chemical formulae

CHEMICAL FORMULAE

The chemical formula of an element shows the symbol representing that element and the number of atoms making up that element.

H2 O2 N2

CH4CH3CH2OH

CuCO3

CuSO4.5H2O

1. Moleculer formula

2. Empirical formula

3. Structural formula

Chemical formulae for molecular elements and molecular/covalent compounds

Chemical formulae for covalent and ionic componds

Chemical formulae for molecular elements and covalent compounds that show the relative arrangement of atoms in a molecule.

The chemical formula of a chemical compound shows the symbols of all elements and the number of atoms of each element in one molecule of the compound.

The molecular formula of a compound is the chemical formula which shows the actual number of atoms of each combining element in one molecule of a compound

1. Molecular formula

Oxygen Gas

O2

One molecule of oxygen contains two oxygen atoms that combined together

Ammonia

NH3

In one molecule of ammonia three hydrogen atoms combined with one nitrogen atom

Trichloromethane CHCl3

One molecule of trichloromethane contains one carbon atom, one hydrogen atom and three chlorine atoms

2. EMPIRICAL FORMULA

Ethane

Molecular formula

C2H6

Emphirical formula

CH3

Glucose

Molecular formula

C6H12O6

Emphirical formula

CH2O

Water H2O

Emphirical formula

H2O

Oxalic Acid

Molecular formula

H2C2O4

Emphirical formula HCO2

The empirical formula of a compound is the chemical formula which shows the smallest ratio of atom of each element that made up a compound.

Molecular formula

Na+

ions Cl- ions

Sodium chloride

Chemical formula for an ionic compound is stated in term of its empirical formula. Ionic compounds do not exist in the form of simple molecules. At room conditions all ionic compounds are crystalline solid that consists of positively and negatively charged ions that are held together by a network of electrostatic attractive forces of attraction (ionic bond). “The whole crystalline solid is a molecule”

Compound Empirical formula

Calcium hydroxide

Ca(OH)2

Aluminium sulphate

Al2(SO4)3

Lithium carbonate

Li2CO3

Zinc sulphate ZnSO4

Ammonium sulphate

(NH4)2SO4

3. Structural formula

EthanolMolecular formula

C2H6O

Structural formula

The chemical formulae for molecular elements and covalent compounds that show the relative arrangement of atoms in a molecule. The formula also shows the actual number of atoms of each element present in one molecule.

Cyclohexane

Molecular formula

C6H12

3-methyl-2-pentane

Structural formula

Molecular formula

C6H12

Structural formula

Calculating the Empirical Formula

In general, the empirical formula of a compound can be determine from the masses of the elements that combine together.

Determination of the empirical formula of magnesium oxide:

Mass of the crucible + the lid = 21.02 g

Mass of the crucible + the lid + Mg ribbon = 21.50 g

Mass of the crucible + the lid + magnesium oxide = 21.82 g

Mass of the crucible + the lid = 21.02 g

Mass of the crucible + the lid + Mg ribbon = 21.50 g

Mass of the crucible + the lid + magnesium oxide = 21.82 g

Mass of magnesium burnt = (21.50 – 21.02)g = 0.48 g

Mass of oxygen used = ( 21.82 – 21.50) = 0.32 g

elements Mg O

Mass 0.48 g 0.32 g

Number of moles of atom 0.48/24 = 0.02 mol 0.32/16 = 0.02

The simplest ratio 0.02/0.02 = 1 0.02/0.02 = 1

The empirical formula = MgO

Example 1: 3.12 g metal X reacts with oxygen to form 4.56 g X oxide. Determine the empirical formula of X oxide.(Relatif atomic mass: O = 16, X = 52)

Element X O

Mass (g) 3.12 (4.56-3.12)

= 1.44 g

Number of moles of atom 3.12/52

= 0.06 mol

1.44/16

= 0.09 mol

Simplest ratio of the number of moles

0.06/0.06 = 1 0.09/0.06 = 1.5

2 3

Empirical formula = X2O3

Elements Fe O

Mass, g 1.12 0.48Number of moles 1.12 = 0.02

560.48 = 0.03

16

Divide by the smallest number

0.02 = 1 0.02

0.03 = 1.5 0.02

Ratio of atom 2 3

Empirical Formula : Fe2O3

• A sample of ferum oxide consists of 1.12 g ferum and 0.48g oxygen.Find the empirical formula of this compound.

• [Relative atomic mass : Fe, 56 ; O,16 ]

Example 2:A compound of zinc bromide contains 28.89% zinc. What is the empirical formula of the compound?(R.A.M : Zn, 65 ; Br,80 )

Elements Zn Br

Mass per 100g compound

28.89 100 – 28.89 = 71.11

Number of moles of atoms

28.89 = 0.44

65

71.11 = 0.89

80

Divide by the smallest number

0.44 = 1

0.44

0.89 = 2

0.44

Ratio of atom 1 2

Empirical formula : ZnBr2

• Example 3

The empirical formula of a compound is given as CH2 . If the RMM of the compound is 56, what is the molecular formula? [ RAM C,12 ; H , 1 ].

( Empirical formula ) n = RMM

( CH2 ) n = 56

( 1x 12 + 2 x 1 ) n = 56

14 n = 56

n = 4

Molecular formula (CH2 )4 = C4H8

• Example 2

A compound CxHyOz contains 40% carbon and 53.3% oxygen. If the relative molecular mass of the compound is 180, finds it molecular formula

[ RAM : H , 1 : C , 12 ; O , 16 ]

Element C H O

Mass in 100g 40 40-53.3 = 6.7 53.3

Number of moles 40 =3.33

12

6.7 = 6.7

1

53.3 = 3.33

16

Simplest ratio 3.33 = 1

3.33

6.7 = 2

3.33

3.33 = 1

3.33

Empirical Formula : CH2O

• (Empirical formula ) n = RMM

• (CH2O)n = 180

• (12 + 2 + 16 ) n = 180

• 30n = 180

• n = 180/30 = 6

Molecular formula = (CH2O)6 = C6 H12 O6

CATION ANION

NAME FORMULA NAME FORMULA

Aluminium Al3+ Bromide Br-

Ammonium NH4+ Carbonate CO3

2-

Barium Ba2+ Chromate(VI) CrO42-

Calsium Ca2+ Dichromate(VI) Cr2O72-

Cromium(III) Cr3+ Fluoride F-

Cobalt(II) Co2+ Hidride H-

Copper(II) Cu2+ Hidrogen carbonate

HCO3-

Copper(I) Cu+ Hidrogen sulphate

HSO4-

Iron(II) Fe2+ Hidroxide OH-

Iron(III) Fe3+ Iodide I-

Lihtium Li+ Nitrate NO3-

Manganate(IV) Mn4+ Nitrite NO2-

Manganate (II) Mn2+ Nitride N3-

Magnesium Mg2+ Oxide O2-

Pottasium K+ Manganate(VII)

MnO4-

Tin(II) Sn2+ Peroksida O22-

Stanum(IV) Sn4+ Fosfat PO4 3-

Plumbum(II) Pb2+ Sulfat SO42-

Plumbum(IV) Pb2+ Sulfit SO32-

Vanadium(V) V5+ Sulfida S2-

Zink Zn2+ Tiosianat SCN-

Natrium Na+ Tiosulfat S2O32-

Titanium(IV) Ti4+ Oksalat C2O42-