numericals on mole concept
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
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