Empirical formula

From The School of Biomedical Sciences Wiki
(Difference between revisions)
Jump to: navigation, search
(created page)
 
m (added one more example and fixed some formatting)
Line 1: Line 1:
 The empirical formula of a substance is its simplest whole number ratio. The Molecular Formula is a multiple of the Empirical Formula.
+
The empirical formula of a substance is its simplest whole number ratio between its constituents. The Molecular Formula is a multiple of the Empirical Formula.  
  
To calculate Empirical Formula:
+
For example: 
  
&nbsp;Mass (g) / Relative Atomic Mass (A<sub>r</sub>)&nbsp;&nbsp;→ This will give you the number of moles.<sub></sub>
+
The Molecular formula of butanoic acid is C<sub>4</sub>H<sub>8</sub>O<sub>2</sub>&nbsp;whilst the empirical formula is&nbsp;C<sub><span style="font-size: 11.0667px;">2</span></sub><span style="font-size: 11.0667px;">H<sub>4</sub>O.&nbsp;</span><sub></sub>
  
&nbsp;You then divide the number of moles by the smallest number reulting from your previous calculations, and that will give you the ratio for the empirical formula.
+
<span style="font-size: 11.0667px;" />
  
 +
To calculate Empirical Formula:
  
 +
&nbsp;Mass (g) / Relative Atomic Mass (A<sub>r</sub>)&nbsp;&nbsp;→ This will give you the number of moles.<sub></sub>
  
Example:
+
&nbsp;You then divide the number of moles by the smallest number reulting from your previous calculations, and that will give you the ratio for the empirical formula.
  
You have 9.62 g of Carbon, 1.60 g of Hydrogen and 4.28 g of Oxygen. What is the empirical formula of this CHO substance?
+
<br>
  
C&nbsp;→ 9.62 / 12.01 (A<sub>r</sub>) = 0.80 (number of moles)<sub></sub>
+
Example:
  
H&nbsp;→ 1.60 / 1.01 = 1.60
+
You have 9.62 g of Carbon, 1.60 g of Hydrogen and 4.28 g of Oxygen. What is the empirical formula of this CHO substance?
  
O&nbsp;→ 4.28 / 16.00 = 0.27
+
C&nbsp;→ 9.62 / 12.01 (A<sub>r</sub>) = 0.80 (number of moles)<sub></sub>
  
Smallest value = 0.27, so:
+
H&nbsp;→ 1.60 / 1.01 = 1.60
  
C&nbsp;→ 0.80 / 0.27 = 2.96&nbsp;
+
O&nbsp;→ 4.28 / 16.00 = 0.27  
  
H&nbsp;→ 1.60 / 0.27 = 5.93
+
Smallest value = 0.27, so:
  
O&nbsp;→ 0.27 / 0.27 = 1
+
C&nbsp;→ 0.80 / 0.27 = 2.96&nbsp;
  
 +
H&nbsp;→ 1.60 / 0.27 = 5.93
  
 +
O&nbsp;→ 0.27 / 0.27 = 1
  
Empirical formula uses whole numbers, so you must round the numbers obtained to the closest whole number. This gives us an empirical formula of:
+
<br>
 +
 
 +
Empirical formula uses whole numbers, so you must round the numbers obtained to the closest whole number. This gives us an empirical formula of:  
  
 
C<sub>3</sub>H<sub>6</sub>O
 
C<sub>3</sub>H<sub>6</sub>O

Revision as of 18:40, 1 December 2018

The empirical formula of a substance is its simplest whole number ratio between its constituents. The Molecular Formula is a multiple of the Empirical Formula.

For example: 

The Molecular formula of butanoic acid is C4H8O2 whilst the empirical formula is C2H4O. 

<span style="font-size: 11.0667px;" />

To calculate Empirical Formula:

 Mass (g) / Relative Atomic Mass (Ar)  → This will give you the number of moles.

 You then divide the number of moles by the smallest number reulting from your previous calculations, and that will give you the ratio for the empirical formula.


Example:

You have 9.62 g of Carbon, 1.60 g of Hydrogen and 4.28 g of Oxygen. What is the empirical formula of this CHO substance?

C → 9.62 / 12.01 (Ar) = 0.80 (number of moles)

H → 1.60 / 1.01 = 1.60

O → 4.28 / 16.00 = 0.27

Smallest value = 0.27, so:

C → 0.80 / 0.27 = 2.96 

H → 1.60 / 0.27 = 5.93

O → 0.27 / 0.27 = 1


Empirical formula uses whole numbers, so you must round the numbers obtained to the closest whole number. This gives us an empirical formula of:

C3H6O

Personal tools
Namespaces
Variants
Actions
Navigation
Toolbox