CHEMISTRY C3

Topic 1 – Qualitative Analysis

■ Demonstrate an understanding that analysis may be qualitative or quantitative.

■ Explain why the test for any ion must be unique.■ Purity of drinking water.■ Describe tests to show:

– Aluminium, Calcium, Copper, Iron (II) and Iron (III)– Ammonium plus sodium hydroxide solution– Cl, Br, I using dilute nitric acid and silver nitrate solution.

Water Testing

■ Qualitative: Investigation into what kind of substances are present.■ Quantitative: Investigation into the amount of substances that are

present.

■ Flame tests are qualitative:– If 2 metal ions produce a similar colour, you could get

the wrong identification.– So, you can add acid and silver nitrate, which

forms a precipitate. This can identify some anions in a more clear way.

Flame Test recap:Potassium – lilac

Calcium – brick redCopper – greenSodium - yellow

Testing for Cations using Sodium Hydroxide:Cation Symbol Effect of adding

Sodium HydroxideName of precipitate formed

Aluminium Al³+ Precipitate formedCloudy (white) (limewater)

Aluminium Hydroxide

Calcium Ca²+ Precipitate formed Cloudy (white)

Calcium Hydroxide

Copper Cu²+ Pale blue precipitate formed

Copper Hydroxide

Iron (II) Fe²+ Green precipitate Iron Hydroxide

Iron (III) Fe³+ Precipitate formedCloudy (brown(rust))

Iron Hydroxide

Copper sulfate + sodium hydroxide → copper hydroxide + sodium sulfateCuSO4 + 2NaOH → Cu(OH)2 +Na2SO4

Qualitative tests are carried out before quantitative tests■ 1) If the precipitate formed is brown, the ion present is _______________■ 2) If the precipitate formed is white, the ion could be ______ or _______■ 3) The level of salt in food is: quantitative or qualitative?■ 4) The source of pollution in a river is: quantitative or qualitative?■ 5) aluminium sulfate + sodium hydroxide → __________ + ____________

Qualitative tests are carried out before quantitative tests■ 1) If the precipitate formed is brown, the ion present is Iron (III)■ 2) If the precipitate formed is white, the ion could be Calcium or

Aluminium■ 3) The level of salt in food is: quantitative or qualitative?■ 4) The source of pollution in a river is: quantitative or qualitative?■ 5) aluminium sulfate + sodium hydroxide → aluminium hydroxide +

sodium sulfate

Effects of drinking sea water

Drinking Sea Water

Blood pressure and heart

rate increases

Physiological changes

eg) headaches

Kidney failure

dehydration

Brain damage

Excessive thirst

Halide (the negative ion formed by a halogen atom)Halide Ion Effect of adding

acidified silver nitrate solution

Chloride Cl- White coloured precipitate

Bromide Br- Cream/grey precipitate

Iodide I- Green/yellow precipitate

Method to test for Halide ions:

■ Carefully, pour 1cm³ of each halide into 3 separate test tubes (chloride, bromide, iodide).

■ Add a few drops of acidified silver nitrate solution until a precipitate begins to form.

■ Shake and allow to settle.■ Record results.

■ LiI + AgNO3 → LiNO3 + AgI– Lithium iodide + _____ _______ → _______ nitrate + silver iodide

■ KBr + AgNO3 → KNO3 + AgBr– Potassium bromide + silver _____ → potassium nitrate + ______

________

■ NaCl + AgNO3 → NaNo3 + AgCl– Sodium chloride + _______ → sodium _____ + silver ________

■ LiI + AgNO3 → LiNO3 + AgI– Lithium iodide + silver nitrate → lithium nitrate + silver iodide

■ KBr + AgNO3 → KNO3 + AgBr– Potassium bromide + silver nitrate → potassium nitrate + silver

bromide■ NaCl + AgNO3 → NaNo3 + AgCl

– Sodium chloride + silver nitrate → sodium nitrate + silver chloride

Testing for ammonium ionsObservations after warming with sodium hydroxide solution.Smell Effect on moist red

litmus paper.A Ammonia → pungent Blue/purple

B No smell No change

C Ammonia → pungent Blue/purple

D Ammonia → pungent Blue/purple

Therefore, Fertiliser B is not an ammonium ion.

Method to test for ammonium ions:■ Pour each pre-made fertilisers into 4 separate test tubes and label A,

B, C, D.■ Add 1cm³ of sodium hydroxide solution and shake.■ Using test tube clamps, hold over a Bunsen burner.■ As soon as it begins to bubble, pour into a beaker and smell it. If it

smells of ammonia (a hair dye smell) then hold moist red litmus paper over the beaker, it should turn blue/purple

■ Record results.

■ To practice this, find a sheet based on Ions in our water and writing chemical equations (C3.2d sheet and C3.4b sheet).

Water solutes

■ Hard water = high in dissolved minerals, specifically calcium and magnesium ions. It doesn’t lather easily with soap.

■ Soft water = is treated water in which the only cation (+) is sodium and forms a good lather with soap.

Hard water

Hard water makes scum

• Reacts with soap to make a nasty precipitate called scum.

• Hard water is caused by Ca²+ and Mg²+ ions.

Magnesium sulfate

(MgSO4) dissolves in

water.

• So does calcium sulfate

Calcium carbonate

• Calcium carbonate exists as chalk, limestone or marble.

• It can react with acid rain to form hydrogencarbonate Ca(HCO3)2.

• It is soluble in water and releases Ca²+.

My results

Sample Amount of soap needed

Distilled 5ml Softer

A 20% 13ml

B 60% 20ml

C 100% 25ml

D 140% 30ml Harder

Method to test hard/soft water

■ Measure 25cm³ of a sampled concentration of water into a boiling tube or conical flask.■ Pour 5ml of the soap solution into the flask or boiling tube using a pipette.■ Fit a bung and shake well■ Allow to settle■ If a lather has not formed, add another 5ml of soap solution and repeat until a lather

forms.■ Record the amount of soap solution needed.

■ To conclude, the pure water forms a lather easier because it is softer and doesn’t contain magnesium or calcium ions. Whereas, the 140% concentrated sample was harder to form a lather as it contained more magnesium and calcium ions.

calculations

■ M = mass of solute (g)■ C = concentration (g dm-3)■ V = volume (dm3)■ 1 dm3 = 1 litre = 1000cm3

– Eg) 250cm3 = 0.25dm3– Eg) 50mg = 0.05g

Worked example

■ What mass of sodium chloride is in 300cm3 of solution with a concentration of 12gdm-3?– 300cm3 = 0.3dm3– M = c x v– 0.3 x 12 = 3.6g

■ There are 0.4g of calcium ions dissolved in 2 litres of water. What is the concentration?– C = m ÷ v– C = 0.4 ÷ 2– C = 0.2 gdm-3

Hard and soft water

■ Magnesium ions and calcium ions make up hard water.– They react with the stearate in the soap.

■ What are the problems with hard water?

Limescalescum

Limescale and scum buildUp on appliances like

Kettles causing hassle.

Temporary hardness

■ Temporary hardness = property of hard water that can be removed by boiling the water.

■ So, this is a thermal decomposition reaction forming insoluble calcium carbonate which forms a limescale precipitate.

■ Permanent hard water = property of hard water that cannot be removed by boiling the water.– has no dissolved hydrogencarbonate ions to decompose – so when

it is heated there is no change and the water stays hard.– Permanent water contains Ca and Mg chlorides and/or sulfates,

which become more soluble as the temperature increases.

Method to test how temporary and permanent hard water are affected by heat

■ Put 25cm3 of temporary hard water into a boiling tube and label A1.■ Put 25cm3 of temporary hard water into a boiling tube and label B1.■ Put 25cm3 of permanent hard water into a boiling tube and label A2.■ Put 25cm3 of permanent hard water into a boiling tube and label B2.■ Add soap to both A’s and shake.■ Heat both B’s, cool, then add soap and shake.■ Record results.

Ion exchange column■ Both types of hard water can be softened by passing them through an

ion exchange column.■ This swaps the calcium ions and magnesium ions in the water for

sodium ions.■ Sodium ions do not cause hardness in water.

■ An ion exchange column is packed with tiny plastic beads made of ‘resin’ (polymer).

■ When hard water is passed through the column, positively charged Ca and Mg ions in the water swap with positively charged Na ions which are weakly attached to the resin.

■ The swap of ions makes the water softer.

TESTS TO REMEMBER!

1) Flame tests

■ Sodium ions – yellow/orange■ Potassium ions – lilac■ Calcium – brick red■ Copper – blue/green

2) Carbonates

■ Acid + carbonate → salt + water + carbon dioxide

3) Sulfate ions

■ Add dilute HCL + Barium chloride solution to form a white precipitate.

4) Chloride ions

■ Add dilute nitric acid + silver nitrate solution

5) Positive ions

■ Add a few drops of sodium hydroxide solution to your mystery compound.

■ Calcium – white precipitate■ Copper – blue precipitate■ Iron (II) – green precipitate■ Iron (III) – brown precipitate■ Aluminium – white precipitate at first but then it re-dissolves to form a

colourless solution.

6) Halides

■ Add nitric acid + silver nitrate solution.■ Chloride – white■ Bromide – cream■ Iodide - yellow

Particles and moles

■ Diamond is made up of carbon atoms. How much diamond is there?– How many carbon atoms are there in 12g of Diamond?

– 6.02x10^23

– How do we know??!!

Avogadro’s number

■ The amount of a substance can be measured in grams, number of particles, or number of moles.

■ The mass of an element equal to its atomic mass in grams always contain 6.02x10^23 atoms.

■ This is Avogadro’s number.

Your turn

■ How many atoms are there in 32g of sulphur?

Your turn

■ How many atoms are there in 32g of sulphur?

■ = sulphur = mass number = 32■ Therefore the mass and the mass number is equal so avogadros

number is the answer!

■ 6.02x10^23

Particles and moles

■ Mole = the quantity of a substance which is equivalent to its relative atomic/formula mass in grams.

■ Eg) RFM (Mr) of water is 18 (H2O = 1+1+16).– 18g of water contains an Avogadro’s number of water molecules,

so 1 mole of water has a mass of 18g.

Which of these contains 1 mole of particles?■ 8g of oxygen

■ 20g of magnesium

■ 44g of carbon dioxide

■ 40g of potassium

Which of these contains 1 mole of particles?■ 8g of oxygen

– 0=16 (6.02x10^23) ÷ 2 = 3.01x10^23 = 0.5M ■ 20g of magnesium

– Mg=24 20g=0.83M■ 44g of carbon dioxide

– CO2=44 44=44=1M■ 40g of potassium

– K=39 39 ÷ 40 = 0.975M

Calculation

■ How many moles are there in 88g of CO2?– CO2 = 44g = 1M 88g = 2M

■ How much mass is there in 162M of water?– H2O = 18 18x162 = 2916g

■ How many moles are there in 80g of calcium?– Ca = 40 = 2M

Solutions

■ Concentrations can be given in: gdm-3 or moldm-3.■ You can calculate moldm-3 using this equation:

Your turn

■ Seawater contains 30g of NaCl in every 1dm3. calculate the concentration.

Your turn

■ Seawater contains 30g of NaCl in every 1dm3. calculate the concentration.

■ 30 ÷ 58.5 = 0.512820mdm3

Preparing soluble salts

■ If soluble salts are prepared from an acid and an insoluble reactant:– Excess of the reactant can be added to ensure that all the acid is

used up.– The excess reactant can be removed by filtration.– The remaining solution is only salt and water.

Sodium hydroxide + hydrochloric acid → sodium chloride + water

Copper oxide + hydrochloric acid → copper chloride + water = neutralisation

Making copper sulfate (CuSo4)

■ Copper oxide + hydrochloric acid → copper chloride + water■ =a soluble salt made form an insoluble base.

■ 1) pour 20cm3 of sulphuric acid into a beaker.■ 2) warm over Bunsen burner whilst adding excess copper oxide to the solution.■ 3) remove from heat and pour into a filter over an evaporating dish.■ 4) it should drip out as a blue solution.■ 5) heat the evaporating dish until the liquid boils.■ 6) leave to cool.

Experiment explanation

■ You know the reaction took place because it changed colour (black to blue).

■ You don’t know that the reaction was ever complete, though,■ The acid was warmed to catalyse the experiment.

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

Titrations

■ Titrations are used to find out concentrations.■ An acid-based titration is a neutralisation reaction where Hydrogen

ions from an acid react with hydroxide ions from a soluble base (alkali).

■ H+ + OH- → H2O

■ Titrations allow you to find out exactly how much acid is needed to neutralise a quantity of alkali (or vise versa).

Method ■ 1) add 25cm3 of alkali to a conical flask, along with 2 or 3 drops of indicator.■ 2) the indicator used depends on the strengths of the acid and alkali:

– Phenolphthalein = used for a weak acid + a strong alkali.– Methyl orange = used for a strong acid + a weak alkali.– If both the acid and the alkali are strong, then any acid based indicator

can be used.■ 3) fill a burette with acid, below eye level.■ 4) using a burette, add the acid to the alkali a small amount at a time – giving

the conical flask a regular swirl. Go slowly when you think the end-point (colour change) is about to be reached.

■ 5) the indicator changes colour when all the alkali has been neutralised.– Eg) phenolphthalein is pink in alkalis, but colourless in acids.

■ 6) record the volume of acid used to neutralise the alkali.

Titration step-by-step calculations.

Titration step-by-step calculations.■ 25cm3 of NaOH solution was titrated against 0.1Moldm-3 HCL. An

average of 20cm3 of the acid was needed to react completely. What is the concentration of the NaOH solution?

Titration step-by-step calculations.■ 25cm3 of NaOH solution was titrated against 0.1Moldm-3 HCL. An

average of 20cm3 of the acid was needed to react completely. What is the concentration of the NaOH solution?

■ Step 1: Number of moles of HCL acid = concentration of HCL x Volume used– Number of moles of HCL acid = 0.1 x (20÷100) = 0.002mol

Titration step-by-step calculations.■ 25cm3 of NaOH solution was titrated against 0.1Moldm-3 HCL. An average of

20cm3 of the acid was needed to react completely. What is the concentration of the NaOH solution?

■ Step 1: Number of moles of HCL acid = concentration of HCL x Volume used– Number of moles of HCL acid = 0.1 x (20÷100) = 0.002mol

■ Step 2: write the balanced equation:– NaOH + HCL → NaCl + H2O– The equation shows that the ratio of NaOH : HCL is 1:1 so the moles are

equal.– Therefore, 0.002:0.002 moles.

Titration step-by-step calculations.■ 25cm3 of NaOH solution was titrated against 0.1Moldm-3 HCL. An average of

20cm3 of the acid was needed to react completely. What is the concentration of the NaOH solution?

■ Step 1: Number of moles of HCL acid = concentration of HCL x Volume used– Number of moles of HCL acid = 0.1 x (20÷100) = 0.002mol

■ Step 2: write the balanced equation:– NaOH + HCL → NaCl + H2O– The equation shows that the artio of NaOH : HCL is 1:1 so the moles are equal.– Therefore, 0.002:0.002 moles.

■ Step 3: concentration of NaOH = moles of NaOH ÷ volume of NaOH– Concentartion of NaOH = 0.002 ÷ (25÷1000) – Concentration of NaOH = 0.002 ÷ 0.025 = 0.08moldm-3

Electrolysis

■ 1) electricity comes from a battery, providing a direct current.■ 2) It requires a liquid to conduct electricity (an electrolyte).■ 3) the electricity is applied by two electrodes.■ 4) this breaks down the compound into its component parts (often as a gas).

■ O – Oxidation■ I – Is■ L – loss of electron■ R – reduction■ I – is■ G – gain of electron

Making ions move with electrolysis (this method is only for coloured ions)■ 1) connect a DC supply to a slide and filter paper.■ 2) put a crystal on the top of the filter paper and slide and add water

(drops).■ 3) the solution should be drawn towards the negative (cathode) to

prove it’s a positive ion.

Half equations

■ Normally: 2NaCl → 2Na + Cl2

■ ½ equation: Cl- → Cl2 + 2e- (oxidation)■ ½ equation: 2Na+ + 2e- → 2Na (reduction)

Practice these: (normally to ½ equations)■ 1) 2LiI → Li + I2

Practice these: (normally to ½ equations)■ 1) 2LiI → Li + I2

– Li + e- → Li (cathode)– 2I → I2 + 2e- (anode)

Electrolysis of saltsObservations Anode (+) Cathide (-) Solution left

Sodium sulfate Bubbles at both electrodes

O2- H2+ Sodium sulfate

Sodium chloride Fizzing, bubbling, clouding

Cl2- H2+ Sodium hydroxide

Copper chloride Bubbles and visible copper

Cl2- Cu2+ Water

Copper sulfate Copper at cathode, bubbles at anode, pieces of visible metal.

O2- Cu2+ Sulphuric acid

Rule

■ At the cathode: Always Hydrogen will be formed, unless the metal is less reactive than hydrogen.

■ At the anode: Always Oxygen unless a halide is present.

Electroplating

■ Electrolyte = solution you are breaking down in electrolysis or electroplating.■ Electroplating = covering one metal with a thin layer of another metal, using

electrolysis.

■ Anion = -■ Cation = +■ Anode = +■ Cathode = -■ Cathode (-) attract cations (+).

Electroplating has many uses:

■ Jewellery and decorative items:– With metals like gold and silver.– Improves appearance.

■ Cooking utensils and cutlery:– Stop them corroding.– With unreactive metals which don’t corrode easily (eg nickel or

chromium).

■ Electrolysis = 2 products■ Electroplating = one product transferred to the cathode.

Copper extraction

■ 2CuCO3 → 2Cu + 3CO2■ Why is copper important to us?

– Used In pipes , electrical generators and motors.■ Industrial extraction of copper (smelting):

– Copper-rich ores →– Cooper can be extracted from these ores using heat in a furnace →– This is smelting→– Cooper then purified using electrolysis.

■ Purification of copper using electrolysis:– -------------------------------------------------------------------

■ Ore = a naturally occurring solid material from which a metal or valuable mineral can be extracted profitably.

■ Extraction = separation of a compound.■ Impure = containing more than one material other than the intended

pure material.■ Smelting = an extraction method using heat, usually also resulting in

oxidation.

Search the ‘Haber process’

Molar volume of gas

■ Another Avogadro’s law = one mole of any gas occupies 24dm3.

■ 1 mole at any gas at room temperature (25°) and normal atmospheric pressure (1 atmosphere), has a volume of 24dm3.

■ Vol of gas = (mass of gas ÷ Mr of gas) x 24

Reversible/irreversible reactions

■ Reactants → products = irreversible.■ Reactants products = forwards reversible■ Products reactants = backwards reversible■ Reversible reaction = a chemical recation that can work in both

directions.

Industrial manufacture of ammonia■ Pressure = 200 atmosphere■ Temperature = 450°C■ Catalyst = iron.■ Ammonia is used to make nitrogenous fertilisers.■ N2 + 3H2 2NH3 ■ N2 = from the air and 3H2 = from natural gases.■ Not all the nitrogen and hydrogen will convert to ammonia. The reaction reaches a

dynamic equilibrium.■ 1) air liquefied under pressure

– Optimum temp = 450°C = forwards reaction– Optimum pressure = 200 atmospheres = forward reaction– Iron catalyst– Exothermic reaction = heat produced.

■ Increasing the temperature will speed up the rate of reaction, making it turn into a backwards reaction which they don’t want, so they must be accurate with not surpassing or exceeding the temperature too much.

Ammonia

■ Cleaning■ Very strong alkali■ Explosives■ 85% used for nitrogenous fertilisers■ Can cause eutrophication.

Dynamic equilibrium

■ = the reactions are taking place in both directions at exactly the same rate, so there is no overall affect.

■ This occurs in a closed system, where no reactants or products can escape.

■ Forwards = exothermic = releases heat■ Backwards = endothermic = takes in heat.■ Catalysts = decrease the activation energy.

Le Chantelier’s Principle

■ Rule = any change made to a reaction which is in equilibrium, will result in the equilibrium position moving to minimise the change made.

■ Exothermic – High temp = backwards– Low temp = forwards

■ Endothermic– High temp = forwards– Low temp = backwards

■ Same for pressure

Questions

■ Define the dynamic equilibrium.■ What will a higher pressure do to the equilibrium yield of ammonia?■ What would a lower temperature do to the equilibrium yield?■ What is a catalyst used for?■ What is the minimum volume of Hydrogen required to convert

1000dm3 of nitrogen into ammonia?

Questions

■ Define the dynamic equilibrium.– = rate of reactions in both direction occur at exactly the same rate.

■ What will a higher pressure do to the equilibrium yield of ammonia?– = increase the yield of ammonia.

■ What would a lower temperature do to the equilibrium yield?– = Increase the equilibrium yield.

■ What is a catalyst used for?– = to increase the rate of reaction.

■ What is the minimum volume of Hydrogen required to convert 1000dm3 of nitrogen into ammonia?– N2 + 3H2 → 2NH3– 1 : 3– 1000 x 3 = 3000dm3.

Alcoholic drinks and ethanol production■ 1 unit = 10cm3 of pure ethanol.

– Average amount that can be processed in one hour.

■ Ethanol catalyst ethene + steam.

Making ethanol

From ethane (cracking crude oil)

Fermentation with yeast

Fermentation with yeast

■ Sugar yeast ethanol + carbon dioxide■ Yeast contains enzymes which convert sugar to ethanol.■ The optimum temperature is 37°C because Zymase works best at this

temperature (if it is too hot the enzyme will denature, but if too cold the reaction will be too slow.

■ The optimum pH is pH 4 as Zymase likes slightly acidic conditions.■ No oxygen should enter the reaction, as this converts ethanol to vinegar

(ethanoic acid).■ When a concentration of 10-15% is reached, the fermentation stops because the

enzyme becomes denatured by the ethanol.■ This process is too slow for large-scale production.

■ Ethanol boils at a lower temperature than water = fractional distillation.

Hydration of ethene

■ Ethene reacts with steam to make ethanol by hydration (adding water)>

■ C2H4 + H2O → C2H5OH■ Optimum temperature = 350°C■ Optimum pressure = 60-70 atmospheres■ Catalyst = phosphoric acid (HPO3).

C = C H

CHH

H

Adv. / Disadv.Advantage Disadvantage

Fermentation Uses a renewable source.

The quality of the ethanol produced isn’t a high standard.Its expensive to concentrate and purify.Less space for farming.Takes a long time.

Ethene 100% pure.Little/no waste products.Quick, continuous, cheap.

Crude oil is non-renewable.High energy costs to maintain high temperature and pressure.

Homologous series

■ A group of similar compounds (with the same general formula and similar properties, but have different number of carbon atoms). Eg) alkanes, alkenes and alcohols. (alkane = CnH2n+2 alkene = CnH2n alcohols = CnH2n+OH)

■ Methane = one carbon atom■ Ethane = two carbon atoms.■ Propane = three carbon atoms.

■ Alcohols have an ‘-OH’ functional group and end in ‘-ol’.– Eg) methanol– Eg) ethanol– Eg) propanol.

Functional groups

■ =the reactive part of the molecule.■ In alcohols it is the –OH group (hydroxyl group). The rest of the

hydrocarbon chain is saturated and hence unreactive.■ Methanol = CH3OH■ Ethanol = C2H5OH ------------------------------------------------ =

O - H - C – C - H

H

H H

H

Alcohol

■ Alcohols combust with oxygen to produce carbon dioxide and water.■ Alcohol is flammable, good fuels and have a clearer flame than

hydrocarbons.liquid methanol Ethanol Propanol hexaneNo. of carbon atoms.

1 2 3 6

Test 1: mixing with water.

Mixed. Pretty mixed. Most mixed but if left will separate to two layers.

Two layers.

Test 2: how does it burn?

Orange flame.Easily.

Blue ish flame Blue flame, burns very well.

Extremely easily ignited. But doesn’t last as long. Orange flame.

Test 3: reaction with sodium.

Fizzing Fizzing fizzing

Ethanoic acid

■ = in vinegar.■ If wine or beer is left open, the ethanol is oxidised to ethanoic acid.■ Ethanol + oxygen → vinegar + water■ C2H5OH + O2 → CH3COOH +H2O■ This reaction is also used for the commercial production of vinegar.■ Vinegar can be used for flavouring and the preservation of food.

Carboxylic acids■ They are weak acids that have a –COOH- functional group (carboxyl) and end in –anoic

acid- .■ Carboxylic acid formula = CnH2nO2.■ Vinegar:

– Ethanol oxidises into ethanoic acid.– Due to bacteria in an aerobic process, unlike ethanol production which is anaerobic. – Processes up to 15% takes 24 hours.– Used to preserve food ‘pickling’ because bacteria cant live In the acidic conditions.– We pickle savoury but not sweet, we usually use jam for sweet food.

– Ethanol + oxygen → ethanoic acid + water– C2H5OH + O2 → CH3COOH +H2O H - C – C – O -

H

H

H IIO

= Ethanoic acidCH3COOHvinegar

Properties of carboxylic acids

■ A) testing ethanoic acid with universal indicator– = it turns red.

■ B) observations when heated– = the copper oxide formed a blue salt solution.

■ C) Adding magnesium– = hydrogen gas is given off– = fizzing and bubbles

■ D) adding sodium hydrogencarbonate– = carbon dioxide produced– = Fizzing and bubbles and turned limewater cloudy.

Reactions

■ Reaction with a metal– Ethanoic acid + magnesium → hydrogen + magnesium ethanoate– 2CH3COOH + Mg → H2 + (CH3COO)2Mg

– = a salt

■ Reaction with a base– Ethanoic acid + sodium hydroxide → sodium ethanoate + water– CH3COOH + NaOH → (CH3COO)Na + H2O

■ Reaction with a carbonate■ Ethanoic acid + sodium carbonate → sodium ethanoate + carbon dioxide + water■ CH3COOH + NaCO3 → (CH3COO)2Na + CO2 + H2O

Esters ■ Have the functional group –COO- and end in –yl…-oate.■ Theyre formed when an alcohol reacts with a carboxylic acid.■ They have sweet and fruity smell and comes in many flavourings and

perfumes eg) peardrops.■ They are volatile.■ They are made from an esterification reaction:

– Alcohol + carboxylic acid ester + water

H - C – C – O - H

H

H IIO

H – O - C – C - HH

H H

H

H - C – CH

H IIO

O - C – C - HH

H H

H+ H2O

H - C – CH

H IIO

- O - C – C - H

H

H H

H+ H2O

Ethanoic acid

Ethanol

Ethyl ethanoate

Questions

■ 1) ethanol + ethanoic acid ethyl ethanoate + ?■ 2) propanol + ethanoic ? ? + water■ 3) Butanol + ?? Butyl ethanoate + water■ 4) ? + propanoic acid Ethyl ? + water

Questions

■ 1) ethanol + ethanoic acid ethyl ethanoate + ?■ 2) propanol + ethanoic ? ? + water■ 3) Butanol + ?? Butyl ethanoate + water■ 4) ? + propanoic acid Ethyl ? + water

■ 1) water■ 2) acid and propyl ethanoate■ 3) ethanoic acid ■ 4) ethanol and propanoate

method

■ 250cm3 of boiling water into a beaker.■ Put 2cm of ethanol into a test tube.■ Get a pre-prepared test tube of 1cm3 concentrated sulphuric acid.■ Mix the two test tubes together. Stand the test tube in the hot water

beaker for 5 mins.■ collect 50cm3 sodium hydrogencarbonate into a beaker.■ Tip the test tube into the hydrogencarbonate and stir.■ smell. And record results.■

polyesters

■ = polymers made form 2 types of monomer:– A carboxyl group ( from a carboxylic acid)– A hydroxyl group ( from an alcohol).

■ Polyesters can be made into long, thin fibres which can be woven together to make fabrics or drinks bottles.

■ Drinks bottles can be recycled to make fleece, which can be used to make clothing.

Name Formula Structure of a molecule

Methanoic acid HCOOH DRAW

Ethanoic acid CH3COOH DRAW

Propanoic acid CH3CH2COOH DRAW

Fats and oils

■ Ethanol ahs OH and ethanoic acid has OH with a double bond of O.■ Fats:

– Esters– Saturated– Bromine water = stays orange– Solid at room temperature

■ Oils:– Esters– Unsaturated– Decolourises bromine water– Liquid at room temperature

Method to make soap

■ Place 2g of solid fat or 2cm3 of oil into a beaker.■ Add 10cm3 of concentrated sodium hydroxide solution.■ Warm the beaker gently and stir with a glass rod until it boils.■ Boil for 5mins and keep stirring.■ Take off heat and add 10cm3 of distilled water and 5 spatulas of salt.■ Boil for another 2-3mins.■ Leave to cool then filter off the solid soap.■ Wash product with a little distilled water and allow to dry.■ Test the soap with a little water and shake. Did you get a good lather?■ Also test with pH paper.

Soap – glycerol tristearate

■ Boil fats/oils with concentrated alkali.■ Esters break down to form

– An alcohol called glycerol– Sodium stearate (long chain carboxylic acid salts).

■ Sodium hydroxide + glycerol tristearate → sodium stearate + glycerol.■ Concentrated alkali + oil/fat → soap + glycerol

How does soap work?

■ The active part of the soap is the anion (stearate group).■ The head is hydrophilic (water loving) which dissolves in water.■ The tail is hydrophobic (water hating) which dissolves in oily dirt/grease.

■ The hydrophobic tails dissolve in the grease and the hydrophilic head dissolves in the water.

■ Some of the soap anions get beneath the grease and start to lift it off the fabric.

■ Grease leaves the fabric, surrounded by soap anions and mixes with the water.

Turning oil into fats

■ Hydrogenation of ethene:

■ Oils (unsaturated hydrocarbons) are reacted with hydrogen, in the presence of a catalyst.

■ This converts the C = C double bonds in the oils into into C – C single saturated bonds.

■ = catalytic hydrogenation.

C = C H

CHH

H+ H2 →

C - C H

CH

HH

HH