Cambridge International AS Level Chemistry Coursebook Chapter 1: Moles and equations 4

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Chemical formulae and chemical equations

Deducing the formula

The electronic structure of the individual elements in a compound determines the formula of a compound (see

page 33). The formula of an ionic compound is determined by the charges on each of the ions present. The number of positive charges is balanced by the number of negative charges so that the total charge on the compound is zero. We can work out the formula for a compound if we know the charges on the ions. Figure 1.11 shows the charges on some simple ions related to the position of the elements in the Periodic Table. The form of the Periodic Table that we shall be using has 18 groups because the transition elements are numbered as Groups 3 to 12. So, aluminium is in Group 13 and chlorine is in Group 17.

For simple metal ions in Groups 1 and 2, the value of the positive charge is the same as the group number. For a simple metal ion in Group 13, the value of the positive charge is 3+. For a simple non-metal ion in Groups 15 to 17, the value of the negative charge is 18 minus the group number. The charge on the ions of transition elements can vary. For example, iron forms two types of ions, Fe2+ and Fe3+ (Figure 1.12).

Figure 1.11 The charge on some simple ions is related to their position in the Periodic Table.

Figure 1.12 Iron(II) chloride (left) and iron(III) chloride (right). These two chlorides of iron both contain iron and chlorine, but they have different formulae.

Ions that contain more than one type of atom are called compound ions. Some common compound ions that you should learn are listed in Table 1.3. The formula for an ionic compound is obtained by balancing the charges of the ions.

Table 1.3 The formulae of some common compound ions.

worked examples

11 Deduce the formula of magnesium chloride.

Ions present: Mg2+ and Cl–.

For electrical neutrality, we need two Cl– ions for every Mg2+ ion.

(2 × 1–) + (1 × 2+) = 0

So the formula is MgCl2.

12 Deduce the formula of aluminium oxide.

Ions present: Al3+ and O2–.

For electrical neutrality, we need three O2– ions for every two Al3+ ions.

(3 × 2?) + (2 × 3+) = 0

So the formula is Al2O3.

The formula of a covalent compound is deduced from the number of electrons needed to achieve the stable electronic configuration of a noble gas (see page 49). In general, carbon atoms form four bonds with other atoms, hydrogen and halogen atoms form one bond and oxygen atoms form two bonds. So the formula of water, H2O, follows these rules. The formula for methane is CH4, with each carbon atom bonding with four hydrogen atoms. However, there are many exceptions to these rules.

Compounds containing a simple metal ion and non-metal ion are named by changing the end of the name of the non-metal element to -ide.

sodium + chlorine --> sodium chloride

zinc + sulfur -->zinc sulfide

Compound ions containing oxygen are usually called -ates. For example, the sulfate ion contains sulfur and oxygen, the phosphate ion contains phosphorus and oxygen.

Question

11 a Write down the formula of each of the following compounds:

i magnesium nitrate

ii calcium sulfate

iii sodium iodide

iv hydrogen bromide

v sodium sulfide

b Name each of the following compounds:

i Na3PO4 iii AlCl3

ii (NH4)2SO4 iv Ca(NO3)2

Balancing chemical equations

When chemicals react, atoms cannot be either created or destroyed. So there must be the same number of each type of atom on the reactants side of a chemical equation as there are on the products side. A symbol equation is a shorthand way of describing a chemical reaction. It shows the number and type of the atoms in the reactants and the number and type of atoms in the products. If these are the same, we say the equation is balanced. Follow these examples to see how we balance an equation.

Question 12 Write balanced equations for the following reactions.

a Iron reacts with hydrochloric acid to form iron(II) chloride, FeCl2, and hydrogen.

b Aluminium hydroxide, Al(OH)3, decomposes on heating to form aluminium oxide, Al2O3, and water.

c Hexane, C6H14, burns in oxygen to form carbon dioxide and water.

Using state symbols

We sometimes find it useful to specify the physical states of the reactants and products in a chemical reaction. This is especially important where chemical equilibrium and rates of reaction are being discussed (see Chapter 8 and Chapter 9). We use the following state symbols:

(s) solid

(l) liquid

(g) gas

(aq) aqueous (a solution in water).

State symbols are written after the formula of each reactant and product. For example:

ZnCO3(s) +?H2SO4(aq) --> ZnSO4(aq) +?H2O(l) + CO2(g)

Question 13 Write balanced equations, including state symbols, for the following reactions.

a Solid calcium carbonate reacts with aqueous hydrochloric acid to form water, carbon dioxide and an aqueous solution of calcium chloride.

b An aqueous solution of zinc sulfate, ZnSO4, reacts with an aqueous solution of sodium hydroxide. The products are a precipitate of zinc hydroxide, Zn(OH)2, and an aqueous solution of sodium sulfate.

Figure 1.13 The reaction between calcium carbonate and hydrochloric acid.

The equation for this reaction, with all the state symbols, is:

CaCO3(s)? + ?2HCl(aq) -->CaCl2(aq)? + ?CO2(g)? + ?H2O(l)

Balancing ionic equations

When ionic compounds dissolve in water, the ions separate from each other. For example:

NaCl(s)? + ?aq --> Na+(aq)? + ?Cl–(aq)

Ionic compounds include salts such as sodium bromide, magnesium sulfate and ammonium nitrate. Acids and alkalis also contain ions. For example H+ (aq) and Cl–(aq) ions are present in hydrochloric acid and Na+(aq) and OH–(aq) ions are present in sodium hydroxide.

Many chemical reactions in aqueous solution involve ionic compounds. Only some of the ions in solution take part in these reactions.

The ions that play no part in the reaction are called spectator ions.

An ionic equation is simpler than a full chemical equation. It shows only the ions or other particles that are reacting. Spectator ions are omitted. Compare the full equation for the reaction of zinc with aqueous copper(II) sulfate with the ionic equation.

full chemical equation:

Zn(s)? + ?CuSO4(aq) --> ZnSO4(aq)? + ?Cu(s)

with charges:

Zn(s)? + ?Cu2+SO42–(aq) --> Zn2+SO42–(aq)? + ?Cu(s)

cancelling spectator ions:

Zn(s)? + ?Cu2+SO42–(aq) --> Zn2+SO42–(aq)? + ?Cu(s)

ionic equation:

Zn(s)? + ?Cu2+(aq) --> Zn2+(aq)? + ?Cu(s)

In the ionic equation you will notice that:

there are no sulfate ions – these are the spectator ions as they have not changed

both the charges and the atoms are balanced.

The next examples show how we can change a full equation into an ionic equation.

Question 14 Change these full equations to ionic equations.

a H2SO4(aq) +?2NaOH(aq) --> 2H2O(aq) +?Na2SO4(aq)

b Br2(aq)? + ?2KI(aq) --> 2KBr(aq)? + ?I2(aq)

Chemists usually prefer to write ionic equations for precipitation reactions. A precipitation reaction is a reaction where two aqueous solutions react to form a solid

– the precipitate. For these reactions the method of writing the ionic equation can be simplified. All you have to do is:

write the formula of the precipitate as the product

write the ions that go to make up the precipitate as the reactants.

Solutions and concentration

Calculating the concentration of a solution

The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 (one cubic

decimetre) of solution. The solvent is usually water. There are 1000 cm3 in a cubic decimetre. When 1 mole of a compound is dissolved to make 1 dm3 of solution the concentration is 1 mol dm–3.

concentration (mol dm–3)

= number of moles of solute (mol) / volume of solution (dm3)

We use the terms concentrated and dilute to refer to the relative amount of solute in the solution. A solution with a low concentration of solute is a dilute solution. If there is a high concentration of solute, the solution is concentrated.

When performing calculations involving concentrations in mol dm–3 you need to:

change mass in grams to moles

change cm3 to dm3 (by dividing the number of cm3 by 1000).

Figure 1.14 The concentration of chlorine in the water in a swimming pool must be carefully controlled.

We often need to calculate the mass of a substance present in a solution of known concentration and volume. To do this we:

rearrange the concentration equation to: number of moles (mol)

= concentration (mol dm–3) × volume (dm3)

multiply the moles of solute by its molar mass mass of solute (g)

= number of moles (mol) × molar mass (g mol–1)

Question

16 a Calculate the concentration, in mol dm–3, of the following solutions: (Ar values: C = 12.0, H = 1.0, Na = 23.0, O = 16.0)

i a solution of sodium hydroxide, NaOH, containing 2.0 g of sodium hydroxide in 50 cm3 of solution

ii a solution of ethanoic acid, CH3CO2H, containing 12.0 g of ethanoic acid in 250 cm3 of solution.

b Calculate the number of moles of solute dissolved in each of the following:

i 40 cm3 of aqueous nitric acid of concentration 0.2 mol dm–3

ii 50 cm3 of calcium hydroxide solution of concentration 0.01 mol dm–3.

Carrying out a titration

A procedure called a titration is used to determine the amount of substance present in a solution of unknown concentration. There are several different kinds of titration. One of the commonest involves

the exact neutralisation of an alkali by an acid (Figure 1.15).

If we want to determine the concentration of a solution of sodium hydroxide we use the following procedure.

Get some of acid of known concentration.

Fill a clean burette with the acid (after having washed the burette with a little of the acid).

Record the initial burette reading.

Measure a known volume of the alkali into a titration flask using a graduated (volumetric) pipette.

Add an indicator solution to the alkali in the flask.

Slowly add the acid from the burette to the flask, swirling the flask all the time until the indicator changes colour (the end-point).

Record the final burette reading. The final reading minus the initial reading is called the titre. This first titre is normally known as a 『rough』 value.

Repeat this process, adding the acid drop by drop near the end-point.

Repeat again, until you have two titres that are no more than 0.10 cm3 apart.

Take the average of these two titre values.

Your results should be recorded in a table, looking like this:

Figure 1.15? a A funnel is used to fill the burette with hydrochloric acid. b A graduated pipette is used to measure 25.0?cm3 of sodium hydroxide solution into a conical flask. c An indicator called litmus is added to the sodium hydroxide solution, which turns blue. d 12.5?cm3 of hydrochloric acid from the burette have been added to the 25.0?cm3 of alkali in the conical flask. The litmus has gone red, showing that this volume of acid was just enough to neutralise the alkali.

You should note:

all burette readings are given to an accuracy of 0.05 cm3

the units are shown like this 『/ cm3』

the two titres that are no more than 0.10 cm3 apart are 1 and 3, so they would be averaged

the average titre is 34.70 cm3.

In every titration there are five important pieces of knowledge:

1 the balanced equation for the reaction

2 the volume of the solution in the burette (in the example above this is hydrochloric acid)

3 the concentration of the solution in the burette

4 the volume of the solution in the titration flask (in the example above this is sodium hydroxide)

5 the concentration of the solution in the titration flask.

If we know four of these five things, we can calculate the fifth. So in order to calculate the concentration of sodium hydroxide in the flask we need to know the first four of these points.