Topic: Specification experiments

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.

1:07 (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

1:10 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.

1:13 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.

1:31 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.

1:36 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)

To determine the formula of a metal oxide by combustion.

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

          2Mg   +   O2   –>   2MgO

Method:

  • Weigh a crucible and lid
  • Place the magnesium ribbon in the crucible, replace the lid, and reweigh
  • Calculate the mass of magnesium
  • 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
  • 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 after heating – mass of crucible)

It is then possible to use this data to calculate the empirical formula of the metal oxide.

 

To determine the formula of a metal oxide by reduction

Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:

          4CuO   +   CH4  –>   4Cu   +  CO2  +   2H2O

By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.

As an alternative, hydrogen gas could be used instead of methane:

          CuO   +   H2  –>   Cu   +  H2O

It is then possible to use this data to calculate the empirical formula of the metal oxide.

There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.

 

1:58 (Triple only) describe experiments to investigate electrolysis, using inert electrodes, of molten compounds (including lead(II) bromide) and aqueous solutions (including sodium chloride, dilute sulfuric acid and copper(II) sulfate) and to predict the products

Electrolysis: The breaking down of a substance caused by passing an electric current through an ionic compound which is molten or in solution. New substances are formed.

 

Rules for working out elements formed from electrolysis of solutions

Follow these rules to decide which ions in solution will react at the electrodes:

At the cathode

Metal ions and hydrogen ions are positively charged. Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series:

  • The metal will be produced if it is less reactive than hydrogen
  • Hydrogen gas (H2) will be produced if the metal is more reactive than hydrogen

 At the anode

At the anode, the product of electrolysis is always oxygen gas (O2) unless the solution contains a high concentration of Cl, Br­- or I ions, in which case a halogen is produced, e.g. chlorine gas (Cl2), bromine gas (Br2), and iodine gas (I2).

 

ELECTROLYSIS OF MOLTEN IONIC COMPOUNDS

Example: The electrolysis of molten lead bromide (PbBr2)

  • Solid lead bromide is heated and becomes molten. Explanation: ions become free to move.

  • Electrodes attached to a power source are placed in the molten lead bromide. Explanation: these electrodes are made of either graphite or platinum because both conduct electricity and are fairly unreactive.
  • From the diagram, the left-hand electrode becomes positively charged, this is called the anode. The right-hand becomes negatively charged, this is called the cathode. Explanation: delocalised electrons flow from the anode to the cathode. 
  • At the anode a brown gas is given off. This is bromine gas (Br2(g)). Explanation: Negatively charged bromide ions are attracted to the anode (positive electrode). At the anode, bromide ions lose electrons (oxidation) and become bromine molecules.
  • At the cathode a shiny substance is formed. This is molten lead (Pb(l) ). Explanation: Positively charged lead ions are attracted to the cathode (negative electrode). At the cathode, lead ions gain electrons (reduction) and become lead atoms.

Overall Reaction

word equation:                    lead bromide      –>   lead + bromine

chemical equation:              PbBr2(l)              –>   Pb(l) +   Br2(g)

Remember

OILRIG : Oxidation Is the Loss of electrons and Reduction Is the Gain of electrons

PANCAKE : Positive Anode, Negative Cathode

 

 

ELECTROLYSIS OF IONIC SOLUTIONS

The electrolysis of sodium chloride solution (NaCl(aq))

  • Solid sodium chloride is dissolved in water. Explanation: The sodium ions and chloride ions become free to move.

  • The solution also contains hydrogen ions (H+) and hydroxide ions (OH). Explanation: Water is a very weak electrolyte. It ionises very slightly to give hydrogen ions and hydroxide ions:

                              H2O(l) ⇋ H+(aq) + OH(aq)

  • Chloride ions (Cl) and hydroxide ions (OH) are attracted to the anode.
  • Sodium ions (Na+) and hydrogen ions (H+) are attracted to the cathode.
  • At the anode a green gas is given off. This is chlorine gas (Cl2(g)). Explanation: chloride ions lose electrons (oxidation) and form molecules of chlorine. The chloride ions react at the anode instead of the hydroxide ions because the chloride ions are in higher concentration. The amount of chlorine gas produced might be lower than expected because chlorine is slightly soluble in water.

                    Electron half equation:   2Cl(aq)    –>  Cl2 (g) + 2e     

  • At the cathode a colourless gas is given off. This is hydrogen gas (H2(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen. The hydrogen ions react at the cathode because hydrogen is below sodium in the reactivity series.

                    Electron half equation:   2H+(aq) + 2e  –>  H2 (g)

  • The solution at the end is sodium hydroxide (NaOH(aq)).

 

 

The electrolysis of copper sulfate solution (CuSO4(aq))

  • Copper sulfate solution is composed of copper ions (Cu2+), sulfate ions (SO42-), hydrogen ions (H+) and hydroxide ions (OH). 
  • At the cathode a shiny orange layer is formed. This is copper. Explanation: copper ions gain electrons (reduction) and form atoms of copper. The copper ions react at the cathode instead of hydrogen ions because copper is below hydrogen in the reactivity series.

                    Electron half-equation:   Cu2+(aq) + 2e  –>  Cu (s)

  • At the anode, bubbles of gas are given off. This is oxygen gas (O2(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

                    Electron half-equation:   4OH(aq)     –>  O2 (g) + 2H2O(l) + 4e      

 

 

The electrolysis of sulfuric acid (H2SO4(aq))

  • Sulfuric acid is composed of sulfate ions (SO42-), hydrogen ions (H+) and hydroxide ions (OH).
  • At the cathode bubbles of gas are formed. This is hydrogen gas (H2(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen.

                    Electron half-equation:   2H+(aq) + 2e  –>  H2(g)

  • At the anode, bubbles of gas are given off. This is oxygen gas (O2(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

                    Electron half-equation:   4OH(aq)     –>  O2 (g) + 2H2O(l) + 4e      

  • Twice the volume of hydrogen gas is produce compared to oxygen gas. Explanation: from the two half equations, O2 needs 4ebut H2 only needs 2e– as can be seen from the equation

                              2H2O(l)  –>  2H2(g)  +  O2(g)

            There are twice the amount (in moles) of H2 compared to O2

1:60 (Triple only) practical: investigate the electrolysis of aqueous solutions

The diagram shows an electrolytic cell.

The electrolyte is an aqueous solution. For example it might be concentrated sodium chloride, NaCl (aq).

The test tubes over the electrodes must not completely cover them to make sure the ions are free to move throughout the solution.

In the case of NaCl (aq) bubbles of gas will be seen forming at the electrodes. These float up and collect in the test tubes when each gas can be tested to assess its identity.

 

 

2:07 understand how displacement reactions involving halogens and halides provide evidence for the trend in reactivity in Group 7

Group 7 elements are called the Halogens. As you go up group 7 (decreasing atomic number), the elements become more reactive. For example, fluorine is the most reactive and astatine is the least reactive.

 

A more reactive halogen will displace a less reactive halogen, e.g. chlorine will displace bromine:

By reacting a halogen solution with a potassium halide solution and making observations, the order of their reactivity can be deduced:

 Potassium chloride, KCl(aq)Potassium bromide, KBr(aq)Potassium iodide, KI(aq)
Chlorine, Cl2(aq)No changeColourless to orangeColourless to brown
Bromine, Br2(aq)No changeNo changeColourless to brown
Iodine, I2(aq)No changeNo changeNo change

From the above results, chlorine displaces both bromine and iodine, and bromine displaces iodine. Therefore the order of reactivity is: chlorine is more reactive than bromine, which in turn is more reactive than iodine.

2:10 understand how to determine the percentage by volume of oxygen in air using experiments involving the reactions of metals (e.g. iron) and non-metals (e.g. phosphorus) with air

The following 3 experiments can be used to determine that oxygen (O2) makes up approximately 20% by volume of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

 

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron and water are in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

 

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

 

2:12 describe the formation of carbon dioxide from the thermal decomposition of metal carbonates, including copper(II) carbonate

On heating metal carbonates thermal decompose into metal oxides and carbon dioxide.

Observation: green powder (CuCO3) changes to a black powder (CuO)

2:14 Practical: determine the approximate percentage by volume of oxygen in air using a metal or a non-metal

The following 3 experiments can be used to determine that oxygen (O2) makes up approximately 20% by volume of air.

Copper

The copper is in excess and uses up the oxygen to form copper oxide (CuO).

All the oxygen in the air is therefore used up, and so the volume of the air decreases by about 20% (the percentage of oxygen in air).

 

Iron

The iron reacts with the oxygen in the air (rusting).

As long as the iron, oxygen and water are all in excess, the total volume of air enclosed by the apparatus decreases by about a fifth (20%) over several days.

 

Phosphorus

The phosphorus is lit with a hot wire.

It reacts with the oxygen in the air and causes the water level in the bell jar to rise by about 20%.

 

2:21 practical: investigate reactions between dilute hydrochloric and sulfuric acids and metals (e.g. magnesium, zinc and iron)

Metals which are above hydrogen in the reactivity series will react with dilute hydrochloric or sulfuric acid to produce a salt and hydrogen.

metal   +   acid   →   salt   +   hydrogen

For example:

         magnesium   +   hydrochloric acid   →   magnesium chloride   +   hydrogen

         Mg (s)         +         2HCl (aq)         →         MgCl₂ (aq)         +         H₂ (g)

This is a displacement reaction.

Image result for magnesium + hydrochloric acid

There is a rapid fizzing and a colourless gas is produced. This gas pops with a lighted splint, showing the gas is hydrogen.

The reaction mixture becomes warm as heat is produced (exothermic).

The magnesium disappears to leave a colourless solution of magnesium chloride.

If more reactive metals are used instead of magnesium the reaction will be faster so the fizzing will be more vigorous and more heat will be produced.

2:33 (Triple only) describe how to carry out an acid-alkali titration

Titration is used to find out precisely how much acid neutralises a certain volume of alkali (or vice versa).

The diagram shows the titration method for a neutralisation reaction between hydrochloric acid and sodium hydroxide, using phenolphthalein as an indicator. The indicator changes colour when neutralisation occurs.

The conical flask is swirled to mix the solutions each time alkali is added. When reading the burette it is important to be aware that the numbers on the scale increase from top to bottom. Readings are usually recorded to the nearest 0.05cm³ so all readings should be written down with 2 decimal places. The second decimal place is given as a ‘0’ if the level of the solution is on a line, or ‘5’ if it is between the lines. The volume of alkali added is calculated by subtracting the final reading from the initial reading. Various indicators can be used such as phenolphthalein or methyl orange. However universal indicator should not be used since it has a wide range of colours rather than one specific colour change so it would be unclear when the precise endpoint of titration was achieved.

This process is repeated a number of times. The first time it is done roughly to get a good approximation of how much alkali needs to be added. On subsequent attempts, the alkali is added very slowly when approaching the correct volume.

2:37 describe the reactions of hydrochloric acid, sulfuric acid and nitric acid with metals, bases and metal carbonates (excluding the reactions between nitric acid and metals) to form salts

Acid reactions summary

         alkali      +      acid      →      water      +      salt

         base      +      acid      →      water      +      salt

         carbonate      +      acid      →      water      +      salt      +      carbon dioxide

         metal   +   acid   →   salt   +   hydrogen

To assist remembering this list, many pupils find it useful to remember this horrid looking but very effective mnemonic:

         AAWS

         BAWS

         CAWS CoD

         MASH

Acids are a source of hydrogen ions (H⁺) when in solution. When the hydrogen in an acid is replaced by a metal, the compound is called a salt. The name of the salt depends on the acid used. For example if sulfuric acid is used then a sulfate salt will be formed.

Parent acidFormulaSaltFormula ion
sulfuric acidH2SO4sulfateSO42-
hydrochloric acidHClchlorideCl-
nitric acidHNO3nitrateNO3-

 

Acid + Alkali   and   Acid + Base

A base is a substance that can neutralise an acid, forming a salt and water only.

Alkalis are soluble bases. When they react with acids, a salt and water is formed. The salt formed is often as a colourless solution. Alkalis are a source of hydroxide ions (OH⁻) when in solution.

         alkali      +      acid      →      water      +      salt

         base      +      acid      →      water      +      salt

Examples of acid + alkali reactions:

  •          sodium hydroxide   +   hydrochloric acid   →   sodium chloride   +   water
  •          NaOH (aq)         +         HCl (aq)         →         NaCl (aq)         +         H₂O (l)
  •          potassium hydroxide   +   sulfuric acid   →   potassium sulfate   +   water
  •          2KOH (aq)         +         H₂SO₄ (aq)         →         K₂SO₄ (aq)         +         2H₂O (l)

Example of an acid + base reaction:

         CuO (s)         +         H₂SO₄ (aq)         →         CuSO₄ (aq)         +         H₂O (l)

 

Acid + Carbonate

         carbonate      +      acid      →      water      +      salt      +      carbon dioxide

A carbonate is a compound made up of metal ions and carbonate ions. Examples of metal carbonates are sodium carbonate, copper carbonate and magnesium carbonate.

When carbonates react with acids, bubbling is observed which is the carbon dioxide being produced. If the acid is in excess the carbonate will disappear.

Examples of acid + carbonate reactions:

  •          calcium carbonate   +   hydrochloric acid   →   calcium chloride   +   water   +   carbon dioxide
  •          CaCO₃ (s)         +         2HCl (aq)         →         CaCl₂ (aq)         +         H₂O (l)         +         CO₂ (g)
  •          potassium carbonate   +   hydrochloric acid   →   potassium chloride   +   water   +   carbon dioxide
  •          K₂CO₃ (aq)         +         2HCl (aq)         →         2KCl (aq)         +         H₂O (l)         +         CO₂ (g)

 

Acid + Metal

         metal   +   acid   →   salt   +   hydrogen

Metals will react with an acid if the metal is above hydrogen in the reactivity series.

When metals react with acids, bubbling is observed which is the hydrogen being produced. If the acid is in excess the metal will disappear.

Examples of acid + metal reactions:

  •          magnesium   +   sulfuric acid   →   magnesium sulfate   +   hydrogen
  •          Mg (s)         +         H₂SO₄ (aq)         →         MgSO₄ (aq)         +         H₂ (g)
  •          aluminium   +   hydrochloric acid   →   aluminium chloride   +   hydrogen
  •          2Al (s)         +         6HCl (aq)         →         2AlCl₃ (aq)         +         3H₂ (g)
  •          copper   +   hydrochloric acid   →   no reaction (since copper is below hydrogen in the reactivity series)

2:40 (Triple only) describe an experiment to prepare a pure, dry sample of a soluble salt, starting from an acid and alkali

Titration Method:

Preparing pure dry crystals of sodium chloride (NaCl) from hydrochloric acid (HCl) and sodium hydroxide (NaOH)

Before the salt preparation is carried out using the below method, the volume of acid that exactly reacts with 25cm3 of the alkali is found by titration using methyl orange indicator.

StepExplanation
Pipette 25cm3 of alkali (NaOH) into a conical flaskAccurately measures the alkali (NaOH)
Do not add indicatorPrevents contamination of the pure crystals with indicator
Using the titration values, titrate the known volume acid (HCl) into conical flask containing alkaliExactly neutralises all of the alkali (NaOH)
Transfer to an evaporating basin & heat the solutionForms a hot saturated solution (NaCl(aq))
Allow the solution to cool so that hydrated crystals formSodium chloride is less soluble in cold water
Remove the crystals by filtration and wash with distilled waterRemoves any impurities
Dry by leaving in a warm placeEvaporates the water

(Note – This process could be reversed with the acid in the pipette and the alkali in the burette)

How to select the right method for preparing a salt:

2:41 (Triple only) describe an experiment to prepare a pure, dry sample of an insoluble salt, starting from two soluble reactants

Precipitation Method:

Preparing pure dry crystals of silver chloride (AgCl) from silver nitrate solution (AgNO3) and potassium chloride solution (KCl)

StepExplanation
Mix the two salt solutions together in a beakerForms a precipitate of an insoluble salt (AgCl)
Stir with glass rodMake sure all reactants have reacted
Filter using filter paper and funnelCollect the precipitate (AgCl)
Wash with distilled waterRemoves any the other soluble salts (KNO3)
Dry by leaving in a warm placeEvaporates the water

2:42 practical: prepare a sample of pure, dry hydrated copper(II) sulfate crystals starting from copper(II) oxide

Excess Solid Method:

Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)

StepExplanation
Heat acid (H2SO4) in a beakerSpeeds up the rate of reaction
Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rodNeutralises all the acid
Filter the mixture using filter paper and funnelRemoves any excess copper oxide
Heat the filtered solution (CuSO4)Hot saturated solution forms
Allow the solution to cool so that hydrated crystals formCopper(II) sulfate is less soluble in cold water
Remove the crystals by filtration and wash with distilled waterRemoves any impurities
Dry by leaving in a warm placeEvaporates the water

 

2:43 (Triple only) practical: prepare a sample of pure, dry lead(II) sulfate

Objective: prepare a pure, dry sample of lead (II) sulfate (PbSO₄).

Preparing a pure, dry sample of lead (II) sulfate (PbSO₄) from lead (II) nitrate solution (Pb(NO₃)₂) and sodium sulfate solution (Na₂SO₄).

      Pb(NO₃)₂ (aq)      +      Na₂SO₄ (aq)      →        PbSO₄ (s)      +      2NaNO₃ (aq)

  1. Mix similar volumes lead nitrate solution and sodium sulfate solution in a beaker. The precise volumes do not matter since any excess will be removed later.
  2. A white precipitate of lead (II) sulfate will form.
  3. The reaction mixture is filtered.
  4. The residue left on the filter paper is washed with distilled water several times to remove impurities.
  5. The residue is then moved to a warm oven to dry.

 

2:44 describe tests for these gases: hydrogen, oxygen, carbon dioxide, ammonia, chlorine

Tests for gases

GasTestResult if gas present
hydrogen (H2)Use a lit splintGas pops
oxygen (O2)Use a glowing splintGlowing splint relights
carbon dioxide (CO2)Bubble the gas through limewaterLimewater turns cloudy
ammonia (NH3)Use red litmus paperTurns damp red litmus paper blue
chlorine (Cl2)Use moist litmus paperTurns moist litmus paper white (bleaches)

2:45 describe how to carry out a flame test

A flame test is used to show the presence of certain metal ions (cations) in a compound.

  • A platinum or nichrome wire is dipped into concentrated hydrochloric acid to remove any impurities.
  • The wire is dipped into the salt being tested so some salt sticks to the end.
  • The wire and salt are held in a non-luminous (roaring) bunsen burner flame.
  • The colour is observed.

 

2:46 know the colours formed in flame tests for these cations: Li⁺ is red, Na⁺ is yellow, K⁺ is lilac, Ca²⁺ is orange-red, Cu²⁺ is blue-green

When put into a roaring bunsen burner flame on a nichrome wire, compounds containing certain cations will give specific colours as follows.

IonColour in flame test
lithium (Li⁺)red
sodium (Na⁺)yellow
potassium (K⁺)lilac
calcium (Ca²⁺)orange-red
copper (II) (Cu²⁺)blue-green

2:47 describe tests for these cations: NH₄⁺ using sodium hydroxide solution and identifying the gas evolved, Cu²⁺, Fe²⁺ and Fe³⁺ using sodium hydroxide solution

Describe tests for the cation NH4+, using sodium hydroxide solution and identifying the ammonia evolved

 

Describe tests for the cations Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution

First, add sodium hydroxide (NaOH), then observe the colour:

2:48 describe tests for these anions: Cl⁻, Br⁻ and I⁻ using acidified silver nitrate solution, SO₄²⁻ using acidified barium chloride solution, CO₃²⁻ using hydrochloric acid and identifying the gas evolved

Describe tests for anions: Halide ions (Cl, Br and I)

 

Describe tests for anions: Sulfate ions (SO42)

 

Describe tests for anions: Carbonate ions (CO32-)

2:49 describe a test for the presence of water using anhydrous copper(II) sulfate

Add anhydrous copper (II) sulfate (CuSO4) to a sample.

If water is present the anhydrous copper (II) sulfate will change from white to blue.

3:02 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation

Calorimetry allows for the measurement of the amount of energy transferred in a chemical reaction to be calculated.

 

EXPERIMENT1: Displacement, dissolving and neutralisation reactions

Example: magnesium displacing copper from copper(II) sulfate

Method:

  1. 50 cm3 of copper(II) sulfate is measured and transferred into a polystyrene cup.
  2. The initial temperature of the copper sulfate solution is measured and recorded.
  3. Magnesium is added and the maximum temperature is measured and recorded.
  4. The temperature rise is then calculated. For example:
Initial temp. of solution (oC)Maximium temp. of solution (oC)Temperature rise (oC)
24.256.732.5

Note:  mass of 50 cm3 of solution is 50 g

 

EXPERIMENT2: Combustion reactions

To measure the amount of energy produced when a fuel is burnt, the fuel is burnt and the flame is used to heat up some water in a copper container

Example: ethanol is burnt in a small spirit burner

Method:

  1. The initial mass of the ethanol and spirit burner is measured and recorded.
  2. 100cm3 of water is transferred into a copper container and the initial temperature is measured and recorded.
  3. The burner is placed under of copper container and then lit.
  4. The water is stirred constantly with the thermometer until the temperature rises by, say, 30 oC
  5. The flame is extinguished and the maximum temperature of the water is measured and recorded.
  6. The burner and the remaining ethanol is reweighed. For example:
Mass of water (g)Initial temp of water (oC)Maximum temp of water (oC)Temperature rise (oC)Initial mass of spirit burner + ethanol (g)Final mass of spirit burner + ethanol (g)Mass of ethanol burnt (g)
10024.254.230.034.4633.680.78

The amount of energy produced per gram of ethanol burnt can also be calculated:

3:08 practical: investigate temperature changes accompanying some of the following types of change: salts dissolving in water, neutralisation reactions, displacement reactions and combustion reactions

Calorimetry allows for the measurement of the amount of energy transferred in a chemical reaction to be calculated.

 

EXPERIMENT1: Displacement, dissolving and neutralisation reactions

Example: magnesium displacing copper from copper(II) sulfate

Method:

  1. 50 cm3 of copper(II) sulfate is measured and transferred into a polystyrene cup.
  2. The initial temperature of the copper sulfate solution is measured and recorded.
  3. Magnesium is added and the maximum temperature is measured and recorded.
  4. The temperature rise is then calculated. For example:
Initial temp. of solution (oC)Maximium temp. of solution (oC)Temperature rise (oC)
24.256.732.5

Note:  mass of 50 cm3 of solution is 50 g

 

EXPERIMENT2: Combustion reactions

To measure the amount of energy produced when a fuel is burnt, the fuel is burnt and the flame is used to heat up some water in a copper container

Example: ethanol is burnt in a small spirit burner

Method:

  1. The initial mass of the ethanol and spirit burner is measured and recorded.
  2. 100cm3 of water is transferred into a copper container and the initial temperature is measured and recorded.
  3. The burner is placed under of copper container and then lit.
  4. The water is stirred constantly with the thermometer until the temperature rises by, say, 30 oC
  5. The flame is extinguished and the maximum temperature of the water is measured and recorded.
  6. The burner and the remaining ethanol is reweighed. For example:
Mass of water (g)Initial temp of water (oC)Maximum temp of water (oC)Temperature rise (oC)Initial mass of spirit burner + ethanol (g)Final mass of spirit burner + ethanol (g)Mass of ethanol burnt (g)
10024.254.230.034.4633.680.78

The amount of energy produced per gram of ethanol burnt can also be calculated:

3:09 describe experiments to investigate the effects of changes in surface area of a solid, concentration of a solution, temperature and the use of a catalyst on the rate of a reaction

The rate of a chemical reaction can be measured either by how quickly reactants are used up or how quickly the products are formed.

The rate of reaction can be calculated using the following equation:

The units for rate of reaction will usually be grams per min (g/min)

 

An investigation of the reaction between marble chips and hydrochloric acid:

Marble chips, calcium carbonate (CaCO3) react with hydrochloric acid (HCl) to produce carbon dioxide gas. Calcium chloride solution is also formed.

Using the apparatus shown the change in mass of carbon dioxide can be measure with time.

As the marble chips react with the acid, carbon dioxide is given off.

The purpose of the cotton wool is to allow carbon dioxide to escape, but to stop any acid from spraying out.

The mass of carbon dioxide lost is measured at intervals, and a graph is plotted:

 

Experiment to investigate the effects of changes in surface area of solid on the rate of a reaction:

The experiment is repeated using the same mass of chips, but this time the chips are larger, i.e. have a smaller surface area.

Since the surface area is smaller, the rate of reaction is less.

Both sets of results are plotted on the same graph.

If instead the chips were smashed into powder (and again same mass of chips used) the surface area would be much larger and so the rate of reaction higher (steeper line on graph).

 

Experiment to investigate the effects of changes in concentration of solutions on the rate of a reaction:

The experiment is again repeated using the exact same quantities of everything but this time with half the concentration of acid. The marble chips must however be in excess. The reaction with the half the concentration of acid happens slower and produces half the amount of carbon dioxide.

 

Experiment to investigate the effects of changes in temperature on the rate of a reaction:

The experiment is once again repeated using the exact same quantities of everything but this time at a higher temperature. The reaction with the higher temperature happens faster.

 

Experiment to investigate the effects of the use of a catalyst on the rate of a reaction:

Hydrogen peroxide naturally decomposes slowly producing water and oxygen gas.

Manganese (IV) oxide can be used as a catalyst to speed up the rate of reaction.

The rate of reaction can be measured by measuring the volume of oxygen produced at regular intervals using a gas syringe.

Both sets of results are plotted on the same graph.

 

 

 

 

 

 

 

Experiment to investigate the reaction between varying concentrations of sodium thiosulfate and hydrochloric acid

Sodium thiosulfate (Na2S2O3) and hydrochloric acid (HCl) are both colourless solutions. They react to form a yellow precipitate of sulfur.

     sodium thiosulfate   +   hydrochloric acid    →     sodium chloride   +   sulfur dioxide   +   sulfur   +   water

    Na2S2O3(aq)         +         2HCl(aq)           →           2NaCl(aq)         +         SO2(g)         +         S(s)         +         H2O(l)

 

To investigate the effects of changes in concentration of sodium thiosulfate on the rate of a reaction, the conical flask is placed above a cross. The reaction mixture is observed from directly above and the time for a cross to disappear is measured. The cross disappears because a precipitate of sulfur is formed.

In order to change the concentration of sodium thiosulfate, the volumes of sodium thiosulfate and water are varied (see results table). However the total volume of solution must always be kept the same as to ensure that the depth of the solution remains constant.

In this reaction, sulfur dioxide gas (SO2), which is poisonous is produced therefore the experiment must be carried out in a well ventilated room.

The results are recorded in the table below and then plotted onto a graph.

Volume of Na2S2O3(aq) (cm3)Volume of water (cm3)Concentration of Na2S2O3(aq) (mol/dm3)Time taken for cross to disappear (s)Rate of reaction (s-1) (1/time)
5000.10450.0222
40100.08600.0167
30200.06800.0125
20300.04130.0769
10400.022550.0039

The graph shows that the rate of reaction is directly proportional to the concentration.

The experiment can also be repeated to show how temperature affects the rate of reaction.

In this experiment the concentration of sodium thiosulfate is kept constant but heated to range of different temperatures.

As a rough approximation, the rate of reaction doubles for every 10oC temperature rise.

 

3:15 practical: investigate the effect of changing the surface area of marble chips and of changing the concentration of hydrochloric acid on the rate of reaction between marble chips and dilute hydrochloric acid

The rate of a chemical reaction can be measured either by how quickly reactants are used up or how quickly the products are formed.

The rate of reaction can be calculated using the following equation:

The units for rate of reaction will usually be grams per min (g/min)

 

An investigation of the reaction between marble chips and hydrochloric acid:

Marble chips, calcium carbonate (CaCO3) react with hydrochloric acid (HCl) to produce carbon dioxide gas. Calcium chloride solution is also formed.

Using the apparatus shown the change in mass of carbon dioxide can be measure with time.

As the marble chips react with the acid, carbon dioxide is given off.

The purpose of the cotton wool is to allow carbon dioxide to escape, but to stop any acid from spraying out.

The mass of carbon dioxide lost is measured at intervals, and a graph is plotted:

 

Experiment to investigate the effects of changes in surface area of solid on the rate of a reaction:

The experiment is repeated using the exact same quantities of everything but using larger chips. For a given quantity, if the chips are larger then the surface area is lesson. So reaction with the larger chips happens more slowly.

Both sets of results are plotted on the same graph.

 

Experiment to investigate the effects of changes in concentration of solutions on the rate of a reaction:

The experiment is again repeated using the exact same quantities of everything but this time with half the concentration of acid. The marble chips must however be in excess. The reaction with the half the concentration of acid happens slower and produces half the amount of carbon dioxide.

 

3:16 practical: investigate the effect of different solids on the catalytic decomposition of hydrogen peroxide solution

Oxygen (O2) is made in the lab from hydrogen peroxide (H2O2) using manganese(IV) oxide (MnO2) as a catalyst.

 

 

Different catalysts could be used to investigate which is the most effective in decomposing hydrogen peroxide. Examples of other substances which could be tested are:

  • Manganese dioxide
  • Liver
  • Potato
  • Potassium iodide
  • Copper oxide
  • Sodium chloride

Only some of these are effective catalysts when used with hydrogen peroxide. If a substance is not a catalyst, there will be no bubbles of oxygen produced. For other substances, such as liver which is a very effective catalyst in the decomposition of hydrogen peroxide, bubbles of oxygen will be produced quickly.

 

 

 

4:27 describe the reactions of alkenes with bromine, to produce dibromoalkanes

Alkenes react with bromine water. UV light is not required for this reaction.

The double bond is broken and the bromine atoms are added. This is an addition reaction.

During this reaction there is a colour change from orange to colourless.

For example:

This is how we can test for the presence of an alkene or another type of unsaturated molecule.

 

 

4:31 (Triple only) know that ethanol can be oxidised by: burning in air or oxygen (complete combustion), reaction with oxygen in the air to form ethanoic acid (microbial oxidation), heating with potassium dichromate(VI) in dilute sulfuric acid to form ethanoic acid

1) Ethanol can be oxidised by complete combustion. With excess oxygen the complete combustion of ethanol (C₂H₅OH) in air produces carbon dioxide and water:

C₂H₅OH (l)         +         3O₂ (g)         →         2CO₂ (g)         +         3H₂O (l)

 

2) Ethanol can be oxidised in air in the presence of microorganisms (‘microbial oxidation’) to form ethanoic acid (CH₃COOH).

 

3) Ethanol can be oxidised by heating with the oxidising agent potassium dichromate(VI) (K₂Cr₂O₇) in dilute sulfuric acid (H₂SO₄).

In the equation below, [O] means oxygen from an oxidising agent.

CH₃CH₂OH         +         2[O]         →         CH₃COOH         +         H₂O

This mixture starts orange but when the reaction happens turns green which indicates the presence of Cr³⁺ ions which are formed when the potassium dichromate(VI) is reduced.

4:43 (Triple only) practical: prepare a sample of an ester such as ethyl ethanoate

Heating a mixture of ethanoic acid and ethanol produces a liquid called ethyl ethanoate. A few drops of concentrated sulfuric acid must be added for the reaction to work. The sulfuric acid acts as a catalyst.

ethanoic acid           +           ethanol           ⇋           ethyl ethanoate           +           water

CH₃COOH (l)         +         CH₃CH₂OH (l)         ⇋         CH₃COOCH₂CH₃ (l)         +         H₂O (l)

The ethyl ethanoate produced is an ester.

The reaction is called esterification.

The reaction can also be described as a condensation reaction because water is made when two molecules join together.

Notice that the reaction is reversible. Pure reactants are used to maximise the yield of ethyl ethanoate. Pure ethanoic acid is called glacial ethanoic acid.

Select a set of flashcards to study:

     Terminology

     Skills and equipment

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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