Chapter 9.6Finding the formula of a

Compound

How can we work out the formula of a compound?

We can conduct experiments to find out the formula of a

compound.

1. First, we find out the mass of the reactants taking

part in the reaction.

2. Next, we work out the relative numbers of moles of

the reactants used.

Example: Working Out the Formula of Magnesium

OxideTo work out the formula of magnesium oxide produced by the combustion of magnesium, the following apparatus is used.

Magnesium ribbon

Clay triangle

Tripod stand

lid

crucible

Magnesium ribbon

Clay triangle

Tripod stand

lid

crucible

Procedure 1. Weigh a crucible together with the lid. Put a coil of magnesium ribbon in it and weigh again.

2. Put the lid on the crucible and heat the crucible gently. When the magnesium catches fire (you will see a white glow through the crucible), heat it more strongly.

Magnesium ribbon

Clay triangle

Tripod stand

lid

crucible

3. Use a pair of tongs to lift the lid slightly from time to time to allow air in. Quickly replace the lid to make sure that magnesium oxide formed does not escape.

4. When the burning is complete, allow the crucible to cool completely. Weigh the crucible together with the lid and the magnesium oxide in it.

Sample resultsMass of crucible + lid = 26.52 gMass of crucible + lid + magnesium = 27.72 gMass of crucible + lid + magnesium oxide = 28.52 g

CalculationsMass of magnesium = 27.72 − 26.52 = 1.20 gMass of magnesium oxide produced = 28.52 − 26.52 = 2.00 gMass of oxygen reacted = 2.00 − 1.20 = 0.80 g

Deriving the FormulaElement Mg O

Mass (from experiment) 1.20 g 0.80 g

Relative atomic mass 24 16

Number of moles

Molar ratio (divide by the smallest number from the previous row)

1.20

24= 0.05

0.05

0.05= 1

0.05

0.05= 1

0.80

16= 0.05

Step 1: List the mass of the element.

Deriving the Formula

Molar ratio (divide by the smallest number from the previous row)

Number of moles

1624Relative atomic mass

0.80 g1.20 gMass (from experiment)

OMgElement

1.20

24= 0.05

0.05

0.05= 1

0.05

0.05= 1

0.80

16= 0.05Step 2: State the Ar of the element.

Deriving the Formula

Step 3: Derive the number of moles by dividing the mass with Ar.

Molar ratio (divide by the smallest number from the previous row)

Number of moles

1624Relative atomic mass

0.80 g1.20 gMass (from experiment)

OMgElement

1.20

24= 0.05

0.05

0.05= 1

0.05

0.05= 1

0.80

16= 0.05

Step 4: Obtain the molar ratio.

Deriving the Formula

The empirical formula is MgO.

Molar ratio (divide by the smallest number from the previous row)

Number of moles

1624Relative atomic mass

0.80 g1.20 gMass (from experiment)

OMgElement

1.20

24= 0.05

0.05

0.05= 1

0.05

0.05= 1

0.80

16= 0.05

Empirical Formula

• simplest formula of a compound• shows the types of element present

in the compound And

• the simplest ratio of the number of the different type of atoms in it

Worked Example 1

Element Cu OStep 1 Mass of element 8 1Step 2: Ar 64 16Step 3: No. of Moles

(Mass/ Ar)8 = 0.125

64 1 = 0.0625

16

Step 4: Molar ratio (Divide by smallest

number)

0.125 = 2 0.0625

0.0625 = 1 0.0625

A sample of an oxide of copper contains 8g of copper combined with 1g of oxygen. Find the empirical formula of the compound.

Empirical formula of the compound is Cu2O1

or usually written as this Cu2O.

Worked Example 2

• A compound has the following percentage composition: Sodium 32.4 %, Sulphur 22.6%, Oxygen 45.0 %

Element Na S OStep 1 % composition by mass 32.4 22.6 45.0Step 2 Ar 23 32 16Step 3 No. of Moles 32.4 = 1.4

2322.6 = 0.7

32 45.0 = 2.8 16

Step 4 Molar ratio ( divide by smallest number)

1.4 = 2 0.7

0.7 =1 0.7

2.8 = 4 0.7

Empirical formula: Na2SO4

Molecular Formula

Molecular Formula

Its actual formula is P4O10. We call this the molecular formula.

The empirical formula of phosphorus(V) oxide as determined by experiment is P2O5.

Molecular Formula

• shows the exact number of atoms of each element in a molecule

• Is sometimes the same as the empirical formula, i.e. in actual the compound exists in the simplest ratio

• Example: water (H2O) and ammonia (NH3)

Molecular Formula

• However most of the compounds do not have a molecular formula that is similar to the empirical formula.

• If the compound’s molecular formula and empirical formula are different, the molecular formula is just a multiple of the empirical formula

• For example, the molecular formula and empirical formula of phosphorus(V) oxide are P4O10 and P2O5 respectively.

• The multiple is 2.

Molecular Formula

• possible for different compounds to have same empirical formula.

• For example:ethane has a molecular formula of C2H4

propane has a molecular formula of C3H6,

Therefore, they have the same empirical formula which is CH2.

Molecular Formula

• To find the molecular formula of a compound we use this method, n is the multiple to the empirical formula

• n = relative molecular mass of the compound

Mr of the empirical formula

Example 3 Propane has the empirical formula CH2. The relative molecular mass of propane is 42. Find the molecular formula of propane n = relative molecular mass of the compound Mr of the empirical formulan = 42 (12 X 1) + 2 = 3Molecular formular = (Empirical formula ) n

= (CH2)3

Hence the molecular formula for propane is C3H6

= C3H6

Example 4 Compound X contains 40.0% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative

molecular mass is 180. What is the molecular formula of X?

Molar ratio

Number of moles

16112Relative atomic mass

53.36.640.0Percentage in compound

OHCElement

= 3.312

40.0 = 3.316

53.3

= 116

3.52= 2

16

3.52

= 6.61

6.6

= 13.3

3.3

Example 2 (continued)Compound X contains 40.0% carbon, 6.6%

hydrogen and 53.3% oxygen. Its relative molecular mass is 180. What is the molecular formula of X?

The empirical formula of X is CH2O.

=

Mr from empirical formula

Relative molecular mass 180

30= 6

Hence, the molecular formula of X = (CH2O)6 = C6H12O6

Mr of CH2O = 12 + (2 x 1) + 16 = 30

n =