(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 (N_A); 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