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SS1: CHEMISTRY - 1ST TERM

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  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
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    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
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    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
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    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
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    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
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The percentage composition of each element in a compound can be determined by 

1. Calculating the relative molecular mass.

2. Calculating the amount, in percentage, of each element in the compound.

Example I

Calculate the percentage by mass of Nitrogen in trioxonitrate (v) acid. (H = 1, N = 14, O = 16)

Solution:

HNO3

 1 + 14 + 16 x 3.

Relative Molecular Mass = 63, N = 14

% by mass of Nitrogen = \( \frac{14}{63} \: \times \: \frac{100}{1} \\ = \scriptsize 22.2 \% \)

Example II

Calculate the percentage composition of all the components elements in calcium trioxocarbonate (IV) (CaCO3)
(Ca = 40, C = 12, O = 16)

Solution:

Relative molecular mass = 40 + 12 + (16 x 3)  =  100

% by mass of calcium = \( \frac{40}{100} \: \times \: \frac{100}{1} \\ = \scriptsize 40 \% \)

% by mass of carbon = \( \frac{12}{100} \: \times \: \frac{100}{1} \\ = \scriptsize 12 \% \)

% by mass of oxygen = \( \frac{48}{100} \: \times \; \frac{100}{1} \\ = \scriptsize 48 \% \)

Note: 

The results above can be interpreted as follows: 100 g of calcium trioxocarbonate (IV) contains 40g of calcium, 12 g of carbon, and 48g of oxygen. The percentage by mass of any element also represents its reacting mass in forming that compound.

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