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How do we write formulae and work out the quantities that react?

Formulae and reacting quantities: chemical formulae using valency and the data booklet, balanced equations with state symbols, the mole and gram formula mass, and concentration of solutions.

An SQA National 5 Chemistry answer on formulae and reacting quantities, covering writing chemical formulae from valency and the data booklet, balancing equations with state symbols, the mole and gram formula mass, and calculating the concentration of a solution.

Generated by Claude Opus 4.812 min answer

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  1. What this key area is asking
  2. Writing chemical formulae
  3. Balanced equations and state symbols
  4. The mole and gram formula mass
  5. Worked example: mass to moles
  6. Concentration of solutions
  7. Worked example: making moles from concentration and volume
  8. Examples in context
  9. Try this

What this key area is asking

The SQA wants you to write chemical formulae using valency and the data booklet, balance symbol equations and add state symbols, and carry out calculations with the mole, gram formula mass and concentration. This is the most heavily calculated key area in Area 1, and the same skills reappear in titrations, energy and the mole work throughout the course.

Writing chemical formulae

For a compound of two elements, balance the valencies (combining powers) so the positive and negative charges cancel. For compounds containing a group ion, take the formula of the group ion from the data booklet and use brackets if you need more than one.

Balanced equations and state symbols

A balanced equation has equal numbers of each kind of atom on both sides. You balance by putting numbers in front of formulae, never by changing a formula. State symbols show the physical state: (s)(s) solid, (l)(l) liquid, (g)(g) gas, (aq)(aq) dissolved in water.

A worked balance: magnesium burning in oxygen.

2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)

Here the 2 in front of MgO is needed so that the two oxygen atoms in O2\text{O}_2 are matched, and then the 2 in front of Mg balances the magnesium.

The mole and gram formula mass

The key relationship links mass, moles and GFM:

Worked example: mass to moles

Concentration of solutions

The concentration of a solution tells you how many moles of solute are dissolved in each litre of solution, measured in mol l−1\text{mol l}^{-1}.

Always convert a volume in cm3\text{cm}^3 to litres by dividing by 1000 before using this relationship.

Worked example: making moles from concentration and volume

Examples in context

These calculations are the working tools of a chemist. A pharmacist preparing a solution of known concentration uses n=C×Vn = C \times V to weigh out the right mass; a quality-control lab uses GFM to convert a measured mass into moles before comparing it with what an equation predicts. The same mole work scales up to industry, where reacting quantities decide how much raw material is needed to make a target mass of product without waste.

Try this

Q1. Calculate the gram formula mass of sodium hydroxide, NaOH. (Na 23, O 16, H 1.) [1 mark]

  • Cue. GFM=23+16+1=40 g\text{GFM} = 23 + 16 + 1 = 40 \text{ g}.

Q2. Calculate the number of moles in 10 g10 \text{ g} of calcium, Ca (Ca 40). [1 mark]

  • Cue. n=10/40=0.25 moln = 10 / 40 = 0.25 \text{ mol}.

Q3. Calculate the concentration of a solution containing 0.2 mol0.2 \text{ mol} in 500 cm3500 \text{ cm}^3. [2 marks]

  • Cue. V=0.5 lV = 0.5 \text{ l}; C=0.2/0.5=0.4 mol l−1C = 0.2 / 0.5 = 0.4 \text{ mol l}^{-1}.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 2018 style3 marksCalculate the gram formula mass of calcium carbonate, CaCO3, and then calculate the number of moles in 25 g of it. (Use Ca 40, C 12, O 16.)
Show worked answer →

Markers reward the correct GFM, the correct rearrangement, and a value with units.

First find the gram formula mass by adding the masses of all the atoms:

GFM=40+12+(3×16)=40+12+48=100 g\text{GFM} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \text{ g}

Then use the mole relationship rearranged for number of moles:

n=mGFM=25100=0.25 moln = \frac{m}{\text{GFM}} = \frac{25}{100} = 0.25 \text{ mol}

A common error is to miss the three oxygen atoms in the carbonate; the small 3 multiplies the oxygen only.

SQA N5 2020 style3 marksA solution is made by dissolving 0.5 mol of sodium hydroxide in water and making it up to 250 cm cubed. Calculate the concentration of the solution in mol per litre.
Show worked answer →

A 3 mark answer needs the volume converted to litres, the correct relationship, and the answer with units.

First convert the volume from cm cubed to litres by dividing by 1000:

V=2501000=0.25 lV = \frac{250}{1000} = 0.25 \text{ l}

Then use the concentration relationship rearranged for concentration:

C=nV=0.50.25=2 mol l−1C = \frac{n}{V} = \frac{0.5}{0.25} = 2 \text{ mol l}^{-1}

The most common slip is leaving the volume in cm cubed; concentration in mol per litre needs the volume in litres.

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