Conservation of mass means we have to have the same number of atoms of each element on the right and left of the equation. Use a pencil so you can change your mind. It can often be a bit of “trial and error”. Use whole numbers (there are some exceptions to this rule). Start with less common elements. Oxygen and hydrogen are best left until last if you can, especially if want is involved.

1. The thermite reaction Al + Fe 2 O 3  Al 2 O 3 + Fe 2. Combustion of heptane C 7 H 16 + O 2  CO 2 + H 2 O 3. Thermal decomposition of lead nitrate Pb(NO 3 ) 2  PbO + NO 2 + O 2

1. The thermite reaction 2Al + Fe 2 O 3  Al 2 O 3 + 2Fe 2. Combustion of heptane C 7 H 16 + O 2  CO 2 + H 2 O 3. Thermal decomposition of lead nitrate Pb(NO 3 ) 2  PbO + NO 2 + O 2

1. The thermite reaction 2Al + Fe 2 O 3  Al 2 O 3 + 2Fe 2. Combustion of heptane C 7 H O 2  7CO 2 + 8H 2 O 3. Thermal decomposition of lead nitrate Pb(NO 3 ) 2  PbO + NO 2 + O 2

1. The thermite reaction 2Al + Fe 2 O 3  Al 2 O 3 + 2Fe 2. Combustion of heptane C 7 H O 2  7CO 2 + 8H 2 O 3. Thermal decomposition of lead nitrate 2Pb(NO 3 ) 2  2PbO + 4NO 2 + O 2

An important piece of information in a chemical equation is the state of the substance. On your whiteboard write symbols to show the 4 states (s) = solid (l) = liquid (g) = gas (aq) = aqueous/dissolved in water

State symbols are placed after the chemical formula. You will learn common states for certain types of substance over the 2 IB years. H 2 O is a liquid So we would write H 2 O(l) Try adding state symbols to this equation Zn(...) + 2HCl(...)  ZnCl 2 (...) + H 2 (...)

State symbols are placed after the chemical formula. You will learn common states for certain types of substance over the 2 IB years. H 2 O is a liquid So we would write H 2 O(l) Try adding state symbols to this equation Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g)

Ionic substances dissolve in water and the ions dissociate from each other. E.g. NaCl(s) + aq  Na + (aq) + Cl - (aq) If we have a reaction occurring involving dissolved ionic substances it can be useful to show all the ions. Try writing this out using it’s ions. AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq)

Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl(s) + K + (aq) + NO 3 - (aq) AgCl was solid while others were aqueous, this means it is not soluble in water. How could we simplify this further?

When we write an ionic equation we are interested in seeing which ions change. Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl(s) + K + (aq) + NO 3 - (aq) Which species have changed and which have stayed the same? Ag + (aq) and Cl - (aq) have become AgCl(s) K + (aq) and NO3 - (aq) have remained the same.

K + (aq) and NO 3 - (aq) are spectator ions, i.e. They don’t change in a chemical sense. We can remove these from the ionic equation. How will it look now? Ag + (aq) + Cl - (aq)  AgCl(s) Much simpler and we can see the important chemical change.

Try writing an ionic equation for the reaction of HCl with NaOH. Hint: Acids dissociate like ionic substances, giving off H + (aq) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  Na + (aq) + Cl - (aq) + H 2 O(l) Becomes, H + (aq) + OH - (aq)  H 2 O(l)

What is an equilibrium? When you have both the forward AND reverse reaction occurring at the same time and at the same rate. Like; 2SO 2 (g) + O 2 (g) 2SO 3 (g) Or; 3H 2 (g) + N 2 (g) 2NH 3 (g) (Haber process)

n = m/M is a very useful equation but it is somewhat limited. When might it be difficult to use? When dealing with solutions (substance dissolved in a solvent, often water) as recorded mass would include the solvent. When dealing with gases, as they are difficult to weigh.

How can we work out the mass of a liquid? Weigh it (more difficult than with a solid but not too hard) Measure the volume. If the density is known we can use: density = mass/volume

Solute dissolves in solvent. Concentration of a solution… Is the amount of solute, in mol, dissolved per 1dm 3 (1000cm 3 ) of solution.

How many moles of solute are in a dm 3 of solution with the following concentration? 4 mol dm -3 Yep, it’s 4 So, 2dm 3 would contain 8 moles, and 0.5 dm 3 would contain 2, etc.

n = c x V n = amount of substance (mol) c = concentration of solute (mol dm -3 ) V = volume of solution (dm 3 ) Remember to check whether you have cm 3 or dm 3 for the given volume.

DON’T confuse this solution volume with gas volume. You can’t insert it from one equation to the other like you can with n.

The concentration of a solution can be described using the terms, dilute and concentrated. Dilute is a small amount of solute per dm 3. Concentrated is a large amount per dm 3.

You might see the letter M used in concentrations, this means “molar” and refers to a solution with a concentration in mol dm -3. So, 2 mol dm -3 and 2M are the same thing.

Normally we look at molar concentration, i.e. the number of moles per dm 3. Sometimes you will see a concentration with the units g dm -3. This is telling us the mass of a solid dissolve in 1 dm 3 of solvent. How could we get from g dm -3 to mol dm -3 ?