Due to the incredibly small size of the basic particles of matter there
are fantastic numbers of them in the masses used in everyday life. Amedeo
Avogadro is the scientist most credited with work done in this domain.
Syllabus reference S1.4.3
Structure 1.4.3 - Molar mass M has the units g mol–1.
- Solve problems involving the relationships between the number of particles, the amount of substance in moles and the mass in grams.
Guidance
- The relationship n = m/M is given in the data booklet.
Tools and links
- Reactivity 2.1 - How can molar masses be used with chemical equations to determine the masses of the products of a reaction?
Amedeo Avogadro
Avogadro's number
Avogadro defined the number of particles of a substance that have a mass
equivalent to the relative mass of that substance. He found that the number
of atoms in 12 g of carbon was equal to approximately 6.02 x 1023
This number was given his name - the Avogadro constant.
Avogadro's number = 6.02 x 1023
An Avogadro number of particles = 6.02 x 1023 particles
The Avogadro number or constant may be represented by the letter 'L' and
when dealing with this number of particles we say that there is an 'amount'
of 1 mole.
1 mole of any substance contains an Avogadro number of particles of that
substance = 6.02 x 1023 particles
Examples:
1 mole of iron contains 6.02 x 1023 iron atoms
1 mole of chlorine gas contains 6.02 x 1023 chlorine molecules
1 mole of electrons contains 6.02 x 1023 electrons (also
called a Faraday of charge)
Note that the Avogadro constant simply describes a specific number of any
particle, be they atoms, molecules, electrons or even armchairs!
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Worked examples
Q313-01 Calculate the number
of molecules in 35.5g of chlorine gas
Answer
|
Chlorine exists as Cl2 molecules
Relative molecular mass of Cl2 = 2 x 35.5 = 71
35.5g of chlorine = 35.5/71 moles = 0.5 moles
1 mole = 6.02 x 1023 chlorine molecules
0.5 moles = 3.01 x 1023 chlorine
molecules
|
Q313-02 Calculate the number
of atoms in 35.5g of chlorine gas
Answer
|
Chlorine exists as Cl2 molecules and each molecule contains
two atoms
Relative atomic mass of Cl = 35.5
35.5g of chlorine = 1 mole of chlorine atoms
1 mole = 6.02 x 1023 chlorine
atoms
|
Q313-03 Find the total number
of carbon atoms in 18g of glucose C6H12O6
|
Relative molecular mass of glucose = (12 x 6) + (12 x 1) + (6 x 16)
= 180
Therefore 18g of glucose = 18/180 moles = 0.1 moles
1 mole of glucose contains 6.02 x 1023 molecules
And one molecule contains 6 carbon atoms
Therefore 0.1 moles of glucose contains 0.1 x 6.02 x 1023
molecules
Which contains 0.1 x 6.02 x 1023 x 6 carbon atoms = 3.61
x 1023 carbon atoms
|
Q313-04 Calculate the number
of electrons in 23g of sodium
Answer
|
Sodium has a relative atomic mass of 23
Therefore 23g of sodium = 23/23 moles = 1 mole
1 sodium atom contains 11 electrons
Therefore 1 mole of sodium contains 6.02 x 1023 x 11 electrons
1 mole of sodium contains 6.62 x 1024
electrons
|
Q313-05 Calculate the number
of protons in 5.6g of iron.
Answer
|
Iron has a relative atomic mass of 56
Therefore 5.6g = 5.6/56 moles = 0.1 moles of iron atoms
0.1 moles of iron contains 6.02 x 1022 atoms
1 atom contains 26 protons
Therefore 0.1 moles contains 6.02 x 1022 x 26 = 1.57
x 1024 protons
|
Q313-06 Calculate the number
of ions in 5.85g of sodium chloride (Na= 23, Cl= 35.5)
Answer
|
Relative formula mass of NaCl = 23 + 35.5 = 58.5
Therefore 5.85g = 5.85/58.5 moles = 0.1 moles
1 mole of sodium chloride contains 1 mole of sodium ions and 1 mole
of chloride ions = 2 moles of ions
Therefore 0.1 moles of sodium chloride contains 0.2 moles of ions
0.2 moles of ions = 0.2 x 6.02 x 1023 ions = 1.20
x 1023 ions
|
Q313-07 Calculate the number
of water molecules in 24.95g of hydrated copper sulfate crystals, formula CuSO4.5H2O
(Cu=63.5, S=32, O=16, H=1)
Answer
|
Relative formula mass of hydrated copper sulfate = 63.5 + 32 + (4
x 16) + (5 x 18) = 249.5
Moles of hydrated copper sulfate = 24.95/249.5 = 0.1 moles
1 moles of CuSO4.5H2O contains 5 moles of water
molecules
Therefore 0.1 moles of CuSO4.5H2O contains
0.1 x 5 moles of water molecules = 0.5 moles
0.5 moles of water molecules = 0.5 x 6.02 x 1023 water
molecules = 3.01 x 1023 water
molecules
|
Q313-08 Calculate the number
of electrons lost when 5.5 grams of manganese 2+ ions are oxidised to manganate(VII)
ions (Mn=55)
Answer
|
When manganese 2+ ions are oxidised to manganate(VII) ions the equation
is as follows:
Mn2+ + 4H2O →
MnO4- + 8H+ + 5e
1 mole of manganese 2+ ions loses 5 moles of electrons
5.5g of manganese ions = 5.5/55 moles = 0.1 moles of manganese ions
0.1 moles of manganese ions lose 0.1 x 5 moles of electrons
0.1 moles of manganese ions lose 0.1 x 5 x 6.02 x 1023
electrons = 3.01 x 1023 electrons
|
Q313-09 Calculate the number
of sulfur molecules (S8) in 32g of sulfur (S8=256)
Answer
|
Relative molecular mass of sulfur molecules = 8 x 32 = 256
Moles of sulfur molecules = 32/256 = 0.125 moles of molecules
0.125 moles = 0.125 x 6.02 x 1023 molecules = 7.53
x 1022 S8 molecules
|
Q313-10 Calculate the total
number of ions present in 45.3 g of aluminium ammonium sulfate, formula Al(NH4)(SO4)2.12H2O
(Al=27, N=14, H=1, S=32, O=16)
Answer
|
Relative formula mass of Al(NH4)(SO4)2.12H2O
= 27 + (14 + 4) + 2(32 + 64) + 12(2 + 16) = 453
Moles of aluminium ammonium sulfate = 45.3/453 = 0.1
1 formula unit of Al(NH4)(SO4)2.12H2O
contains 4 ions [1 x Al3+, 1 x NH4+,
and 2 x SO42- ions]
Therefore 0.1 moles contains 0.4 moles of ions
0.4 moles of ions = 0.4 x 6.02 x 1023 ions = 2.41
x 1023 ions
|
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Now test yourself
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