Back to Course

SS1: CHEMISTRY - 1ST TERM

0% Complete
0/0 Steps
  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1Topic
    |
    1 Quiz
  4. Separation Techniques II | Week 4
    5Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5Topics
    |
    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
    |
    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
    |
    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
    |
    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3Topics
    |
    1 Quiz
Lesson Progress
0% Complete

Empirical Formula indicates the ratios of the atoms present in the substance. The empirical formula of a compound can be calculated from the percentage composition of the compound.

Determination of Empirical Formula from the percentage composition

1. The percentage composition of each element in the compound or the mass of each element in the compound will be provided.

2. Divide the percentage composition by the relative atomic masses.

3. Divide the result obtained in (2) above by the smallest value.

Example I

A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen. Calculate its Empirical Formula. (Na = 23, C = 12, O = 16)

Solution:

NaC0
Step I43.411.345.3
Step II \( \frac{43.4}{23}\\ \scriptsize = 1.89 \)\( \frac{11.3}{12} \\ \scriptsize = 0.9416 \)\( \frac{45.3}{16}\\ \scriptsize = 2.83125 \)
Step III \( \frac{1.89}{0.9416} \)\( \frac{0.9416}{0.9416} \)\( \frac{2.83125}{0.9416} \)
213

= Na2C1O3

Empirical Formula = Na2CO3

back-to-top
error: