Topic 5 - Formulae, Equations, + Amounts of Substance
In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include determining a simple empirical formula such as MgO or CuO, determining the number of moles of water of crystallisation in a
salt such as Epsom salts, performing a wide range of titrations involving different indicators, preparing salts.
Mathematical skills that could be developed in this topic include converting between units such as cm 3 and dm3, using standard form with the
Avogadro constant, rearranging formulae for calculating moles in solids and in solutions, calculating atom economy, dealing with percentage errors.
Within this topic, students first encounter core practicals and can consider ideas of measurement uncertainty, evaluating their results in terms of
systematic and random errors. They can also consider how the concept of atom economy is useful to help chemists make decisions so that reactions
can be made more efficient in terms of resources.
1. know that the mole (mol) is the unit for amount of a substance
Give the definition of a mole.
1 mole is the amount of a substance that contains 6.02x1023 particles of that substance.
2. be able to use the Avogadro constant, L, (6.02 × 10 23 mol-1) in calculations
What mass of aluminium contains 1.505x1023 atoms?
1.505x.02x1023 = 6.75g
A defence contractor is having a sale – one billion atoms of weapons-grade titanium are now only 1p!!
How much would you have to pay for a milligram of this titanium?
1bn = 1x109 atoms
1x.02x1023 = 1.66…x10-15 mol
1mg = 1x10-3g
n Ti = 1x10-.9 = 2.087x10-5 mol
Value = 2.087x10-.66x10-15
Value = 1.257x1010p or £125,680,734 (great deal imo)
3. know the definition of the molar mass
Give the definition of molar mass.
The molar mass of a substance is the mass per mole of the substance in g/mol or gmol -1.
4. know what is meant by the terms ‘empirical formula’ and ‘molecular formula’
What is the difference between empirical and molecular formula?
Molecular formulae show the real mole ratios of elements in a substance.
Empirical formulae are the simplest whole number ratios of that formula (e.g N2H4 -> NH2).
, 5. be able to use experimental data to calculate
i. empirical formulae
1.24g of phosphorus burns in air to form 2.84g of its oxide. Find its empirical formula.
2.84g – 1.24g = 1.6g of oxygen used mol = mass / mr and ratio = mol / smallest mol
mol of P = 1.24/31 = 0.04mol mol of O = 1.6/16 = 0.1mol
ratio of P = 0.04/0.04 = 1 ratio of O = 0.1/0.04 = 2.5
empirical formula = P2O5
Quinone is an organic compound containing only C, H, and O.
What is the empirical formula of the compound if 0.105g of the compound gives off 0.257g of CO 2 and
0.035g of H2O when burned?
n C = n CO2 = 0.257/44 = 0.00584mol n H = 2(n H2O) = 2(0.035/18) = 0.00388mol
mass C = 0.00584 x 12 = 0.07g mass H = 0.00388 x 1 = 0.00388g
therefore mass O = mass CHO – (mass C + mass H) = 0.105 – (0.07 + 0.00388) = 0.03112g
n O = 0.03112/16 = 0.001945mol
C:H:O = 0.00584 : 0.00388 : 0.001945 (divide all by 0.001945) C:H:O = 3 : 2 : 1 -> C3H2O empirical
Given a molecular weight of about 108g/mol, what is quinone’s molecular formula?
Atomic mass = 12x3 + 16 + 2 = 54 Real weight = 108g/mol -> 108/54 = 2
Therefore molecular formula = C6H4O2
A sample of gypsum (hydrated calcium sulfate) contains 79% of calcium sulfate by mass.
Calculate the value of x in CaSO4.xH2O. [Mr of CaSO4 = 136] [Mr of H2O = 18]
79 grams calcium sulfate per 100g gypsum. Therefore, we have 21 grams of water per 100g of gypsum.
Mol = mass / Mr
Mol calc = 79/136 Mol water = 21/18
Mol calc = 0.581 Mol water = 1.167
0..581 = 1 1..581 = 2.008
Therefore, x = 2 (CaSO4.2H2O)
Suggest why this value may be lower, or higher, than the actual value for the water of crystallisation.
Calculated value is lower than actual: substance not heated to constant mass – i.e. not all the water was
evaporated from the gypsum.
Calculated value is higher than actual: Some crystals were lost whilst heating (e.g. they jumped out of the
crucible).
2
In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include determining a simple empirical formula such as MgO or CuO, determining the number of moles of water of crystallisation in a
salt such as Epsom salts, performing a wide range of titrations involving different indicators, preparing salts.
Mathematical skills that could be developed in this topic include converting between units such as cm 3 and dm3, using standard form with the
Avogadro constant, rearranging formulae for calculating moles in solids and in solutions, calculating atom economy, dealing with percentage errors.
Within this topic, students first encounter core practicals and can consider ideas of measurement uncertainty, evaluating their results in terms of
systematic and random errors. They can also consider how the concept of atom economy is useful to help chemists make decisions so that reactions
can be made more efficient in terms of resources.
1. know that the mole (mol) is the unit for amount of a substance
Give the definition of a mole.
1 mole is the amount of a substance that contains 6.02x1023 particles of that substance.
2. be able to use the Avogadro constant, L, (6.02 × 10 23 mol-1) in calculations
What mass of aluminium contains 1.505x1023 atoms?
1.505x.02x1023 = 6.75g
A defence contractor is having a sale – one billion atoms of weapons-grade titanium are now only 1p!!
How much would you have to pay for a milligram of this titanium?
1bn = 1x109 atoms
1x.02x1023 = 1.66…x10-15 mol
1mg = 1x10-3g
n Ti = 1x10-.9 = 2.087x10-5 mol
Value = 2.087x10-.66x10-15
Value = 1.257x1010p or £125,680,734 (great deal imo)
3. know the definition of the molar mass
Give the definition of molar mass.
The molar mass of a substance is the mass per mole of the substance in g/mol or gmol -1.
4. know what is meant by the terms ‘empirical formula’ and ‘molecular formula’
What is the difference between empirical and molecular formula?
Molecular formulae show the real mole ratios of elements in a substance.
Empirical formulae are the simplest whole number ratios of that formula (e.g N2H4 -> NH2).
, 5. be able to use experimental data to calculate
i. empirical formulae
1.24g of phosphorus burns in air to form 2.84g of its oxide. Find its empirical formula.
2.84g – 1.24g = 1.6g of oxygen used mol = mass / mr and ratio = mol / smallest mol
mol of P = 1.24/31 = 0.04mol mol of O = 1.6/16 = 0.1mol
ratio of P = 0.04/0.04 = 1 ratio of O = 0.1/0.04 = 2.5
empirical formula = P2O5
Quinone is an organic compound containing only C, H, and O.
What is the empirical formula of the compound if 0.105g of the compound gives off 0.257g of CO 2 and
0.035g of H2O when burned?
n C = n CO2 = 0.257/44 = 0.00584mol n H = 2(n H2O) = 2(0.035/18) = 0.00388mol
mass C = 0.00584 x 12 = 0.07g mass H = 0.00388 x 1 = 0.00388g
therefore mass O = mass CHO – (mass C + mass H) = 0.105 – (0.07 + 0.00388) = 0.03112g
n O = 0.03112/16 = 0.001945mol
C:H:O = 0.00584 : 0.00388 : 0.001945 (divide all by 0.001945) C:H:O = 3 : 2 : 1 -> C3H2O empirical
Given a molecular weight of about 108g/mol, what is quinone’s molecular formula?
Atomic mass = 12x3 + 16 + 2 = 54 Real weight = 108g/mol -> 108/54 = 2
Therefore molecular formula = C6H4O2
A sample of gypsum (hydrated calcium sulfate) contains 79% of calcium sulfate by mass.
Calculate the value of x in CaSO4.xH2O. [Mr of CaSO4 = 136] [Mr of H2O = 18]
79 grams calcium sulfate per 100g gypsum. Therefore, we have 21 grams of water per 100g of gypsum.
Mol = mass / Mr
Mol calc = 79/136 Mol water = 21/18
Mol calc = 0.581 Mol water = 1.167
0..581 = 1 1..581 = 2.008
Therefore, x = 2 (CaSO4.2H2O)
Suggest why this value may be lower, or higher, than the actual value for the water of crystallisation.
Calculated value is lower than actual: substance not heated to constant mass – i.e. not all the water was
evaporated from the gypsum.
Calculated value is higher than actual: Some crystals were lost whilst heating (e.g. they jumped out of the
crucible).
2
