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(Chemistry Ch-1) 5. Mole Concept and Molar Masses




  • 1 mole of any substance can be defined as:
  • Amount of a substance that contains as many particles (atoms, molecules or ions) as there are atoms in 12 g of the 12C isotope
  • Avogadro number or Avogadro constant (NA); equal to 6.022 × 1023 particles
  • Example − 1 mole of oxygen atoms = 6.022 × 1023 atoms
1 mole of carbon dioxide molecules = 6.022 × 1023 molecules
1 mole of sodium chloride = 6.022 × 1023 formula units of sodium chloride
Molar mass of a substance can be defined as:
  • Mass of one mole of a substance in grams
  • Numerically equal to atomic/molecular/formula mass in u.
  • Example − Molar mass of CO2 = 44.011 g mol−1
Molar mass of NaCl = 58.5 g mol−1
Examples
1. What number of moles contains 3.011 × 1023 molecules of glucose?
Solution:
1 mole of glucose is equivalent to 6.022 × 1023 molecules of glucose.
Hence, 3.011 × 1023 molecules of glucose will be present in
mol = 0.5 mol (of glucose)
Thus, 0.5 mole of glucose contains 3.011 × 1023 molecules of glucose.
2. What is the mass of a mole of fluorine molecule?
Solution:
1 mole of fluorine molecule contains 6.022 × 1023 molecules and weighs 38 g.
Therefore, mass of a fluorine molecule =g
= 6.31 × 10−23 g
Percentage Composition
Mass percent of an element = 
Example
What is the mass percent of oxygen in potassium nitrate? (Atomic mass of K = 39.10 u, atomic mass of N = 14.007 u, atomic mass of O = 16.00 u)
Solution:
Atomic mass of K = 39.10 u (Given)
Atomic mass of N = 14.007 u (Given)
Atomic mass of O = 16.00 u (Given)
Therefore, molar mass of potassium nitrate (KNO3)
= 39.10 + 14.007 + 3(16.00)
= 101.107 g
Therefore, mass percent of oxygen in KNO3
= 47.47% (approx)
  • Empirical formula and molecular formula:
Empirical formula
Molecular formula
Represents the simplest whole number ratio of various atoms present in a compound
Represents the exact number of different types of atoms present in a molecule of a compound
  • Empirical formula is determined if mass % of various elements are known.
  • Molecular formula is determined from empirical formula if molar mass is known.


Example
A compound contains 92.26% carbon and 7.74% hydrogen. If the molar mass of the compound is 26.038 g mol−1, then what are its empirical and molecular formulae?
Solution:
Mass percent of carbon (C) = 92.26% (Given)
Mass percent of hydrogen (H) = 7.74% (Given)
Number of moles of carbon present in the compound = 
= 7.68 mol
Number of moles of hydrogen present in the compound 
= 7.68 mol
Thus, in the given compound, carbon and hydrogen are present in the ratio C : H = 7.68 : 7.68
= 1 : 1
Therefore, the empirical formula of the compound is CH.
Empirical formula mass of CH = (12.011 + 1.008)g
= 13.019 g
Molar mass of the compound = 26.038 g (Given)
Therefore, n = 
=
= 2
Hence, the molecular mass of the compound is (CH)n, i.e., (CH)2 or C2H2.

Interconversion among number of moles, mass and number of molecules

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