Chemistry c3

Chemistry c3 chapter 5 

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  • Created by: bini
  • Created on: 17-05-12 18:42

Flame Tests
Identifying Group 1 and Group 2 metals is a piece of cake when we burn them, as they tend to have unique flames which we can associate with the elements. We call these testsflame tests. We perform a flame test by putting a small amount of a compound to be tested in a platinum wire loop which has been dipped in hydrochloric acid, and then we hold the substance over a blue Bunsen flame. The flame should show a particular colour which can be used to identify the unknown substance.

  • lithium will burn a bright red flame
  • sodium will burn a golden yellow flame
  • potassium will burn a lilac flame
  • calcium will burn a brick red flame
  • barium will burn a green flame

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Testing for Positive Ions
Another test for unknown substances is to test the reactions with sodium hydroxide solution. Aluminium, calcium and magnesium ions all form a white precipitate when they react with sodium hydroxide (NaOH). Hence adding it, if a white precipitate forms, we know its one of those three. To find out which one, we can add more and more NaOH – because eventually aluminium ions dissolve in it. If it dissolves, it’s aluminium, otherwise it’s either calcium or magnesium. To find out which of those two it is, we can use their flame tests – calcium burns with a brick red flame, whereas magnesium produces no special flame.

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Other metal ions produce coloured precipitates when they have NaOH added to them. Adding sodium hydroxide solution to:

  • a substance with copper(II) ions produces a light blue precipitate
  • a substance with iron(II) ions produces a “dirty” green precipitate
  • a substance with iron(III) ions produces a red-brown precipitate

As well as this, NaOH can be used to detect if ammonium ions ( NH4+ ) are present in an unknown substance. Ammonium ions react with NaOH to form ammonia and water:

NH4+ (aq) + OH- (aq)  NH3 (aq) + H2O (l)

To test for ammonium ions, we add NaOH to a solution of an unknown substance. If ammonium ions are present, ammonia (as well as water) forms. When we warm the solution, ammonia is then given off as a gas. We can detect ammonia gas using damp red litmus which should turn blue as ammonia is an alkaline gas.

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Testing for Negative Ions
There are a number of different tests we can use to detect negative ions. Each type of negative ion has its own test…

Testing for carbonates: If we add a dilute acid (e.g. hydrochloric) to a carbonate, it fizzes and produces carbon dioxide gas. We can test for carbon dioxide gas using limewater, and if it fizzes and produces the gas, we know it’s a carbonate. There are two particular metal carbonates which are giveaways, however, which makes it slightly easier to detect them. For example, copper carbonate is a green substance which when heated decomposes to give black copper oxide and carbon dioxide. Also, when zin carbonate, a white substance, is heated, it forms the lemon-yellow zinc oxide and carbon dioxide. These two reactions are shown below:

CuCO3 (s) → CuO (s) + CO2 (g)

ZnCO3 (s) → ZnO (s) + CO2 (g)

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Testing for halides: When we add dilute nitric acid and silver nitrate solution to an unknown solution, the appearance of a precipitate tells us what halide ion is present. Chlrodie ions give a white precipitate; bromide ions give a cream precipitate and iodide ions give a pale yellow precipitate. The ionic equation for this is, where X- is the halide ion:Ag+ (aq) + X- (aq) → AgX (s)

Testing for sulphates: Adding hydrochloric acid followed by a barium chloride solution to sulphate ions in solution produces a white precipitate (barium sulphate, an insoluble salt). The ionic equation for this is shown:Ba2+ (aq) + SO42- (aq) → BaSO4 (s)

Testing for nitrates: The test for ammonia (see Testing for Positive Ions above) is used again here. We add sodium hydroxide to a solution of the unknown substance and gently warm it. If no ammonia is detected, we add some aluminium powder. This reduces the nitrate ions to ammonium ions. These react with the sodium hydroxide to produce ammonia gas, which is given off. This is detected using damp red litmus which will turn blue.

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Testing for Carbon=Carbon Double Bonds
200 years ago, the Swedish scientist Jöns Jakob Berzelius decided to categorise all substances depending on their behaviour when heated. Chemicals which burned or charred when heated came mainly from living things, so were called organic substances. Other substances melted or vaporised when heated, and returned to their original state when cooled – these were inorganic substances. However, nowadays we refer to anything “organic” as based on the element carbon.

Unsaturated hydrocarbons will react with bromine water to give a colourless compound (see Crude Oil, C1). Unsaturated hydrocarbons contain carbon-carbon double bonds(C=C), so this is a good test to detect them. It is the basis for detecting C=C bonds in unsaturated oils and fats. If we want to do this, the oil is titrated against an iodine solution (iodine solution reacts in the same way with C=C bonds as bromine water). The iodine number is then calculated – based on the number of iodine molecules needed to react with all of the C=C bonds in one molecule of fat.

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Combustion Analysis
We can work out the empirical formula (see Chemical Calculations, C2) of an organic compound by burning it and measuring the amounts of the formed products. For example, an organic substance A contains hydrogen and carbon. A sample of A is burnt in an excess of oxygen, producing 1.80g of water and 3.52g of carbon dioxide. To work out the empirical formula for A we would:

Firstly, calculate the moles of carbon dioxide:                                                        The relative atomic mass of carbon dioxide is 12 + (2 x 16) = 44g                                Amount of carbon dioxide = 3.52 ÷ 44 = 0.08 moles                                                Then calculate the moles of water: The relative atmoic mass of water is (2 x 1) + 16 = 18g.  Amount of water = 1.80 ÷ 18 = 0.1 mole                                                      Each molecule of carbon dioxide formed requires one carbon atom from a molecule of A. So for every mole of carbon dioxide formed, A must contain one mole of carbon atoms                                                                                                               Amount of carbon atoms in sample of A = 0.08 moles                                           A must contain two moles of hydrogen atoms The Amount of hydrogen atoms in sample of A = 0.1 x 2 = 0.2 moles  So A contains carbon atoms and hydrogen atoms in the ratio:   0.08 : 0.20 = 2:5The empirical formula of A is then C2H5 

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