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: '''Empirical formulae''' are calculated from the amount of [[atom]]s in a [[Chemical Reaction|chemical reaction]].
: The number of [[atom]]s can be found if you know the [[mass]] of different [[element]]s and the [[Relative Atomic Mass|relative atomic mass]] of the [[element]]s in the [[Chemical Reaction|reaction]].
===Finding the Empirical Formula===
{| class="wikitable"
| style="height:20px; width:200px; text-align:center;" |'''Number of Atoms'''
| style="height:20px; width:200px; text-align:center;" |'''Picture'''
| style="height:20px; width:200px; text-align:center;" |'''Structural Diagram'''
|-
| style="height:20px; width:200px; text-align:center;" |100 [[atom]]s of [[Hydrogen]] [[Chemical Reaction|react]] completely with 50 [[atom]]s of [[Oxygen]].
|[[File:SkeletalFormulaButane.png|center|200px]]
|[[File:BallandStickButan12diol.png|center|200px]]
|-
| style="height:20px; width:200px; text-align:center;" |
The [[ratio]] of [[Hydrogen]] [[atom]]s to [[Oxygen]] [[atom]]s
H:O
100:50
2:1
So the '''empirical formula''' is H<sub>2</sub>O
| style="height:20px; width:200px; text-align:center;" |
In this [[diagram]] there are 4 [[Carbon]] [[atom]]s, 10 [[Hydrogen]] [[atom]]s.
The [[ratio]] of [[atom]]s is:
C:H
4:10
2:5
So the '''empirical formula''' is C<sub>2</sub>H<sub>5</sub>
| style="height:20px; width:200px; text-align:center;" |
In this [[diagram]] there are 4 [[Carbon]] [[atom]]s, 10 [[Hydrogen]] [[atom]]s and 2 [[Oxygen]] [[atom]]s.
The [[ratio]] of [[atom]]s is:
C:H:O
4:10:2
2:5:1
So the '''empirical formula''' is C<sub>2</sub>H<sub>5</sub>O
|}