Chemical formulae and chemical equations (Section 1e)

Chemical formulae and chemical equations

1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification 

Word equations show the reactants and products of a reaction, using their names as opposed to their symbols. For example;

Hydrogen + Oxygen > Water 

Balanced equations also show the reactants and products of a reaction, but use symbols to do so. In order for an equation to be balanced, there must be an equal number of each element on both sides of the equation. To balance an equation, you can put numbers in front of the element to alter the equation. For example;

• Hydrogen + Oxygen > Water 

         H + O > H2O 

There is one hydrogen element on the left hand side, and two on the right hand side. There is one oxygen element on the left hand side, as well as on the right hand side. So, there needs to be 2 hydrogen elements on the left hand side. 

2H + O > H2O

The equation is now balanced as there are two hydrogen elements on both sides and one oxygen element on both sides. 

• magnesium + oxygen > magnesium oxide

          Mg + O2 > MgO

There is one magnesium element on the left hand side, and one on the right hand side also. There are two oxygen elements on the left hand side, but only one on the right hand side, so there needs to be two on the right hand side.

      Mg + O2 > 2MgO 

There are now two oxygen elements on both sides of the equation, but there is one magnesium element on the left and two on the right. In order for the equation to be fully balanced, there must be two on both sides. 

2MgO + O2 > 2MgO 

There are two magnesium elements on both sides, and also two oxygen elements on both sides. Therefore, the equation is now balanced. 

1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to 
represent solids, liquids, gases and aqueous solutions respectively 

State symbols are used in equations to show what physcial state the reactants and products are in. They are shown in brackets after the symbol of the reactant or product. 
(s) = solid 
(l) = liquid 
(g) = gas 
(aq) = aqueous (dissolved in water.)

1.23 understand how the formulae of simple compounds can be obtained
experimentally, including metal oxides, water and salts containing water of

crystallisation

If you weigh the compound, then remove an element from the compound through a reaction, and weigh it again after doing this, you will have the weight of a particular element of this compound. Say the compound is XY and the removed element is X, then the weight of the other element will be; XY-X=Y To work out the formulae of each element, you divide the weight (in grams) by the Ar of the element. 

For example; if X = 64 with an Ar of 16
                             64/16 = 4

 and if Y = 33g with an Ar of 11,
                 33/11 = 3

So, the formulae of the compound would be: X4Y3.


1.24 calculate empirical and molecular formulae from experimental data

The empirical formula of a compound is the simplest ratio of different elements in the compound. The molecular formula of a compound tells you the exact number of atoms of each element in a single molecule. 

To find the empirical formula of a compound; 

• List all the elements that are in the compound 
• Under these, write their respective experimental masses or percentages. 
• Divide each mass or percentage by the Ar (relative atomic mass) of that element.
• Turn the numbers you get from this into a simpler ratio by multiplying or dividing to get a whole number, easier number etc.
• Simply the ratio to it's simplest form - this is the empirical formula of the compound.

To find the molecular formula of a compound; 

• Find the relative atomic mass of the empirical formula.
• Divide the given relative molecular mass by the relative atomic mass to get the number of empirical units in the molecule. 
• The molecular formula will then be the empirical formula x the number of empirical units from previous step. 

1.25 calculate reacting masses using experimental data and chemical equations

In order to do this, you need to; 

• Write out the balanced equation of the reaction.
• Work out the relative formula mass for the two parts of the equation you need. 
• Divide to get one, then multiply by the number in the question.

Example; What mass of magnesium oxide is produced when 60g of magnesium is burnt in air?

1) The balanced equation of this reaction is; 2Mg + O2 > 2MgO 

2) The relative formula masses of the two needed bits (2Mg and 2MgO); 

2Mg; 2 x 24 = 48 
2MgO; 2 x (24 + 16) = 80

3) Divide to get one, multiply to get all. 

          48g of Mg ....... 80g of MgO 

  (/48)                                                 (/48)
          1g of Mg ........ 1.67g of MgO

(x60)                                                   (x60) 
           60g of Mg ....... 100g of MgO 

Therefore, 60g of magnsium will produce 100g of magnesium oxide. 

1.26 calculate percentage yield

The yield of a reaction is the mass of product. For example, the mass of magnesium oxide in the example above is the yield of that reaction. Masses calculated like the one above are called theoretical yields. 

percentage yield = (actual yield (grams)/theoretical yield (grams)) x 100 

1.27 carry out mole calculations using volumes and molar concentrations.

Concentration = number of moles/volume 
Moles = concentration x volume 
Volume = number of moles/concentration 

(If the question provides the volume in cm³, divide by 1000 to get the volume in dm³.) 


Example: How many moles of sodium chloride are in 250cm³ of a 3 mol/dm³ solution?

250cm³ = 0.25dm³. (250/1000=0.25) 
Moles = concentration x volume 
           = 3 x 0.25 
            = 0.75 moles. 

Relative formula masses and molar volumes of gases (Section 1d)

Relative formula masses and molar volumes of gases

1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar)

The relative formula mass of a compound is the relative atomic masses of the atoms it contains added together. (number of this element within molecule x relative atomic mass of element) + (number of this element within molecule x relative atomic mass of element.) 

For example, to find the relative formula mass (Mr) of MgCl2 (magnesium chloride);

There is one element of Mg within molecule, and the atomic mass = 24.

There are 2 elements of Cl within molecule, and atomic mass = 35.5

So, relative formula mass = (1 x 24) + (2 x 35.5)

                                      = 24 + 71

                                      = 95.

1.17 understand the use of the term mole to represent the amount of substance 

The mole is the amount of substance that contains 6 x 10²³ of that substance. When you have exactly that number of atoms or molecules, they weigh the same number of grams as the relative atomic mass of that element, or relative formula mass of that compound. So, as carbon has an Ar of 12, one mole of carbon will weigh exactly 12g. Also, as carbon dioxide has an Mr of 44, one mole of carbon dioxide will weigh exactly 44g. 

1.18 understand the term mole as the Avogadro number of particles (atoms, molecules, formulae, ions or electrons) in a substance 

The number 6 x 10²³  is known as Avogadro's number and means that a mole refers to the Avogadro's number of particles in a substance. These particles can be atoms, molecules, formulae, ions or electrons.

1.19 carry out mole calculations using relative atomic mass (Ar) and relative 

formula mass (Mr) 

The formula for finding the number of moles in a given mass is as follows; 

Number of moles = Mass in g of element or compound/Mr of element or compound 

Example, How many moles are there in 66g of carbon dioxide?
Mr of carbon dioxide (CO2) = (1 x 12) + (16 x 2)
                                         = 12 + 32 
                                          = 44
Number of moles = mass in grams/Mr 
                          = 66/44
                          = 1.5 moles.

So, there are 1.5 moles in 66g of carbon dioxide.

1.20 understand the term molar volume of a gas and use its values 
(24 dm3 and 24,000 cm3) at room temperature and pressure (rtp) in 

calculations. 

The molar volume of a gas is the volume that one mole of gas fills. One mole of any gas will always occupy 24dm3 (24,000cm3) at room temperature and pressure. RTP = 25 degrees celsius, and 1 atmosphere.) 

Therefore, you can use the following formula to convert the number of moles or mass of any gas to a volume;

Volume (dm3) = moles of gas x 24 

Atomic Structure - (Section 1c)

Atomic Structure 

1.9 understand that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons orbiting in shells. 

The centre of an atom is known as the 'nucleus.' It consists of protons and neutrons, and has a positive charge because of the protons. Nearly the whole mass of the atom is concentrated in the nucleus, however it is tiny in size compared to the rest of the atom. The nucleus is surrounded by orbiting shells, or energy levels,  in which electrons lie. They are negatively charged, have virtually no mass and despite the fact they cover a lot of space, they are tiny.

1.10 recall the relative mass and relative charge of a proton, neutron and electron. 

Proton = Relative mass of 1, relative charge of +1.

Neutron, Relative mass of 1, relative charge of 0.

Electron, Relative mass of 0, relative charge of -1. 

1.11 understand the terms atomic number, mass number, isotopes and relative atomic mass

Atomic number = the number of protons (and therefore electrons as these two vales are the same) within an atom. 

Mass number = the total number of protons and neutrons within an atom.

Isotopes = Different atomic forms of the same element, which have the same number of protons but a different number of neutrons. 

Relative atomic mass = the average mass of all the isotopes of an element. This is calculated from the masses and relative abundances of all the isotopes of a certain element. It is often given by the symbol Ar. 

1.12 calculate the relative atomic mass of an element from the relative abundances of its isotopes. 

In order to find the relative atomic mass of an element from the relative abundances of its isotopes, 
1) Multiply the mass of each isotope by its relative abundance 
2) Add these together.
3) Divide by the sum of the relative abundances. 

For example., work out the relative atomic mass of chlorine. 

Relative mass of isotopes = 35, 47.
Respective to above, relative abundances = 3,1.

1) Multiply the mass of each isotope by its relative abundance - (35x3), (37x1)
2) Add these together - 105 + 37 = 142
3) Divide by sum of relative abundances =  142/(3+1) = 142/4 
                                                                  = 35.5

So, the relative atomic mass of chlorine = 35.5

1.13 understand that the Periodic Table is an arrangement of elements in order of atomic number.

The elements of the Periodic Table are shown in order of increasing atomic number. Going horizontally and to the right, each element has an atomic number than the previous element. This can be seen in the periodic table, with the numbers above the symbols displaying each elements atomic number. 

1.14 deduce the electronic configurations of the first 20 elements from their positions in the periodic table. 

The number of electrons in an element is shown by the atomic number of the element. So, if an element has an atomic number of 16 (sulphur), then it has 16 electrons. To work out the electronic configuration, you count up to 16, keeping in mind the number of electrons allowed per each shell in an element. As there are 2 electrons allowed in the first shell, and eight in the following two shells, sulphur will have an electronic configuration of 2.8.6

You can check this is correct using the elements' position in the periodic table. The vertical columns in the periodic table are known as groups. The number group an element is in tells you how many electrons that element has in its outer shell. Sulphur is in group 6 of the periodic table, and will therefore have six electrons in it's outer shell. The horizontal columns in the periodic table are known as periods. The number period an element is in tells you how many energy levels the element has. Sulphur is in period three of the periodic table, and will therefore have three energy levels, so the electronic configuration above is correct. 

1.15  deduce the number of outer electrons in a main group element from its position in the Periodic Table. 
The number of the group (vertical column) an element is in corresponds to the number of electrons there are in it's outer shell.For example, Boron is in group 3 of the Periodic Table and therefore has 3 electrons in its outer shells. 

Atoms - Continued

Atoms - Continued

1.7  describe experimental techniques for the separation of mixtures, including simple distillation, fractional distillation, filtration, crystallisation and paper chromatography

i) Simple distillation 
As the components of a mixture are not chemically bonded, they can be separated by physical means. Simple distillation is used to separate out a liquid from a solution. It can only be used to separate things with very different boiling points. Things with similar boiling points must be separated using fractional distillation. 

Process;
1) The solution is heated, causing the part of said solution with the lowest boiling point to evaporate.
2) The vapour will then be cooled until it condenses and turns back into a liquid, when it will be collected. 
3) The remainder of the solution will be left behind in the flask or container.

The process of simple distillation.

ii) Fractional distillation

Fractional distillation is used to separate a mixture of liquids with very different boiling points. 

Fractional distillation of crude oil 

1) The crude oil is heated until most of it has turned into gas. The gases then enter a fractionating column while the bitumen (liquid part) is drained off at the bottom.

2) There is a temperature gradient within this column, in which it is hot at the bottom and gets gradually cooler as you go up. When a certain substance of the crude oil reaches the part of the column where the temperature is lower than it's own boiling point, they condense and turn back into a liquid. 

3) When the first liquid has been collected, you raise the temperature until the next one reaches the top and condenses also. 

Diagram showing the 

iii) Filtration

Filtration is used to separate an insoluble solid from a liquid. It is the process of separating an undissolved solid from a mixture of that solid and a liquid/solution. It is done through simply pouring the solution into a beaker through a barrier, often filter paper, which will leave the solid in the paper as the liquid solution travels past the filter paper into the beaker it is being poured into. 

Process of filtration. 

iv) Crystallisation 

Crystallisation is used to separate a soluble solid from a solution. It is the process of separating a dissolved solid from a solution. 

Process; 

1) Pour the solution into an evaporating dish

2) Heat the solution slowly, this will cause some of the solvent to evaporate and the solution will become more concentrated. Stop heating the solution when crystals start to form. 

3) Remove the dish from the heat, leaving it in a warm place to allow for the the rest of the solvent to evaporate. This will encourage large crystals to form.

4) Lastly, dry the product. This can be done by using a drying oven or a desiccator (contains chemicals that remove water from the surroundings.) 

Diagram showing the process of crystallisation. 

v) Paper chromatography

Paper chromatography is used to separate out mixtures of dyes. It works because different dyes move up the paper at different rates.

Process; 

1) Draw a line near the bottom of a sheet of filter paper, using a pencil as pencil marks are insoluble and will not react with the solvent. 

2) Add spots of each different dye to the line at regular intervals.

3) Roll up the sheet of paper and put in the beaker of solvent. However, make sure the dyes are not touching the solvent as they could dissolve in it if this happens.

4) Place a lid on the top of the container to stop the solvent evaporating. 

5) The solvent will then seep up the paper, carrying the dyes with it.

6) Each different dye that is present within the solvent will move up the paper at a different rate and form a mark in a different place. 

7) The end result will be a pattern of spots known as a chromatogram. 

1.8 explain how the information from chromatograms can be used to identify the composition of a mixture. 

The information from chromatograms can be compared to reference materials in order to identify the composition of a mixture. The reference materials will be chromatograms for some dyes that you think may be in the ink. If you compare these, you will see what dyes match up and therefore what dyes are present in the composition of your mixture.