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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
    |
    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
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The relative molecular mass of an element or a compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of carbon-12.

\(\scriptsize Relative \:Molecular \:Mass = \normalsize \frac{mass \: of \: 1 \: molecule \: of \: substance}{\frac{1}{12} \: \times \: mass \: of \: Carbon-12 } \)

The relative molecular mass of an element or compound is the sum of the relative atomic mass of the atoms in one molecule of that substance.

Example

Calculate the relative molecular mass (RMM) of the following compounds.

(a) CaCO3 (b) HNO3 (c) KCl.

(Ca = 40, C = 12, O = 16, H = 1, N =14, K = 39, Cl = 35.5)

Solution:

(a) CaCO3

40 + 12 + (16 x 3)

40 + 12 + 48 = 100

(b) HNO3

1 + 14 + (16 x 3)

15  + 48 =   63

(c) KCl

39 + 35.5 = 74.5

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