Standard level
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 ref: S1.4.3Structure 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!
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 |
Now test yourself