer to carefully control of temperature of the column, keeping it at 78oC, only the alcohol remains as vapour all the way up to the top of the column and passes into the condenser.

The alcohol vapours then condense back into a liquid.

 

 

Filtration

This method is used to separate an insoluble solid from a liquid. For example: separating sand from a mixture of sand and water.

The mixture is poured into the filter paper. The sand does not pass through and is left behind (residue) but the water passes through the filter paper and is collected in the conical flask (filtrate).

 

 

Crystallisation

This method is used to obtain a salt which contains water of crystallisation from a salt solution. For example: hydrated copper sulfate crystals (CuSO4.5H2O(s)) from copper sulfate solution (CuSO4(aq)).

Gently heat the copper sulfate solution in an evaporating basin until a hot saturated solution forms. Leave in a warm place to allow the hydrated copper sulfate crystals to form. Remove the crystals by filtration and wash with distilled water. Dry by leaving in a warm place.

If instead the solution is heated until all the water evaporates, you would produce a powder of anhydrous copper sulfate (CuSO4(s)).

 

Paper chromatography

This method can be used to separate the parts of a mixture into their components. For example, the different dyes in ink can all be separated and identified.

The coloured mixture to be separated (e.g. a food dye) is dissolved in a solvent like water or ethanol and carefully spotted onto the chromatography paper on the baseline, which is drawn in pencil so it doesn’t ‘run or smudge’.

The paper is carefully dipped into the solvent and suspended so the baseline is above the liquid solvent, otherwise all the spots would dissolve in the solvent. The solvent is absorbed into the paper and rises up it as it soaks into the paper. The choice of solvent depends on the solubility of the dye. If the dye does not dissolve in water then normally an organic solvent (e.g. ethanol) is used.

As the solvent rises up the paper it will carry the dyes with it. Each different dye will move up the paper at different rates depending on how strongly they stick to the paper and how soluble they are in the solvent.

1:11 understand how a chromatogram provides information about the composition of a mixture

Paper chromatography can be used to investigate the composition of a mixture.

A baseline is drawn on the paper. The mixture is spotted onto the baseline alongside known or standard reference materials. The end of the paper is then put into a solvent which runs up the paper and through the spots, taking some or all of the dyes with it.

Different dyes will travel different heights up the paper.

The resulting pattern of dyes is called a chromatogram.

In the example shown, the mixture is shown to contain the red, blue and yellow dyes. This can be seen because these dots which resulted from the mixture have travelled the same distance up the paper as have the red, blue and yellow standard reference materials.

1:12 understand how to use the calculation of Rf values to identify the components of a mixture

When analysing a chromatogram, the mixture being analysed is compared to standard reference materials by measuring how far the various dyes have travelled up the paper from the baseline where they started.

For each dye, the Rf value is calculated. To do this, 2 distances are measured:

  • The distance between the baseline and the dye
  • The distance between the baseline and the solvent front, which is how far the solvent has travelled from the baseline

The Rf value is calculated as follows:

 R_f=\frac{distance\:of\:dye\:from\:baseline }{distance\:of\:solvent\:front\:from\:baseline}

If the Rf value of one of the components of the mixture equals the Rf value of one of the standard reference materials then that component is know to be that reference material. 

Note that because the solvent always travels at least as far as the highest dye, the Rf value is always between 0 and 1.

Dyes which are more soluble will have higher Rf values than less soluble dyes. In other words, more soluble dyes move further up the paper. The extreme case of this is for insoluble dyes which don’t move at all (Rf value = 0). The other aspect affecting how far a dye travels is the affinity that dye has for the paper (how well it ‘sticks’ to the paper).

1:13 practical: investigate paper chromatography using inks/food colourings

  1. A pencil line (baseline) is drawn 1cm from the bottom of the paper. Pencil will not dissolve in the solvent, but if ink were used instead it might dissolve and interfere with the results of the chromatography.
  2. A spot of each sample of dye is dropped at different points along the baseline.
  3. The paper is suspended in a beaker which contains a small amount of solvent. The bottom of the paper should be touching the solvent, but the baseline with the dyes should be above the level of the solvent. This is important so the dyes don’t simply dissolve into the solvent in the beaker.
  4. A lid should cover the beaker so the atomosphere becomes saturated with the solvent. This is so the solvent does not evaporate from the surface of the paper.
  5. When the solvent has travelled to near the top of the paper, the paper is removed from the solvent and a pencil line drawn (and labelled) to show the level the solvent reached up the paper. This is called the solvent front.
  6. The chromatogram is then left to dry so that all the solvent evaporates.

Common solvents are water or ethanol. The choice of solvent depends on whether most of the dyes are soluble in that solvent.

1:15 know the structure of an atom in terms of the positions, relative masses and relative charges of sub-atomic particles

An atom consists of a central nucleus, composed of protons and neutrons.

This is surrounded by electrons, orbiting in shells (energy levels).

Atoms are neutral because the numbers of electrons and protons are equal.

 
 MassCharge
Proton1+1
Neutron10
Electronalmost 0-1

1:16 know what is meant by the terms atomic number, mass number, isotopes and relative atomic mass (Aᵣ)

Atomic number: The number of protons in an atom.

Mass number: The number of protons and neutrons in an atom.

Isotopes: Atoms of the same element (same number of protons) but with a different number of neutrons.

Relative atomic mass (Ar): The average mass of an atom compared to 1/12th the mass of carbon-12.

1:17 be able to calculate the relative atomic mass of an element (Aᵣ) from isotopic abundances

75% of chlorine atoms are the type 35Cl (have a mass number of 35)

25% of chlorine atoms are of the type 37Cl (have a mass number of 37)

In order to calculate the relative atomic mass (Ar) of chlorine, the following steps are used:

  1. Multiply the mass of each isotope by its relative abundance
  2. Add those together
  3. Divide by the sum of the relative abundances (normally 100)

    \[ A_r = \frac{( (35 \times 75) + (37 \times 25) )}{100} \]

    \[ A_r = 35.5 \]

 

Example question:

A sample of bromine contained the two isotopes in the following proportions: bromine-79 = 50.7% and bromine-81 = 49.3%.

Calculate the relative atomic mass (Ar) of bromine.

    \[ A_r = \frac{( (79 \times 50.7) + (81 \times 49.3) )}{100} \]

    \[ A_r = 79.99 \]

 

1:18 understand how elements are arranged in the Periodic Table: in order of atomic number, in groups and periods

The elements in the Periodic Table are arranged in order of increasing atomic number.

 

Image result for periodic table groups and periods

Columns are called Groups. They indicate the number of electrons in the outer shell of an atom.

Rows are called Periods. They indicate the number of shells (energy levels) in an atom.

1:19 understand how to deduce the electronic configurations of the first 20 elements from their positions in the Periodic Table

Electrons are found in a series of shells (or energy levels) around the nucleus of an atom.

Each energy level can only hold a certain number of electrons. Low energy levels are always filled up first.

Rules for working out the arrangement (configuration) of electrons:

Example – chlorine (Cl)

1) Use the periodic table to look up the atomic number. Chlorine’s atomic number (number of protons) is 17.

2) Remember the number of protons = number of electrons. Therefore chlorine has 17 electrons.

3) Arrange the electrons in levels (shells):

  • 1st shell can hold a maximum of 2
  • 2nd can hold a maximum of 8
  • 3rd can also hold 8

Therefore the electron arrangement for chlorine (17 electrons in total) will be written as 2,8,7

4) Check to make sure that the electrons add up to the right number

The electron arrangement can also be draw in a diagram.

Electron arrangement for the first 20 elements:

1:20 understand how to use electrical conductivity and the acid-base character of oxides to classify elements as metals or non-metals

Metals

  • conduct electricity
  • have oxides which are basic, reacting with acids to give a salt and water

 

Non – Metals

  • do not conduct electricity (except for graphite)
  • have oxides which are acidic or neutral

 

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

Word equation:

     sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

     Na + H2O –> NaOH + H2

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

     2Na + 2H2O –> 2NaOH + H2

If asked for an equation, the chemical equation must be given.

 

State symbols are used to show what physical state the reactants and products are in.

State symbolsPhysical state
(s)Solid
(l)Liquid
(g)Gas
(aq)Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).

     2Na(s) + 2H2O(l) –> 2NaOH(aq) + H2(g)

1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)

Relative formula mass (Mr) is mass of a molecule or compound (on a scale compared to carbon-12).

It is calculated by adding up the relative atomic masses (Ar) of all the atoms present in the formula.

Example:

The relative formula mass (Mr) for water (H2O) is 18.

Water                     = H2O

Atoms present      = (2 x H) + (1 x O)

Mr                           = (2 x 1) + (1 x 16) = 18

1:27 know that the mole (mol) is the unit for the amount of a substance

In Chemistry, the mole is a measure of amount of substance (n).

The abbreviation for mole is mol.

The mass of 1 mole of a substance is the relative formula mass (Mr) of the substance in grams.

Example:

The Mr of water is 18.

Therefore the mass of 1 mol of water equals 18 g.

1:28 understand how to carry out calculations involving amount of substance, relative atomic mass (Aᵣ) and relative formula mass (Mᵣ)

The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:

    \[Amount(mol) = \frac{mass(g)}{M_r} \]

Example 1:

Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO2).

Step 1: Calculate the relative formula mass (Mr) of carbon dioxide (CO2).

    \[Carbon\,dioxide = CO_2 \]

    \[Atoms\,present = (1 \times C) + (2 \times O) \]

    \[M_r = (1 \times 12) + (2 \times 16) = 44 \]

Step 2: Use the formula to calculate the amount in moles.

    \[Amount = \frac{mass(g)}{M_r} \]

    \[Amount = \frac{8.8}{44} \]

    \[Amount = 0.2 mol \]

Example 2:

Calculate the mass of 2 mol of copper(II) sulfate (CuSO4).

Step 1: Calculate the relative formula mass (Mr) of copper(II) sulfate (CuSO4).

    \[copper(II)\,sulfate = CuSO_4 \]

    \[Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O) \]

    \[M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5 \]

Step 2: Rearrange the formula to calculate the mass.

    \[mass = amount \times M_r \]

    \[mass = 2 \times 159.5 \]

    \[mass = 319g\]

1:29 calculate reacting masses using experimental data and chemical equations

Example: When calcium carbonate (CaCO3) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.

     CaCO3     –>         CaO      +      CO2

Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO3)

    \[M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16) \]

    \[M_r(CaCO_3)= 100\]

    \[Amount = \frac{mass(g)}{M_r} \]

    \[Amount = \frac{25}{100} \]

    \[Amount = 0.25 mol \]

Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO3.

This step involves using the ratio of CaCO3 to CaO from the chemical equation.

     CaCO3     –>         CaO      +      CO2

From the equation you can see that the ratio of CaCO3 to CaO is 1:1.

Therefore if you have 0.25 mol of CaCO3 this will produced 0.25 mol of CaO.

Step 3: Calculate the mass of 0.25 mol of CaO.

    \[M_r(CaO)=(1 \times 40) + (1 \times 16) \]

    \[M_r(CaO)= 56\]

    \[Mass(CaO)=amount \times M_r \]

    \[Mass(CaO)= 0.25 \times 56\]

    \[Mass(CaO)= 14g\]

 

A simple format for laying out this method can be used.

Example: What mass of ammonia (NH3) is formed when 7 g of nitrogen (N2) is combined with hydrogen (H2).

     

 

1:30 calculate percentage yield

Yield is how much product you get from a chemical reaction.

The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.

In most chemical reactions, however, you rarely achieved your theoretical yield.

For example, in the following reaction:

     CaCO3     –>            CaO         +           CO2

You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO3.

However, what if the actual yield is only 48 g of CaO.

By using the following formula, the % yield can be calculated:

    \[yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product} \]

    \[yield= \frac{48}{56} \]

    \[yield=0.86 \]

    \[\% yield=86\% \]

1:31 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

Finding the formula of a metal oxide experimentally

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.

Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).

Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
   (mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
   (lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
   (this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
   (this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)

 

Finding the formula of a salt containing water of crystallisation

When some substances crystallise from solution, water becomes chemically bound up with the salt.  This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula  CuSO4.5H2O  which formula indicates that for every CuSO4 in a crystal there are five water (H2O) molecules.

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO4.5H2O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO4). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.

1:32 know what is meant by the terms empirical formula and molecular formula

The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.

The molecular formula shows the actual number of atoms of each type of element in a molecule.

   For example, for ethane:

        Molecular formula = C2H6

        Empirical formula = CH3

Here are some more examples:

NameMolecular formulaEmpirical Formula
penteneC5H10CH2
buteneC4H8CH2
glucoseC6H12O6CH2O
hydrogen peroxideH2O2HO
propaneC3H8C3H8

Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.

 

1:33 calculate empirical and molecular formulae from experimental data

Calculating the Empirical Formula

Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?

Step 1: Put the symbols for each element involved at the top of the page.

Step 2: Underneath, write down the masses of each element combining.

Step 3: Divide by their relative atomic mass (Ar).

Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.

Step 5: Use this to give the empirical formula.

(If your ratio is 1:1.5 then multiple each number by 2

If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)

 

Calculating the Molecular Formula

If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.

Example: A compound has the empirical formula CH2, and its relative formula mass is 56. Calculate the molecular formula.

Step 1: Calculate the relative mass of the empirical formula.

Step 2: Find out the number of times the relative mass of the empirical formula goes into the Mr of the compound.

Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.

 

Empirical formula calculations involving water of crystallisation

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula  BaCl2.nH2O  where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.

Example: If you heated hydrated barium chloride (BaCl2.nH2O) in a crucible you might end up with the following results.

     

Step 1: Calculate the mass of the anhydrous barium chloride (BaCl2) and the water (H2O) driven off.

     

Step 2: Use the empirical formula method to find the value of n in the formula.

     

1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

Concentration is a measurement of the amount of substance per unit volume.

In Chemistry, concentration is measured in mol/dm3 (read as moles per cubic decimetre).

The following formula allows for the interconversion between a concentration (in mol/dm3), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm3).

    \[amount(mol) = concentration(mol/dm^3) \times volume(dm^3) \]

or

    \[concentration(mol/dm^3) = \frac{amount(mol)}{volume(dm^3)} \]

Note: 1 dm3 = 1000 cm3

For example, to convert 250 cm3 into dm3:

    \[250cm^3 = \frac{250}{1000} \]

    \[250cm^3 = 0.25dm^3 \]

 

Example 1:     0.03 mol of sodium carbonate (Na2CO3) is dissolved in 300 cm3 of water. Calculate the concentration of the solution.

Step 1:  Convert the volume of water from cm3 into dm3.

    \[volume = 300cm^3 \]

    \[volume = \frac{300}{1000}dm^3 \]

    \[volume = 0.300dm^3 \]

Step 2: Use the molar concentration formula to calculate of the concentration of the solution.

    \[concentration = \frac{amount}{volume} \]

    \[concentration = \frac{0.03}{0.300} \]

    \[concentration = 0.1 mol/dm^3 \]

 

Example 2:     Calculate the amount, in moles, of 25cm3 of hydrochloric acid (HCl) with a concentration of 2 mol/dm3.

Step 1: Convert the volume of HCl from cm3 into dm3.

    \[volume = 25cm^3 \]

    \[volume = \frac{25}{1000}dm^3 \]

    \[volume = 0.025dm^3 \]

Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.

    \[amount= concentration \times volume \]

    \[amount = 2 \times 0.025 \]

    \[amount = 0.050 mol \]

 

Titration calculations, example 1

The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.

For example, a titration problem will look like this:

A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution (NaOH) required 23.50 cm3 of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm3 of the acid.

The chemical equation for this reaction is:

                                 NaOH(aq)               +              HCl(aq)      –>    NaCl(aq)     +     H2O(l)

volume                    25.0cm3                                  23.50cm3

concentration         0.100mol/dm3

It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.

 

Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).

First convert the volume of NaOH from cm3 into dm3.

    \[volume = 25cm^3 \]

    \[volume = \frac{25}{1000}dm^3 \]

    \[volume = 0.025dm^3 \]

Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.

    \[amount= concentration \times volume \]

    \[amount = 0.100 \times 0.025 \]

    \[amount = 0.0025 mol \]

 

Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).

From the chemical equation the ratio of NaOH to HCl is 1:1

Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.

 

Step 3: Calculate the concentration of the hydrochloric acid (HCl).

Convert the volume of HCl from cm3 into dm3.

    \[volume = 23.5cm^3 \]

    \[volume = \frac{23.5}{1000}dm^3 \]

    \[volume = 0.0235dm^3 \]

Use the molar concentration formula to calculate of the concentration of the HCl.

    \[concentration = \frac{amount}{volume} \]

    \[concentration = \frac{0.0025}{0.0235} \]

    \[concentration = 0.106 mol/dm^3 \]

 

Titration calculations, example 2

25.0 cm3 of sodium carbonate solution (Na2CO3) of unknown concentration was neutralised by 30.0 cm3 of 0.100 mol/dm3 nitric acid (HNO3).

          Na2CO3(aq)      +      2HNO3(aq)      –>      2NaNO3(aq)      +      CO2(g)      +      H2O(l)

Calculate the concentration, in mol/dm3 of the sodium carbonate solution.

 

Step 1: Calculate the amount, in moles, of nitric acid (HNO3).

    \[volume = 30cm^3 \]

    \[volume = \frac{30}{1000}dm^3 \]

    \[volume = 0.030dm^3 \]

    \[amount= concentration \times volume \]

    \[amount = 0.100 \times 0.030 \]

    \[amount = 0.0030 mol \]

 

Step 2: Deduce the amount, in moles of sodium carbonate (Na2CO3).

Using the equation:

          Na2CO3(aq)      +      2HNO3(aq)      –>      2NaNO3(aq)      +      CO2(g)      +      H2O(l)

Ratio Na2CO3:HNO3 = 1:2

    \[amount=\frac{0.0030}{2} = 0.0015mol\]

 

Step 3: Calculate the concentration of sodium carbonate (Na2CO3).

    \[volume = 25cm^3 \]

    \[volume = \frac{25}{1000}dm^3 \]

    \[volume = 0.025dm^3 \]

    \[concentration = \frac{amount}{volume} \]

    \[concentration = \frac{0.0015}{0.025} \]

    \[concentration = 0.06 mol/dm^3 \]

 

Converting mol/dm3 into g/dm3

Concentration can also be expressed in g/dm3 (grams per cubic decimetre).

Therefore mol/dm3 can be converted into g/dm3.

Example:     Convert 0.06 mol/dm3 of sodium carbonate (Na2CO3) into g/dm3

Step 1: calculate the relative formula mass (Mr) of sodium carbonate (Na2CO3).

    \[M_r= (2 \times 23) + (1 \times 12) (3 \times 16) = 106 \]

Step 2: Recall the formula giving the relationship between mass, amount and formula mass.

    \[mass= amount(in\,moles) \times M_r \]

therefore

    \[mass/dm^3= moles/dm^3 \times M_r \]

    \[mass/dm^3= 0.06 \times 106\]

    \[mass/dm^3= 6.36 g/dm^3 \]

1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))

The molar volume of a gas is the volume that one mole of any gas will occupy.

1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm3 or 24,000 cm3.

Note: 1 dm3 = 1000 cm3

The following formulae allows for the interconversion between a volume in dm3 or cm3 and a number of moles for a given gas:

    \[Amount(mol) = \frac{volume(dm^3)}{24} \]

    or

    \[Amount(mol) = \frac{volume(cm^3)}{24000} \]

 

Example 1:

Calculate the amount, in moles, of 12 dm3 of carbon dioxide (CO2).

    \[Amount = \frac{volume(dm^3)}{24} \]

    \[Amount = \frac{12}{24} \]

    \[Amount = 0.5mol \]

 

Example 2:

Calculate the volume at rtp in cubic centimetres (cm3), of 3 mol of oxygen, (O2).

    \[Volume = amount \times 24000 \]

    \[Volume = 3 \times 24000 \]

    \[Volume = 72000cm^3 \]

 

1:36 practical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.

Example
When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
   (mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
   (lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
   (this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
   (this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)

1:38 know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)

Name of IonFormulaCharge
SulfateSO42--2
CarbonateCO32--2
NitrateNO3--1
HydroxideOH--1
AmmoniumNH4++1
Silver ionAg++1
Zinc ionZn2++2

1:39 write formulae for compounds formed between the ions listed in 1:38

Writing the electron configuration of an atom allows you to work out the electron configuration of the ion and therefore the charge on the ion.

 

Examples:

Atom = Mg

Electron configuration = 2,8,2

remove the two electrons from the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = Mg2+ 

[2,8]2+

 

Atom = O

Electron configuration = 2,6

add two electrons to the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = O2-  [2,8]2-

 

1:40 draw dot-and-cross diagrams to show the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7 only outer electrons need be shown

Sodium chloride, NaCl

 
  
Magnesium chloride, MgCl2

 
Potassium oxide, K2O

 

Calcium oxide, CaO

 
Aluminium oxide, Al2O3
 
 
Magnesium nitride, Mg3N2

1:46 understand how to use dot-and-cross diagrams to represent covalent bonds in: diatomic molecules, including hydrogen, oxygen, nitrogen, halogens and hydrogen halides, inorganic molecules including water, ammonia and carbon dioxide, organic molecules containing up to two carbon atoms, including methane, ethane, ethene and those containing halogen atoms

1:47 explain why substances with a simple molecular structures are gases or liquids, or solids with low melting and boiling points. The term intermolecular forces of attraction can be used to represent all forces between molecules

 

Carbon dioxide (CO2) has a simple molecular structure. This just means that it is made up of molecules.

Within each molecule are atoms bonded to each other covalently. These covalent bonds inside the molecules are strong.

However, between the molecules are weak forces of attraction that require little energy to break. These forces are not covalent bonds. This is why simple molecular substances such as carbon dioxide have a low boiling point.

So when carbon dioxide changes from a solid to a gas, for example, that process can be represented as:

CO₂ (s) → CO₂ (g)

Notice that even though there has been a dramatic change of state from solid to gas, the substance before and after the change is always made up of carbon dioxide molecules. During the change of the state the covalent bonds within each molecule remain unbroken. Instead it is the weak forces of attraction between the molecules which have been overcome.

 

1:48 explain why the melting and boiling points of substances with simple molecular structures increase, in general, with increasing relative molecular mass

Larger molecules tend to have higher boiling points.

This is because larger molecules (molecules with more mass) have more forces of attraction between them. These forces, although weak, must be overcome if the substance is to boil, and larger molecules have more attractions which must be overcome.

1:50 explain how the structures of diamond, graphite and C60 fullerene influence their physical properties, including electrical conductivity and hardness

Allotropes are different forms of the same element. Three different allotropes of carbon are shown here as examples: diamond, graphite and C60 fullerene.

 

Diamond is made up of only carbon atoms, where each of those atoms has a strong covalent bonds to 4 other carbon. Every one of carbon’s 4 outer electrons is involved in one of these strong covalent bonds.

Diamond is extremely hard because it is a giant covalent structure with many strong covalent bonds.

Because it is hard, diamond is used in high speed cutting tools, eg diamond-tipped saws.

 

 

 

Graphite is also made of only carbon atoms, and is also a giant structure, but it is formed of layers where each carbon atom has a strong covalent bond to 3 other carbons. This means each carbon atom has one electron not involved in a covalent bond, and these electrons form a sea of delocalised electrons between the layers. Each layer is a giant structure, with weak forces of attraction between the layers.

Graphite is used as a lubricant and in pencils because it is soft and slippery.

Graphite is soft and slippery because it has weak forces of attraction between layers.

These layers can easily slide over each other.

 

 

C60 fullerene (also known as a buckyball) is also made of only carbon atoms, but it forms molecules of 60 carbon atoms. The molecule has weak intermolecular forces of attraction between them which take little energy to overcome. Hence C60 fullerene has a low melting point, and it is soft.

C60 fullerene cannot conduct electricity. Although in each molecule every carbon is only covalently bonded to 3 others and the other electrons are delocalised, these electrons cannot jump between different molecules.

 

 

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     Terminology

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Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals