Ch - 3. Avogadro's Law

Key to Avogadro's Principle

Avogadro's Principle

• Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
• So, at constant temperature and pressure, the volume of a sample of gas is proportional to the number of molecules in the gas sample.
• Since 1 mole = 6.02 x 10²³ molecules, at constant temperature and pressure, the volume, V, of a sample of gas is proportional to the moles of gas, n, in the sample.

V/n = constant

V₁/n₁ = V₂/n₂

• If the quantity of gas increases, then at the same temperature and pressure the volume the gas occupies must also increase.
• If the quantity of gas decreases, then at the same temperature and pressure the volume the gas occupies must also decrease.

Examples

Changing Quantity of Gas

10 moles of carbon dioxide gas has a volume of 245L at 25^o and 1atm pressure.
If 5 moles of the carbon dioxide is removed at the same temperature and pressure, what volume will the gas now occupy?

V1 = 245L	V2 = ?L
n1 = 10mol	n2 = 5mol

V1 n1

=

V2 n2

V2 =

V1n2 n1

=

245 x 5 10

= 122.5L

Changing Volume of Gas

At 0^oC and 1atm pressure, 0.5mol of oxygen gas has a volume of 11.2L.
If the volume of gas is expanded to 22.4L by allowing more oxygen gas into the system while maintaining the same temperature and pressure, what is the final quantity of gas in moles?

V1 = 11.2L	V2 = 22.4L
n1 = 0.5mol	n2 = ?mol

V1 n1

=

V2 n2

n2 =

V2n1 V1

=

22.4 x 0.5 11.2

= 1mol

Comparing Different Gases at the Same Temperature and Pressure

At a certain temperature and pressure, a 500mL flask contains 25 mol of nitrogen gas.
A different flask at the same temperature and pressure contains 100 mol of helium gas.
How many moles of helium are in the second flask?

V1 = 0.500L	V2 = 0.100L
n1 = 25mol	n2 = ?mol

V1 n1

=

V2 n2

n2 =

V2n1 V1

=

0.100 x 25 0.500

= 5mol

 

Definitions of a mole

Key Concepts

• Mole is abbreviated to mol and given the symbol n
• 1 mole contains the same number of particles as there are in 12g of carbon-12 atoms by definition.
  This number is called Avogadro's number or Avogadro's constant (N_A) and is equal to 6.022 x 10²³ particles.
• 1 mole of a pure substance has a mass in grams equal to its molecular mass (MM) [also known as molecular weight (MW) or formula mass (FM) or formula weight (FW)].
  This is often referred to as the molar mass.
• 1 mole of an ideal gas has a volume of:

    22.4 litres (22.4L) at S.T.P.
    [Standard Temperature and Pressure, 0^oC (273K) and 101.3kPa (1 atm)]

    24.47 litres (24.47L) at S.L.C
    [Standard Laboratory Conditions, 25^oC (298K) and 101.3kPa (1atm)]

Examples

Avogadro's Number (N_A)

• 1 mole of atoms or molecules contains 6.022 x 10²³ atoms or molecules
      eg, 1 mole of helium atoms (He) contains 6.022 x 10²³ helium atoms (He)
• To find the number of atoms in a known number of moles,
      multiply the moles by 6.022 x 10²³
      eg, 2 moles of helium atoms (He) contains 2 x 6.022 x 10²³
      = 1.2044 x 10²⁴ helium atoms. (He)
• To find the moles of atoms,
      divide the number of atoms by 6.022 x 10²³
      eg, If we have 4.2154 x 10²³ neon atoms, how many moles of neon atoms are there?
      Moles of neon atoms = (4.2154 x 10²³) ÷ (6.022 x 10²³)
      = 0.7 mol
• 1 mole of molecules contains 6.022 x 10²³ molecules.
      eg, 1 mole of oxygen molecules (O₂) contains 6.022 x 10²³ oxygen molecules (O₂).
• To find the moles of molecules,
      multiply the number of molecules by 6.022 x 10²³
      eg, ½ mole of oxygen molecules (O₂) contains ½ x 6.022 x 10²³
      = 3.011 x 10²³ oxygen molecules (O₂).
• To find the number of molecules,
      divide the moles of molecules by 6.022 x 10²³
      eg, If we have 6.022 x 10²¹ chlorine molecules (Cl₂), how many moles of chlorine molecules are there?
      Moles of chlorine molecules = (6.022 x 10²¹) ÷ (6.022 x 10²³)
      = 0.01 mol
• 1 mole of molecules does not necessarily contain 1 mole of atoms of each element in the formula
      e.g.  1 mole of HCl WILL contain 1 mole of hydrogen atoms (H) and 1 mole of chlorine atoms (Cl)
      eg, 1 mole of HCl contains 6.022 x 10²³ hydrogen atoms and 6.022 x 10²³ chlorine atoms
      eg, 5 moles of oxygen molecules (O₂) contains 5 x 2 = 10 moles of oxygen atoms (O)
      eg, 5 moles of oxygen molecules contains 10 x 6.022 x 10²³ = 6.022 x 10²⁴ oxygen atoms.
      eg, 1 mole of ammonia molecules (NH₃) will contain 1 mole of nitrogen atoms (N) and 3 moles of hydrogen atoms (H)
      1 mole of ammonia molecules contains 6.022 x 10²³ nitrogen atoms and 3 x 6.022 x 10²³ = 1.8066 x 10²⁴ hydrogen atoms

Molar Mass

• 1 mole of a pure substance has a mass in grams equal to its molecular mass (MM).
      eg, 1 mole of Helium (a monatomic gas with the formula He) has a mass equal to its relative atomic mass, 4.003g
      eg, 1 mole of hydrogen gas (a diatomic gas with the formula H₂) has a mass equal to 2 x 1.008 = 2.016g
      eg, 1 mole of ammonia gas (NH₃) has a mass equal to 14.01 + (3 x 1.008) = 17.034g
      eg, 1 mole of water (H₂O) has a mass equal to (2 x 1.008) + 16.00 = 18.016g

Ideal Gas Volumes

• at S.T.P [0^oC (273K), 101.3kPa (1 atm)], an ideal gas has a volume of 22.4L
• To find the volume of a certain number of moles of gas, multiply the moles by 22.4L
      eg, What is the volume of 2.5 moles of chlorine gas at S.T.P?
      Volume of chlorine gas = 2.5 x 22.4 = 56.0L
• To find the moles of a certain volume of gas, divide the volume by 22.4L
      eg, How many moles of argon are in 3.36L of argon gas at S.T.P?
      moles of argon gas = 3.36 ÷ 22.4 = 0.15 mol
• at S.L.C [25^oC (298K), 101.3kPa (1 atm)], an ideal gas has a volume of 24.47L
      eg, To find the volume of a certain number of moles of gas, multiply the moles by 24.47L
      What is the volume of 0.2 moles of hydrogen sulfide gas at S.L.C?
      Volume of hydrogen sulfide gas = 0.2 x 24.47 = 4.894L
• To find the moles of a certain volume of gas, divide the volume by 24.47L
      eg, How many moles of carbon monoxide are in 70.5L of carbon monoxide gas at S.L.C?
      moles of argon gas = 70.5 ÷ 24.47 = 2.881 mol

Mass-Mole Calculations (n = mass ÷ MM)

Key Concepts

• 1 mole of a pure substance has a mass equal to its molecular mass (MM)
      So 2 moles would have a mass = 2 x MM
      3 moles would have a mass = 3 x MM etc
• This leads to the formula: mass = n x MM
      mass is in grams,
      n = moles of pure substance,
      MM = molecular mass of the pure substance
• This formula can be rearranged to give the following:
      n = mass ÷ MM
      MM = mass ÷ n

Examples

1. mass = n x MM

Calculate the mass of 0.25 moles of water

• mass (in grams) = n x MM
• n = 0.25mol
• MM of water (H₂O) = (2 x 1.008) + 16.00 = 18.016g mol^-1
• mass = 0.25 x 18.016 = 4.504g

2. n = mass ÷ MM

Calculate the moles in 124.5g of oxygen gas.

• n = mass ÷ MM
• mass = 124.5g
• MM of O₂ = 2 x 16.00 =32.00g mol^-1
• n = 124.5 ÷ 32.00 = 3.89mol

3. MM = mass ÷ n

Calculate the molecular mass of a pure substance if 1.75 moles of the substance has a mass of 29.792g

• MM = mass ÷ n
• mass = 29.792g
• n = 1.75mol
• MM = 29.792 ÷ 1.75 = 17.024g mol^-1

Ø      Dalton’s atomic theory was published in 1803.

Ø      First four law of chemical combination was explained using dalton’s atomic theory.

Ø      John Jacob Berzelius in 1811 proposed that under same temperature and pressure equal volume of all gases contain same number of atom.

Ø      After the moderation of Berzelius conception, Amadeo Avogadro first introduced the conception of molecule in 1811.

Ø      In substances there are two type of smallest particle as dalton’s atomic theory as simple atom and compound atom.

Ø      The above conception of the kinds of smallest particle is corrected by Avogadro who distinguished the atom and molecule.

Ø      Simple atom is corrected to atom and compound atom to molecule.

Ø      Amount of any substance in which 6.023× 10²³ number of chemical entity remain is called mole.

Ø      In gram atomic mass of any substances remains 6.023× 10²³ number of that atom.

Ø      In gram molecular mass of any substances remains 6.023× 10²³ number of that molecule.

Ø      Volume is occupied the 1 gram molecular mass of any gas at STP is called molar volume.

Ø      Molar volume of any gas is 22.4 litre or 22400 cc.

Ø      Mass of one atom =  

Ø      Mass of one molecule =

Ø      Number of atom in 1 gram of

 any metal or carbon =

Ø      Number of molecule in 1 gram any compound =

Ø      Vapour density is the relative density of gas.

Ø      All elementary gases are di-atomic. For example ; Cl₂, H₂, O₂, N₂, F₂, Br₂, I₂ etc.

Ø      Identify diatomic molecules from the following: (i) HCl (ii) P₄ (iii) He (iv) O₃ (v) H₂S (vi) CO

Ø      HCl, CO are diatomic.