An acid is a proton donor.
A base is a proton acceptor.
A proton is the same as a hydrogen ion. A good way to think about that is to realise that a hydrogen atom is just one proton and zero neutrons surrounded by only one electron. If that atom becomes an ion by the removal of the electron, then only one proton is left.
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 acid | Formula | Salt | Formula ion |
|---|---|---|---|
| sulfuric acid | H2SO4 | sulfate | SO42- |
| hydrochloric acid | HCl | chloride | Cl- |
| nitric acid | HNO3 | nitrate | NO3- |
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:
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:
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:
A base is a substance that neutralises an acid by combining with the hydrogen ions in them to produce water.
A base usually means a metal oxide, a metal hydroxide or ammonia.
Alkalis are bases which are soluble in water.
Some metal oxides are soluble in water and react with it to form solutions of metal hydroxides. For example:
Na₂O (s) + H₂O (l) → 2NaOH (aq)
Apart from this and other group 1 oxides (such as potassium oxide) most other metal oxides are not soluble in water.
One exception is calcium oxide which does dissolve slightly in water to form calcium hydroxide (known as limewater):
CaO (s) + H₂O (l) → Ca(OH)₂ (aq)
Ammonia is another base. Ammonia reacts with water to form ammonium ions and hydroxide ions:
NH₃ (aq) + H₂O (l) ⇋ NH₄⁺ (aq) + OH⁻ (aq)
All the solutions produced here contain hydroxide ions (OH⁻) so they are all alkalis.
Excess Solid Method:
Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)
| Step | Explanation |
|---|---|
| Heat acid (H2SO4) in a beaker | Speeds up the rate of reaction |
| Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rod | Neutralises all the acid |
| Filter the mixture using filter paper and funnel | Removes any excess copper oxide |
| Gently heat the filtered solution (CuSO4) | To evaporate some of the water |
| until crystals form on a glass rod | Shows a hot saturated solution formed |
| Allow the solution to cool so that hydrated crystals form | Copper sulfate less soluble in cold solution |
| Remove the crystals by filtration | Removes crystals |
| Dry by leaving in a warm place | Evaporates the water |
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.
| Step | Explanation |
|---|---|
| Pipette 25cm3 of alkali (NaOH) into a conical flask | Accurately measures the alkali (NaOH) |
| Do not add indicator | Prevents contamination of the pure crystals with indicator |
| Using the titration values, titrate the known volume acid (HCl) into conical flask containing alkali | Exactly neutralises all of the alkali (NaOH) |
| Transfer to an evaporating basin & heat the solution | Forms a hot saturated solution (NaCl(aq)) |
| Allow the solution to cool so that hydrated crystals form | Sodium chloride is less soluble in cold water |
| Remove the crystals by filtration and wash with distilled water | Removes any impurities |
| Dry by leaving in a warm place | Evaporates 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:
Precipitation Method:
Preparing pure dry crystals of silver chloride (AgCl) from silver nitrate solution (AgNO3) and potassium chloride solution (KCl)
| Step | Explanation |
|---|---|
| Mix the two salt solutions together in a beaker | Forms a precipitate of an insoluble salt (AgCl) |
| Stir with glass rod | Make sure all reactants have reacted |
| Filter using filter paper and funnel | Collect the precipitate (AgCl) |
| Wash with distilled water | Removes any the other soluble salts (KNO3) |
| Dry by leaving in a warm place | Evaporates the water |
Excess Solid Method:
Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)
| Step | Explanation |
|---|---|
| Heat acid (H2SO4) in a beaker | Speeds up the rate of reaction |
| Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rod | Neutralises all the acid |
| Filter the mixture using filter paper and funnel | Removes any excess copper oxide |
| Gently heat the filtered solution (CuSO4) | To evaporate some of the water |
| until crystals form on a glass rod | Shows a hot saturated solution formed |
| Allow the solution to cool so that hydrated crystals form | Copper sulfate less soluble in cold solution |
| Remove the crystals by filtration | Removes crystals |
| Dry by leaving in a warm place | Evaporates the water |
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)
Test for hydrogen gas (H2)
Test for carbon dioxide (CO2)
Tests for gases
| Gas | Test | Result if gas present |
|---|---|---|
| hydrogen (H2) | Use a lit splint | Gas pops |
| oxygen (O2) | Use a glowing splint | Glowing splint relights |
| carbon dioxide (CO2) | Bubble the gas through limewater | Limewater turns cloudy |
| ammonia (NH3) | Use red litmus paper | Turns damp red litmus paper blue |
| chlorine (Cl2) | Use damp litmus paper | Turns damp litmus paper white (bleaches) |
A flame test is used to show the presence of certain metal ions (cations) in a compound.
Properties of the platinum or nichrome wire is:
When put into a roaring bunsen burner flame on a nichrome wire, compounds containing certain cations will give specific colours as follows.
| Ion | Colour in flame test |
|---|---|
| lithium (Li⁺) | red |
| sodium (Na⁺) | yellow |
| potassium (K⁺) | lilac |
| calcium (Ca²⁺) | orange-red |
| copper (II) (Cu²⁺) | blue-green |
Describe tests for the cations Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution
First, add sodium hydroxide (NaOH), then observe the colour:
Describe tests for anions: Carbonate ions (CO32-)
Describe tests for anions: Halide ions (Cl–, Br– and I–)
Underneath are the tests for:
chloride ion test, bromide ion test, iodide ion test, sulfate ion test, carbonate ion test
Describe tests for anions: Sulfate ions (SO42–)
Describe tests for anions: Carbonate ions (CO32-)
Add anhydrous copper (II) sulfate (CuSO4) to a sample.
If water is present the anhydrous copper (II) sulfate will change from white to blue.
If the sample is pure water it will boil at 100oC
Exothermic: chemical reaction in which heat energy is given out.
Endothermic: chemical reaction in which heat energy is taken in.
(So, in an exothermic reaction the heat exits from the chemicals so temperature rises)
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:
| Initial temp. of solution (oC) | Maximium temp. of solution (oC) | Temperature rise (oC) |
|---|---|---|
| 24.2 | 56.7 | 32.5 |
Note: mass of 50 cm3 of solution is 50 g
The cup used is polystyrene because:
polystyrene is an insulator which reduces heats loss
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:
| 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) |
|---|---|---|---|---|---|---|
| 100 | 24.2 | 54.2 | 30.0 | 34.46 | 33.68 | 0.78 |
The amount of energy produced per gram of ethanol burnt can also be calculated:
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:
| Initial temp. of solution (oC) | Maximium temp. of solution (oC) | Temperature rise (oC) |
|---|---|---|
| 24.2 | 56.7 | 32.5 |
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:
| 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) |
|---|---|---|---|---|---|---|
| 100 | 24.2 | 54.2 | 30.0 | 34.46 | 33.68 | 0.78 |
The amount of energy produced per gram of ethanol burnt can also be calculated:
The symbol ΔH is used to represent the change in heat (or enthalpy change) of a reaction.
ΔH is measured in kJ/mol (kilojoules per mole).
The change in heat (enthalpy change) can be represented on an energy level diagram. ΔH must also labelled.
In an exothermic reaction, the reactants have more energy than the products.
Energy is given out in the form of heat which warms the surroundings.
ΔH is given a negative sign, because the reactants are losing energy as heat, e.g ΔH = -211 kJ/mol.
In an endothermic reaction, the reactants have less energy than the products.
Energy is taken in which cools the surroundings.
ΔH is given a positive sign, because the reactants are gaining energy, e.g ΔH = +211 kJ/mol.
During chemical reactions, the bonds in the reactants must be broken, and new ones formed to make the products.
Breaking bonds need energy and therefore is described as endothermic.
Energy is released when new bonds are made and therefore is described as exothermic.
If bonds are both broken and made during chemical reactions, why can a reaction overall be describe as either exothermic or endothermic?
Example: hydrogen reacts with oxygen producing water. Overall energy is released and therefore the reaction is exothermic.
The reaction is exothermic because the energy needed to break the bonds is less than the energy released in making new bonds.
If a reaction is endothermic then the energy needed to break the bonds is more than the energy released in making new bonds.
Each type of chemical bond has a particular bond energy. The bond energy can vary slightly depending what compound the bond is in, therefore average bond energies are used to calculate the change in heat (enthalpy change, ΔH) of a reaction.
Example: dehydration of ethanol
Note: bond energy tables will always be given in the exam, e.g:
| Bond | Average bond energy in kJ/mol |
|---|---|
| H-C | 412 |
| C-C | 348 |
| O-H | 463 |
| C-O | 360 |
| C=C | 612 |
So the enthalpy change in this example can be calculated as follows:
| Breaking bonds | Making bonds | ||
|---|---|---|---|
| Bonds | Energy (kJ/mol) | Bonds | Energy (kJ/mol) |
| H-C x 5 | (412 x 5) = 2060 | C-H x 4 | (412 x 4) = 1648 |
| C-C | 348 | C=C | 612 |
| C-O | 360 | O-H x 2 | (463 x 2) = 926 |
| O-H | 463 | ||
| Energy needed to break all the bonds | 3231 | Energy released to make all the new bonds | 3186 |
Enthalpy change, ΔH = Energy needed to break all the bonds - Energy released to make all the new bonds ΔH = 3231 – 3186 = +45 kJ/mol (ΔH is positive so the reaction is endothermic) |
|||
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:
| Initial temp. of solution (oC) | Maximium temp. of solution (oC) | Temperature rise (oC) |
|---|---|---|
| 24.2 | 56.7 | 32.5 |
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:
| 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) |
|---|---|---|---|---|---|---|
| 100 | 24.2 | 54.2 | 30.0 | 34.46 | 33.68 | 0.78 |
The amount of energy produced per gram of ethanol burnt can also be calculated:
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) |
|---|---|---|---|---|
| 50 | 0 | 0.10 | 45 | 0.0222 |
| 40 | 10 | 0.08 | 60 | 0.0167 |
| 30 | 20 | 0.06 | 80 | 0.0125 |
| 20 | 30 | 0.04 | 13 | 0.0769 |
| 10 | 40 | 0.02 | 255 | 0.0039 |
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.
Increasing the surface area of a solid increases the rate of a reaction.
Increasing the concentration of a solution increases the rate of a reaction.
Increasing the pressure of a gas increases the rate of a reaction.
Increasing the temperature increases the rate of a reaction.
Using a catalyst increases the rate of a reaction.
Increasing the surface area of a solid:
Increasing the concentration of a solution or pressure of a gas:
Increasing the temperature:
A catalyst is a substance that increases the rate of a reaction, but is chemically unchanged at the end of the reaction.
Catalyst: A substance that speeds up a chemical reaction while remaining chemically unchanged at the end of
the reaction.
A catalyst is not used up in a reaction.
A catalyst speeds up a reaction by providing an alternative pathway with lower activation energy.
Below is a diagram showing the reaction profile for the reaction of hydrogen with oxygen, which is EXOTHERMIC:
The activation energy is the minimum amount of energy required to start the reaction.
For an exothermic reaction, the products have less energy than the reactants. The difference between these energy levels is ΔH.
For an exothermic reaction, more energy is released when bonds are formed than taken in when bonds are broken.
Below is a diagram showing the reaction profile for the thermal decomposition of calcium carbonate, which is ENDOTHERMIC:
The activation energy is the minimum amount of energy required to start the reaction.
For an endothermic reaction, the products have more energy than the reactants. The difference between these energy levels is ΔH.
For an endothermic reaction, more energy is taken in to break bonds than is released when new bonds are formed.
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.
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:
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.
Some reactions are reversible. This is indicated by the symbol ⇌
For example:
Dehydration of copper(II) sulfate
Heating ammonium chloride
On heating, white solid ammonium chloride decomposes forming ammonia and hydrogen chloride gas. On cooling, ammonia and hydrogen chloride react to form a white solid of ammonium chloride:
A dynamic equilibrium can only be achieved in a sealed container so substances cannot escape.
Features of a reaction mixture that is in dynamic equilibrium:
A catalyst is a substance which increases the rate of reaction without being chemically changed at the end of the reaction.
A reversible reaction is one where the forward reaction and the backward reaction happen simultaneously. For example:
3H₂ + N₂ ⇋ 2NH₃
In such a reaction a catalyst speeds up both the forward and the backward reactions. Hence, although the system will reach dynamic equilibrium more quickly, the addition of a catalyst will not affect the position of equilibrium.
In a reversible reaction the position of the equilibrium (the relative amounts of reactants and products) is dependent on the temperature and pressure of the reactants.
If the conditions of an equilibrium reaction are changed, the reaction moves to counteract that change.
Therefore by altering the temperature or pressure the position of the equilibrium will change to give more or less products.
Adding a catalyst does not affect the position of the equilibrium.
If a change in conditions moves equilibrium to the right, the yield of the substances on the right is increased.
Changing the temperature:
All reactions are exothermic in one direction and endothermic in the other way.
For this reaction the enthalpy change, ΔH is negative therefore the forward reaction is exothermic:
CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH = –91 kJ/mol
If temperature is decreased the position of the equilibrium will shift to the right because it is an exothermic reaction.
For this reaction the enthalpy change, ΔH is positive therefore the forward reaction is endothermic:
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ΔH = +210 kJ mol–1
If temperature is increased the position of the equilibrium will shift to the right because it is an endothermic reaction.
Key point: an increase (or decrease) in temperature shifts the position of equilibrium in the direction of the endothermic (or exothermic) reaction
Changing the pressure:
Reactions may have more molecules of gas on one side than on the other.
For this reaction there are 2 molecules on the left and 4 molecules on the right:
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ΔH = +210 kJ mol–1
If the pressure is increased the position of the equilibrium will shift to the left because there are fewer molecules on the left-hand side.
For this reaction there are 3 molecules on the left and 1 molecule on the right
CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH = –91 kJ/mol
If the pressure is decreased the position of the equilibrium will shift to the left because there are more molecules on the left-hand side.
Key point: an increase (or decrease) in pressure shifts the position of equilibrium in the direction that produces fewer (or more) moles of gas
Hydrocarbon: A molecule containing only hydrogen and carbon
The molecular formula shows the actual number of atoms of each element in a molecule.
The empirical formula shows the simplest whole number ratio of atoms present in a compound. So the molecular formula is a multiple of the empirical formula.
The general formula shows the relationship between the number of atoms of one element to another within a molecule. Members of a homologous series share the same general formula. The general formula for alkanes is CnH2n+2 and the general formula for alkenes is CnH2n.
A structural formula shows how the atoms in a molecule are joined together.
The displayed formula is a full structural formula which shows all the bonds in a molecule as individual lines.
The terms above are demonstrated with the example of butane.
The terms above are demonstrated with the example of ethene, which contains a double bond.
Isomers are molecules with the same molecular formula but with a different structure.
A functional group is an atom or a group of atoms that determine the chemical properties of a compound.
For example the functional group of an alcohol is the -OH group and that of alkenes is the C=C carbon to carbon double bond.
A Homologous series is a group of substances with:
Isomers are molecules with the same molecular formula but with a different structure.
The names of organic molecules are based on the number of carbon atoms in the longest chain. This chain is the longest consecutive line of carbon atoms, even if this line bends.
| The name is based on the number of carbon atoms in the longest chain |
|---|
| 1 Meth- |
| 2 Eth- |
| 3 Prop- |
| 4 But- |
| 5 Pent- |
| 6 Hex- |
| 7 Hept- |
| 8 Oct- |
| 9 Non- |
| 10 Dec- |
Hydrocarbons are molecules which contain only hydrogen and carbon.
Naming straight-chain alkanes
The simplest hydrocarbons are alkanes. They contain only single bonds, and have “-ane” in the name.
For example, the displayed formula of ethane (C₂H₆) is:
The name “ethane” contains “eth-” because there are 2 carbon atoms in the longest chain, and the name contains “-ane” because the molecule only has single bonds so is an alkane.
Another example is pentane (C₅H₁₂) which has the displayed formula:
The name “pentane” contains “pent-” because there are 5 carbon atoms in the longest chain, and the name contains “-ane” because the molecule only has single bonds, so is an alkane.
Remember, it does not matter if the longest consecutive line of carbons bends around. For example the displayed formula below still shows a very normal molecule of pentane (5 carbons in a row). Pentane is not normally drawn with the longest chain of carbons bent around because it could be confusing.
You might also see the bonds drawn at angles. Don’t worry, the displayed formula below is still pentane, as can be seen by the fact there are 5 carbon atoms in the longest chain, surrounded by hydrogen atoms bonded to the carbon atoms by single bonds.
A shorter way to express the detailed structure of an organic molecule is the structural formula. The structural formula for pentane is CH₃-CH₂-CH₂-CH₂-CH₃, which tells us the same information about the molecule as does the displayed formula, without the hassle of having to draw all the bonds or all the hydrogen atoms.
Naming straight-chain alkenes
Another simple group of hydrocarbons is the alkenes. They contain a carbon-to-carbon double bond, which also means they have two fewer hydrogen atoms than their corresponding alkane. An alkene has “-ene” in its name.
For example, the displayed formula for ethene (C₂H₄) is:
and the displayed formula of propene (C₃H₆) is:
With longer alkene molecules the double bond might appear in different locations of the carbon chain, so the name needs to be a little bit more complicated to be able to describe these differences clearly. A number is added in the middle of the name to indicate at which carbon the double bond starts.
So the displayed formula of pent-1-ene is:
and this is the displayed formula of pent-2-ene:
However, take care that when counting which carbon has the double bond. The numbers start from the end that produces the smallest numbers in the name. For example, this is the displayed formula for pent-1-ene again, but just drawn the other way round. It is still pent-1-ene (you can’t get pent-4-ene):
Naming straight-chain alcohols
We get the same pattern all over again with the group of organic molecules called alcohols, which are recognised by an -OH functional group. For example here is the displayed formula for ethanol, which has 2 carbon atoms in the longest chain:
and here is the displayed formula for butanol:
Summary of naming simple straight-chain organic molecules
The following table summarises the naming of some of the straight-chain alkanes, alkenes and alcohols, giving a name and a molecular formula for each:
| Carbons in longest chain | Alkanes | Alkenes | Alcohols |
|---|---|---|---|
| 1 | methane, CH₄ | - | methanol, CH₄O |
| 2 | ethane, C₂H₆ | ethene, C₂H₄ | ethanol, C₂H₆O |
| 3 | propane, C₃H₈ | propene, C₃H₆ | propanol, C₃H₈O |
| 4 | butane, C₄H₁₀ | butene, C₄H₈ | butanol, C₄H₁₀O |
Naming branched alkanes and alkenes
The naming conventions for organic molecules cover more than the straight chain molecules. Branched molecules are named depending on the number of carbon atoms in the branch. A branch with 1 carbon is called “methyl” and a branch with 2 carbons is called “ethyl”. This is similar to the conventions covered above, plus the “-yl-” bit just says it is a branch.
For example, this is the displayed formula for 2-methyl hexane:
In the name 2-methyl hexane, the number 2 indicates that when counting along the longest carbon chain the methyl branch comes off the second carbon atom. The “methyl” bit of the name says there is one branch of 1 carbon. The “hex” bit of the name says the longest consecutive chain of carbon atoms is 6. The “ane” bit says the molecule has only single bonds.
When counting along the carbon atoms of the longest chain to work out the name, the numbering of carbon atoms starts from the end nearest to the branch. Another way to put this is that the name is given such that the numbers in the name are as low as possible. For example, here is the displayed formula for 4-ethyl octane:
Another example, this time with 2 methyl branches coming off the second and third carbons of the chain, is 2,3-dimethyl hexane. The “di” in the name indicates there are two methyl groups. This is the same way in which “di” indicates there are two oxygen atoms in carbon dioxide.
Another example of how the naming convention works for branches is 2,2-dimethyl hexane:
This naming of branches also applies to alkenes. Here is the displayed formula of 4-methylpent-1-ene:
The molecular formula describes the actual number of each type of each atom in a molecule.
For example, a molecular formula of C₆H₁₂ tells us that in each molecule there are 6 carbon atoms and 12 hydrogen atoms.
However, the molecular formula tells us nothing about how those atoms are arranged. For example, it does not tell us if there any branches of carbon atoms coming off the main carbon-carbon chain, nor how long or how many there might be.
On the other hand, the structural formula and displayed formula of a molecule tell us clearly how the atoms are arranged in that molecule.
This means that if we are given a molecular formula only, there may be several possible structural and displayed formulae all of which could apply for that molecule.
When trying to work out possible structural or displayed formulae from a molecular formula there are several clues:
In a substitution reaction an atom or group of atoms is replaced by a different atom or group of atoms. For example when ethane reacts with bromine gas one of the hydrogen atoms in ethane is substituted by one of the atoms of bromine from within the bromine molecule:
CH₃-CH₃ + Br-Br → CH₃-CH₂Br + H-Br
ethane + bromine → bromoethane + hydrogen bromide
An addition reaction occurs when an atom or group of atoms is added to a molecule without taking anything away. For example when ethene reacts with bromine gas, the product is simply the addition of the two molecules:
CH₂=CH₂ + Br-Br → CH₂Br-CH₂Br
A combustion reaction is another way to say ‘burning’ and is a reaction with oxygen. Combustion of hydrocarbons with excess oxygen gives the products water and carbon dioxide, and also releases heat energy (exothermic reaction). Two examples the combustion of propane and the combustion of butene:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
C₄H₈ + 6O₂ → 4CO₂ + 4H₂O
Crude Oil is a mixture of hydrocarbons.
Crude oil is separated into fractions by the process of fractional distillation.
| Fraction | Use |
|---|---|
| Refinery gases | Bottled gas |
| Gasoline | Fuel for cars |
| Kerosene | Fuel for aeroplanes |
| Diesel Oil | Fuel for lorries |
| Fuel Oil | Fuel for ships |
| Bitumen | Road Surfacing |
The boiling point increases as the number of carbon atoms (chain length) increases.
The viscosity increases as the number of carbon atoms (chain length) increases.
The greater the number of carbon atoms (chain length), the darker in colour that fraction is.
The viscosity of a fluid describes how easily it flows. Water has a low viscosity, it flows very easily. Crude oil has a higher viscosity than water, it does not flow very easily.
| Fractions (in order) | Properties |
|---|---|
| Refinery gases | Smallest molecules. Lowest boiling point. Lowest viscosity. Lightest in colour. |
| Gasoline | |
| Kerosene | |
| Diesel | |
| Fuel oil | |
| Bitumen | Largest molecules. Highest boiling point. Highest viscosity. Darkest in colour. |
