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Higher tier only: This whole lesson is Higher tier. Foundation candidates do not need the mole, but should still be confident with relative formula mass from the previous lesson.
Atoms are unimaginably tiny — a single carbon atom has a mass of about 2×10−23 grams — so chemists never count them one by one. Instead they count them in enormous, fixed-size batches called moles, in the same way that a baker counts eggs in dozens. The mole lets us connect the masses we can actually weigh in the laboratory to the number of particles taking part in a reaction. This lesson, part of Topic C3 of OCR Gateway Science A, defines the mole and the Avogadro constant, shows how to convert between mass and moles using Mr, and uses balanced equations to calculate reacting masses.
By the end of this lesson you should be able to define the mole and state the Avogadro constant, use relative atomic and formula mass, convert between mass and moles, and calculate the mass of a product or reactant from a balanced equation.
A mole is the unit chemists use for amount of substance. One mole of any substance contains the same number of particles — and that number is the Avogadro constant:
6.02×1023 particles per mole
So one mole of carbon atoms contains 6.02×1023 carbon atoms; one mole of water molecules contains 6.02×1023 water molecules. The mole is just a very large counting number, chosen so that the mass of one mole of a substance in grams is numerically equal to its relative atomic or formula mass.
| Substance | Ar or Mr | Mass of 1 mole |
|---|---|---|
| Carbon (C) | 12 | 12 g |
| Water (H2O) | 18 | 18 g |
| Carbon dioxide (CO2) | 44 | 44 g |
| Sodium chloride (NaCl) | 58.5 | 58.5 g |
Exam Tip: The mass of one mole in grams equals the Ar (for an element) or Mr (for a compound). So "the mass of one mole of CO2" is simply its Mr, 44, expressed in grams.
The relative atomic mass (Ar) of an element is taken from the periodic table (it is the larger of the two numbers given for each element). The relative formula mass (Mr) of a compound is the sum of the Ar values of all its atoms.
Using Ar: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Ca = 40, Cu = 64.
These are worth knowing because they appear in reacting-mass calculations again and again.
The central equation of the whole topic links the number of moles, the mass in grams and the relative formula mass:
moles=Mrmass
This can be rearranged two ways:
mass=moles×MrandMr=molesmass
A simple triangle helps: cover the quantity you want, and the triangle shows the calculation.
How many moles are there in 36 g of water (H2O, Mr=18)?
Step 1 — write the equation: moles=Mrmass.
Step 2 — substitute: moles=1836.
Step 3 — calculate: moles=2 mol.
Answer: 2 moles of water.
What is the mass of 0.25 mol of calcium carbonate (CaCO3, Mr=100)?
Step 1 — rearrange: mass=moles×Mr.
Step 2 — substitute: mass=0.25×100.
Step 3 — calculate: mass=25 g.
Answer: 25 g.
Exam Tip: Always check your Mr first and write the three lines — equation, substitution, answer with unit. A wrong Mr wrecks the whole calculation, so it pays to count the atoms carefully (and multiply everything inside any bracket).
A balanced equation tells you the ratio of moles in which substances react. Combine that ratio with moles=Mrmass and you can work out the mass of any reactant or product. The method has four steps every time:
What mass of magnesium oxide is made when 6 g of magnesium burns completely? Equation: 2Mg+O2→2MgO. Use Ar: Mg = 24, O = 16, so Mr(MgO)=40.
Step 1 — moles of Mg: 246=0.25 mol.
Step 2 — mole ratio from the equation: 2Mg:2MgO, which is 1:1.
Step 3 — moles of MgO =0.25 mol.
Step 4 — mass of MgO =0.25×40=10 g.
Answer: 10 g of magnesium oxide.
Check with conservation of mass: the oxygen added has mass 10−6=4 g, which is 324=0.125 mol of O2 — exactly half the moles of Mg, as the 2:1 ratio of Mg to O2 requires. The numbers are consistent.
What mass of calcium oxide is produced by heating 50 g of calcium carbonate? Equation: CaCO3→CaO+CO2. Use Mr(CaCO3)=100 and Mr(CaO)=40+16=56.
Step 1 — moles of CaCO3: 10050=0.5 mol.
Step 2 — ratio CaCO3:CaO=1:1.
Step 3 — moles of CaO =0.5 mol.
Step 4 — mass of CaO =0.5×56=28 g.
Answer: 28 g of calcium oxide.
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