WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

WJEC Chemistry for AS Level

Empirical and molecular formulae We can use calculations involving masses of the elements that combine together to find the formulae of compounds. Empirical formula is the simplest formula showing the simplest whole number ratio of the number of atoms of each element present. Molecular formula shows the actual number of atoms of each element present in the molecule. It is a simple multiple of the empirical formula. Usually the relative formula mass is needed to determine the molecular formula. Empirical formulae can be calculated from known masses or percentage composition data. There are three steps in the calculation: Step 1 Find the amount in moles of each element present (divide by the molar mass). Step 2 Find the ratio of the number of atoms present (divide by the smallest value in step 1). Step 3 Convert these numbers into whole numbers (atoms combine together in whole number ratios).

Key terms Empirical formula  is the simplest formula showing the simplest whole number ratio of the number of atoms of each element present. Molecular formula  shows the actual number of atoms of each element present in the molecule. It is a simple multiple of the empirical formula.

Worked example A compound of carbon, hydrogen and oxygen has a relative molecular mass of 60. The

percentage composition by mass is C 40.0%; H 6.70%; O 53.3%. What is (a) the empirical formula and (b) the molecular formula? (a) C : H : O Molar ratio of atoms

Study point When you divide the percentages by the relevant atomic masses, do not truncate the answers, e.g. 1.25 to 1. The figures provided in the question should give reasonably simple ratios for the empirical formula.

40 12 53.3 16 3.33 6.63 3.33 6.7 1.01

Divide by smallest number 1

2

1

Empirical formula is CH 2 O (b) Mass of empirical formula = 12 + 2.02 + 16 = 30.02 Number of CH 2 O units in a molecule = 60 30.02  = 2 Molecular formula is C 2 H 4 O 2

Knowledge check

8

Find the empirical formula of the compound formed when 1.172g iron forms 3.409g of iron chloride.

Some salts have water molecules incorporated into their structure. These are known as hydrated salts and the water is known as water of crystallisation. If we know the mass of the anhydrous salt and the mass of the water in the hydrated salt we can calculate the number of moles of water in the hydrated salt. Worked example Sodium carbonate can form a hydrate, Na 2 CO 3 . x H 2 O. When 4.64g of this hydrate were heated, 2.12g of the anhydrous salt, Na 2 CO 3 , remained. What is the value of x ? Mass of water in the hydrate = 4.64 − 2.12 = 2.52g Moles Na 2 CO 3  = 2.12 106  = 0.020 Moles H 2 O = 2.52 18.02  = 0.140 Mole ratio of DRAFT H 2 O : Na 2 CO 3 0.140 : 0.020 Divide by smaller number 7 : 1 Value of x = 7 and formula is Na 2 CO 3 .7H 2 O 38