Key Calculations quiz
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The solubility of solids changes as temperature changes. This can be plotted on a solubility curve.
The salts shown on this graph are typical: the solubility increases as temperature increases.
For example, the graph above shows that in 100g of water at 50⁰C the maximum mass of potassium nitrate (KNO₃) which will dissolve is 80g.
However, if the temperature were 80⁰C a mass of 160g of potassium nitrate (KNO₃) would dissolve in 100g of water.
75% of chlorine atoms are the type ^{35}Cl (have a mass number of 35)
25% of chlorine atoms are of the type ^{37}Cl (have a mass number of 37)
In order to calculate the relative atomic mass (A_{r}) of chlorine, the following steps are used:
Example question:
A sample of bromine contained the two isotopes in the following proportions: bromine-79 = 50.7% and bromine-81 = 49.3%.
Calculate the relative atomic mass (A_{r}) of bromine.
This excellent video from Tyler de Witt walks you through what isotopes are, and how the relative abundance of those isotopes can be used to calculate the relative atomic mass of an element.
Relative formula mass (M_{r}) is mass of a molecule or compound (on a scale compared to carbon-12).
It is calculated by adding up the relative atomic masses (A_{r}) of all the atoms present in the formula.
Example:
The relative formula mass (M_{r}) for water (H_{2}O) is 18.
Water = H_{2}O
Atoms present = (2 x H) + (1 x O)
M_{r} = (2 x 1) + (1 x 16) = 18
Here’s an excellent Tyler de Witt video explaining how to calculate the relative formula mass of compounds with:
In Chemistry, the mole is a measure of amount of substance (n).
The abbreviation for mole is mol.
The mass of 1 mole of a substance is the relative formula mass (M_{r}) of the substance in grams.
Example:
The M_{r} of water is 18.
Therefore the mass of 1 mol of water equals 18 g.
The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:
Example 1:
Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO_{2}).
Step 1: Calculate the relative formula mass (M_{r}) of carbon dioxide (CO_{2}).
Step 2: Use the formula to calculate the amount in moles.
Example 2:
Calculate the mass of 2 mol of copper(II) sulfate (CuSO_{4}).
Step 1: Calculate the relative formula mass (M_{r}) of copper(II) sulfate (CuSO_{4}).
Step 2: Rearrange the formula to calculate the mass.
This video shows how to perform calculations involving mass (in grams), amount (in moles) and relative atomic mass:
Example: When calcium carbonate (CaCO_{3}) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.
CaCO_{3} –> CaO + CO_{2}
Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO_{3})
Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO_{3}.
This step involves using the ratio of CaCO_{3} to CaO from the chemical equation.
CaCO_{3} –> CaO + CO_{2}
From the equation you can see that the ratio of CaCO_{3} to CaO is 1:1.
Therefore if you have 0.25 mol of CaCO_{3} this will produced 0.25 mol of CaO.
Step 3: Calculate the mass of 0.25 mol of CaO.
A simple format for laying out this method can be used.
Example: What mass of ammonia (NH_{3}) is formed when 7 g of nitrogen (N_{2}) is combined with hydrogen (H_{2}).
This video steps through a very useful method used to calculate reacting masses.
Yield is how much product you get from a chemical reaction.
The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.
In most chemical reactions, however, you rarely achieved your theoretical yield.
For example, in the following reaction:
CaCO_{3} –> CaO + CO_{2}
You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO_{3}.
However, what if the actual yield is only 48 g of CaO.
By using the following formula, the % yield can be calculated:
Finding the formula of a metal oxide experimentally
The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.
Example: When magnesium is burned in air, it reacts with oxygen (O_{2}) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)
Finding the formula of a salt containing water of crystallisation
When some substances crystallise from solution, water becomes chemically bound up with the salt. This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula CuSO_{4}.5H_{2}O which formula indicates that for every CuSO_{4} in a crystal there are five water (H_{2}O) molecules.
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO_{4}.5H_{2}O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO_{4}). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.
The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.
The molecular formula shows the actual number of atoms of each type of element in a molecule.
For example, for ethane:
Molecular formula = C_{2}H_{6}
Empirical formula = CH_{3}
Here are some more examples:
Name | Molecular formula | Empirical Formula |
---|---|---|
pentene | C_{5}H_{10} | CH_{2} |
butene | C_{4}H_{8} | CH_{2} |
glucose | C_{6}H_{12}O_{6} | CH_{2}O |
hydrogen peroxide | H_{2}O_{2} | HO |
propane | C_{3}H_{8} | C_{3}H_{8} |
Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.
Calculating the Empirical Formula
Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?
Step 1: Put the symbols for each element involved at the top of the page.
Step 2: Underneath, write down the masses of each element combining.
Step 3: Divide by their relative atomic mass (A_{r}).
Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.
Step 5: Use this to give the empirical formula.
(If your ratio is 1:1.5 then multiple each number by 2
If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)
Calculating the Molecular Formula
If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.
Example: A compound has the empirical formula CH_{2}, and its relative formula mass is 56. Calculate the molecular formula.
Step 1: Calculate the relative mass of the empirical formula.
Step 2: Find out the number of times the relative mass of the empirical formula goes into the M_{r} of the compound.
Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.
Empirical formula calculations involving water of crystallisation
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula BaCl_{2}.nH_{2}O where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.
Example: If you heated hydrated barium chloride (BaCl_{2}.nH_{2}O) in a crucible you might end up with the following results.
Step 1: Calculate the mass of the anhydrous barium chloride (BaCl_{2}) and the water (H_{2}O) driven off.
Step 2: Use the empirical formula method to find the value of n in the formula.
In this video, the lovely Tyler de Witt explains Empirical Formulae and Molecular formulae.
(Please excuse the horrid error right at the start where the video shows a diagram of prop-2-ene, but it is labelled ‘ethene’ by mistake.)
This second video then explains how, if given an empirical formula and a molecular mass, you can calculate the molecular formula:
Concentration is a measurement of the amount of substance per unit volume.
In Chemistry, concentration is measured in mol/dm^{3} (read as moles per cubic decimetre).
The following formula allows for the interconversion between a concentration (in mol/dm^{3}), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm^{3}).
or
Note: 1 dm^{3} = 1000 cm^{3}
For example, to convert 250 cm^{3} into dm^{3}:
Example 1: 0.03 mol of sodium carbonate (Na_{2}CO_{3}) is dissolved in 300 cm^{3} of water. Calculate the concentration of the solution.
Step 1: Convert the volume of water from cm^{3} into dm^{3}.
Step 2: Use the molar concentration formula to calculate of the concentration of the solution.
Example 2: Calculate the amount, in moles, of 25cm^{3} of hydrochloric acid (HCl) with a concentration of 2 mol/dm^{3}.
Step 1: Convert the volume of HCl from cm^{3} into dm^{3}.
Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.
Titration calculations, example 1
The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.
For example, a titration problem will look like this:
A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm^{3} of 0.100 mol/dm^{3} sodium hydroxide solution (NaOH) required 23.50 cm^{3} of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm^{3} of the acid.
The chemical equation for this reaction is:
NaOH(aq) + HCl(aq) –> NaCl(aq) + H_{2}O(l)
volume 25.0cm^{3} 23.50cm^{3}
concentration 0.100mol/dm^{3}
It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.
Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).
First convert the volume of NaOH from cm^{3} into dm^{3}.
Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.
Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).
From the chemical equation the ratio of NaOH to HCl is 1:1
Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.
Step 3: Calculate the concentration of the hydrochloric acid (HCl).
Convert the volume of HCl from cm^{3} into dm^{3}.
Use the molar concentration formula to calculate of the concentration of the HCl.
Titration calculations, example 2
25.0 cm^{3} of sodium carbonate solution (Na_{2}CO_{3}) of unknown concentration was neutralised by 30.0 cm^{3} of 0.100 mol/dm^{3} nitric acid (HNO_{3}).
Na_{2}CO_{3}(aq) + 2HNO_{3}(aq) –> 2NaNO_{3}(aq) + CO_{2}(g) + H_{2}O(l)
Calculate the concentration, in mol/dm^{3} of the sodium carbonate solution.
Step 1: Calculate the amount, in moles, of nitric acid (HNO_{3}).
Step 2: Deduce the amount, in moles of sodium carbonate (Na_{2}CO_{3}).
Using the equation:
Na_{2}CO_{3}(aq) + 2HNO_{3}(aq) –> 2NaNO_{3}(aq) + CO_{2}(g) + H_{2}O(l)
Ratio Na_{2}CO_{3}:HNO_{3} = 1:2
Step 3: Calculate the concentration of sodium carbonate (Na_{2}CO_{3}).
Converting mol/dm^{3} into g/dm^{3}
Concentration can also be expressed in g/dm^{3} (grams per cubic decimetre).
Therefore mol/dm^{3} can be converted into g/dm^{3}.
Example: Convert 0.06 mol/dm^{3} of sodium carbonate (Na_{2}CO_{3}) into g/dm^{3}
Step 1: calculate the relative formula mass (M_{r}) of sodium carbonate (Na_{2}CO_{3}).
Step 2: Recall the formula giving the relationship between mass, amount and formula mass.
therefore
Here is a video to help you with concentration calculations. In Chemistry, concentration is know as molarity (i.e. the amount per volume).
Note that in the video, the lovely Tyler de Witt uses the unit “liter” for volume, but the correct international unit (as used by UK exam boards) is decimeters cubed (dm^{3}) which is the same thing as Tyler’s “liter”.
and a follow up video with some more examples:
The molar volume of a gas is the volume that one mole of any gas will occupy.
1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm^{3} or 24,000 cm^{3}.
Note: 1 dm^{3} = 1000 cm^{3}
The following formulae allows for the interconversion between a volume in dm^{3} or cm^{3} and a number of moles for a given gas:
or
Example 1:
Calculate the amount, in moles, of 12 dm^{3} of carbon dioxide (CO_{2}).
Example 2:
Calculate the volume at rtp in cubic centimetres (cm^{3}), of 3 mol of oxygen, (O_{2}).
To determine the formula of a metal oxide by combustion.
Example: When magnesium is burned in air, it reacts with oxygen (O_{2}) to form magnesium oxide (MgO).
2Mg + O_{2} –> 2MgO
Method:
It is then possible to use this data to calculate the empirical formula of the metal oxide.
To determine the formula of a metal oxide by reduction
Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:
4CuO + CH_{4} –> 4Cu + CO_{2} + 2H_{2}O
By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.
As an alternative, hydrogen gas could be used instead of methane:
CuO + H_{2} –> Cu + H_{2}O
It is then possible to use this data to calculate the empirical formula of the metal oxide.
There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.
Titration is used to find out precisely how much acid neutralises a certain volume of alkali (or vice versa).
The diagram shows the titration method for a neutralisation reaction between hydrochloric acid and sodium hydroxide, using phenolphthalein as an indicator. The indicator changes colour when neutralisation occurs.
The conical flask is swirled to mix the solutions each time alkali is added. When reading the burette it is important to be aware that the numbers on the scale increase from top to bottom. Readings are usually recorded to the nearest 0.05cm³ so all readings should be written down with 2 decimal places. The second decimal place is given as a ‘0’ if the level of the solution is on a line, or ‘5’ if it is between the lines. The volume of alkali added is calculated by subtracting the final reading from the initial reading. Various indicators can be used such as phenolphthalein or methyl orange. However universal indicator should not be used since it has a wide range of colours rather than one specific colour change so it would be unclear when the precise endpoint of titration was achieved.
This process is repeated a number of times. The first time it is done roughly to get a good approximation of how much alkali needs to be added. On subsequent attempts, the alkali is added very slowly when approaching the correct volume.
Here are a couple of videos explaining how to perform titration calculations:
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 (^{o}C) | Maximium temp. of solution (^{o}C) | Temperature rise (^{o}C) |
---|---|---|
24.2 | 56.7 | 32.5 |
Note: mass of 50 cm^{3} 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 (^{o}C) | Maximum temp of water (^{o}C) | Temperature rise (^{o}C) | 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:
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) |
Here are a couple of videos explaining how to do bond energy calculations: