Chemistry Module 1, 2 and 3 revision

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The arrangement of an atom

relative charge                             relative mass

protons                                              +ve                                            1

neutrons                                               0                                              0

electrons                                            -ve                                             1

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Formulas

n - amount of substance, mol

m - mass, g

M - molar mass, g mol-1

n =  m     m = n x M   M = m

_                                 _

M                                n

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Empirical Formula : Worked example 1

0.6075g of magnesium + 3.995g of bromine form a compound

[Ar: Mg, 24.3; Br, 79.9]

1. Find the molar ration of atoms   2. Divide by the smallest number (0.025

Mg : Br

0.6075     :    3.9975                     0.025 : 0.05

_____           _____                             1 : 2

24.3               79.9                           3. Empirical formula = MgBr2

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Empirical Forumla : Worked example 2

Na: 74.19%;   O: 25.81%;

[Ar: Na, 23.0; O, 16.0]

100.0g of the compound contains 74.19g of Na and 24.81g of O.

1. Find the molar ratio of atoms   2. Divide by the smallest number (1.613)

Na : O                                        3.226 : 1.613

74.19 : 25.81                                         2 :  1

______    _____                             3. Empirical formula = Na2O

3.226         1.613

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Molecular Formula : Worked Example

Empirical formula = CH2

Relative molecular mass, Mr = 56.0

[Ar: C, 12.0; H, 1.0]

What is the molecular formula?

Empirirical formula mass of CH2 = (12.0 + (1.0 x 2)) = 14.0

No. of CH2 units in a molecule =  56.0    = 4  > Molecular formula = CH2 x 4 = C4H8

____

14.0

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