Amount of Substance
Mole and Avogadro’s constant
A ‘mole’ simply refers to the amount of a certain substance. It's equivalent to 6.02 x 1023 atoms, which chemists like to refer to as Avogadro's number, funnily enough because a guy called Amedeo Avogadro came up with it.
One mole of any element will always weigh the same as its mass number (or RAM). This means that one mole of carbon will weigh 12 g and one mole of calcium will weigh 40 g.
The same thing is true for molecules. Let’s say we have exactly one mole of carbon dioxide gas, CO2. We can determine how much this weighs by working out the relative formula mass (12 + (16 x 2) = 44). So one mole of CO2 weighs 44 g.
Molar gas volume
One mole of a gas always occupies a volume of 24 dm 3 (or 24,000 cm3). This is true of all gases when they are at room temperature (20 oC) and standard pressure (1 atm).
This means that whenever we know the amount of gas (i.e. the moles), or we can calculate the moles from its mass, then we also know its volume. To find the volume in dm3, multiply the number of moles by 24. Remember that to convert between dm3 to cm3, multiply your answer by 1000.
Worked example - molar gas volume
Calculate the volume of 7.52 g of ammonia (NH3) gas at room temperature and pressure.
• Moles = mass / Mr
• 7.52 / 17 = 0.44 mol
• Volume = moles x 24
• 0.44 x 24 = 10.56 dm3
Worked example - using molar ratios to calculate molar gas volume
3.25 g of zinc is reacted with an excess of hydrochloric acid to form zinc chloride and hydrogen gas, as shown in the equation below. Calculate the volume of hydrogen gas (in cm3) formed in this reaction.
• Use the mass of zinc to work out the moles.
• Moles of zinc = 3.25 / 65 = 0.05 mol
• There is a 1 : 1 ratio between zinc and hydrogen. This means that if there are 0.05 moles of zinc reacting, 0.05 moles of hydrogen are formed.
• Volume of hydrogen = moles x 24
• 0.05 x 24 = 1.2 dm3
• 1.2 x 1000 = 1200 cm3
Empirical and molecular formulae
The empirical formula of a compound shows the smallest whole-number ratio of atoms of each element present in the compound. It’s a simplified version of the molecular formula, which gives the actual number of atoms of each element in the compound.
For example, glucose has the molecular formula C6H12O6. So each glucose molecule contains 6 carbon atoms, 12 hydrogen atoms and six oxygen atoms. To work out its empirical formula, divide these numbers by the largest number that they can all be divided by. In this case, we can divide each element by 6, to give the empirical formula CH2O.
You may be asked to work out the empirical formula for a compound when given the percentage composition of different elements in the compound. Read through the worked example below to see how you can do this.
Worked example – calculating empirical formula
A compound contains 40.6% carbon, 5.1% hydrogen and 54.2% oxygen. Calculate its empirical formula.
Water of crystallisation
Water of crystallisation is water that is chemically bonded in a crystal structure. For example, hydrated copper sulfate is a compound of copper sulfate bound to and surrounded by water molecules. The water can be removed through evaporation when heat is applied, creating an anhydrous salt. You may be given the formula of a hydrated salt and be expected to use the molecular mass to calculate the moles of water of crystallisation. You can see how to do this in the worked example below:
Worked example – calculating the formula of a hydrated salt
White crystals of hydrated sodium sulfate, Na2SO4.XH2O were found to have a molar mass of 322.1 g mol-1. Calculate the value of x.
- Work out the molar mass of the anhydrous salt (i.e. everything apart from the water).
- Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142
- Now work out how much of the molar mass is coming from water by subtracting the mass of the anhydrous salt from the mass of the whole hydrated compound.
- 322.1 – 142 = 180.1
- This gives us the total mass of water. To work out how many moles of water there are, divide the mass by the Mr of water.
- 180.1/18 = 10.0056
- X = 10
Molar calculations
The concentration of a solution tells us how much solute is dissolved in a given volume of solvent. A concentrated salt solution will contain a larger mass of salt dissolved in a particular volume of water compared to a dilute salt solution. Concentration can be measured in grams per dm3 (g dm-3) or moles per dm3 (mol dm-3).
A concentration of 1 g dm-3 means that 1 gram of solute is dissolved in every dm3 of solvent. Use the equation concentration = mass/volume to work out the concentration of any given solution.
For example, if I dissolve 25 grams of sodium chloride into a 1000 dm3 of solution, I have made a solution with a concentration of 0.025 g dm-3.
Concentrations in mol dm-3 tell us the amount of solute (in moles) dissolved in every dm3 of solvent. For example, a sodium chloride solution with a concentration of 5 mol dm-3 means that in every dm3 of water there are five moles of dissolved sodium chloride.
We can calculate the concentration of a solution by dividing the moles by the volume. For example, if I have 2 moles of magnesium sulfate and dissolve it in 500 dm3 of water, the concentration of the solution formed would be 2/500 = 0.004 mol dm-3.
We can also work out the concentration of a solution if we know the mass of the solute that we’re dissolving. To do this, we use the moles = mass/Mr equation to convert the mass into moles. We can then use our answer to work out the concentration using the concentration = moles/volume equation. This has been done in the worked example below.
Worked example - using mass of solute to calculate concentration
100 grams of potassium chloride (KCl) is dissolved in 200 dm3 of water. Give the concentration of the resulting solution in mol/dm3.
Mr of potassium = 39, Mr of chlorine = 35.5
• First work out the moles of potassium chloride by dividing the mass by the Mr
• 100 / 74.5 = 1.34 mol
• Then find out the concentration of the solution by dividing the moles by the volume
• 1.34 / 200 = 0.0067 mol/dm3
Concentrations of reacting solutions
If two solutions react completely so that both reactants are completely used up, there must have been equal moles of each reactant. Since moles is equal to concentration x volume, this means that the concentration x volume of solution 1 must be equal to the concentration x volume of solution 2. In other words, C1 x V1 = C2 x V2. This means that if we know the volumes of both solutions and the concentration of one of the solutions is known, the concentration of the other solution can be calculated.
Worked example - calculating concentration of a solution from a reacting solution
A 1 mol dm3 sodium chloride solution with a volume of 500 cm3 completely reacts with 350 cm3 of silver nitrate solution of unknown concentration. A precipitate of silver chloride and aqueous sodium nitrate are formed. Calculate the concentration of the silver nitrate solution.
• Concentration x volume of sodium chloride solution = concentration x volume of silver nitrate solution
• 1 x 500 = ? x 350
• 500 / 350 = 1.43
• Concentration of silver nitrate solution = 1.43 mol dm3
Ideal gas equation
We’ve already learnt that one mole of gas occupies 24 dm3 of volume at room temperature and pressure. Let’s say the temperature or pressure in a certain environment deviates from rtp, then we use the ideal gas equation to work out what volume it occupies.
Worked example - ideal gas equation
What volume will be occupied by 88g of propane gas (C3H8) at a temperature of 30oC and a pressure of 250 kPa?
First, let’s convert everything into the proper units for the ideal gas equation.
• P = 250,000 Pa
• n = mass /Mr so 88 / 44 = 2
• R = 8.314
• T = 30 + 273 = 303 K
I’m now going to put all of those numbers into the equation then rearrange to find v.
PV = nRT
- 250,000 x V = 2 x 8.314 x 303
- 250,000 x V = 5038.3
- V = 5038.3 / 250,000
- V = 0.02 m3
- V = 20 dm3
Percentage yield
The law of conservation states that no atoms are gained or lost in a chemical reaction. In theory, this means that if you start with 10 grams of reactants, you should end up with 10 grams of product. In reality, this is rarely the case for the following reasons:
The reaction might be reversible and won’t go to completion
Some product may be lost when it is separated from the reaction mixture
Some of the reactants may react in a different way to the expected reaction (e.g. if you add a metal with an acid, some of the metal may react with the oxygen in the air before it has chance to react with the acid)
The amount of reactant that is successfully converted into product is known as the yield. The more reactant converted into product, the higher the yield. When we compare this with the maximum amount that can be theoretically obtained, we call this the percentage yield. Percentage yield can be calculated using the equation:
Worked example - percentage yield
In a school laboratory experiment, a teacher combines sulfuric acid with magnesium to produce 6.25 g of magnesium sulfate. If the maximum yield is 8.5 g, calculate the percentage yield in this experiment.
Percentage yield = (actual yield / theoretical yield) x 100
Percentage yield = (6.25 / 8.5) x 100
Percentage yield = 73.5 %
Atom economy
The atom economy of a reaction is a measure of the amount of reactants which are converted into useful products. The fewer waste products there are, the higher the atom economy. It is cheaper and more sustainable to use reactions with a higher atom economy. The percentage atom economy of a reaction is calculated using the formula:
Worked example - atom economy
Ethanol can be manufactured by fermentation of glucose in the presence of yeast. It can also be made through the hydration of ethene. Equations for both reactions are given below:
Calculate the atom economy for each of these reactions.
For the fermentation of glucose:
Mr of desired product (ethanol) = 92
Mr of total products = 92 + 88 = 180
(Mr of desired product / Mr of total product) x 100 = 51%
For the hydration of ethene, ethanol is the only product formed which means that the atom economy is 100% (i.e. all of the reactants are converted into useful product).