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.