contents

Chemistry

9.3 - Processing Data & Calculations

12 min

note 1 of 10

What a calculation question looks like

Most numerical questions on the National 5 Chemistry paper give you data and ask you to work an answer out, often in more than one step.

A 20 cm³ water sample gave a precipitate of silver chloride with a mass of 1.435 g.

(a) Calculate the number of moles of silver chloride, AgCl, present in this sample.

(b) Using your answer to part (a), calculate the concentration, in mol l^-1, of chloride ions in this sample.

How multi-step questions are marked

Each step earns its own credit - Part (a) is marked on its own even though part (b) depends on it.
A wrong value carried forward still earns credit, as long as it's used correctly from then on. A slip in part (a) doesn't cost you part (b) if you divide correctly by the volume there.

The data booklet

Carries the relative atomic masses needed to find a formula mass.
Gives the relationships used across calculations, including $E_{h} = c m Δ T$ , $n = c V$ , and average rate.
You don't need to memorise relative atomic masses or formulas - you need to know which one to choose and how to substitute into it.

note 2 of 10

Formula for amount in moles

$n = \frac{m}{GFM}$

Where:

$n$ = Amount of substance, in moles (mol)
$m$ = Mass, in grams (g)
$GFM$ = Gram formula mass, found by adding the relative atomic masses of every atom in the formula (from the data booklet)

Turning a mass into an amount in moles uses the gram formula mass (GFM).

Worked example - Moles of silver chloride

A 20 cm³ water sample gave a precipitate of silver chloride, AgCl, with a mass of 1.435 g. Calculate the number of moles of AgCl present.

Step 1 - Find the GFM

GFM of AgCl = 108 + 35.5 = 143.5 g mol^-1

Step 2 - Calculate moles

$n = \frac{1.435}{143.5} = 0.01 mol$

note 3 of 10

Formula for concentration

$c = \frac{n}{V}$

Where:

$c$ = Concentration, in moles per litre (mol l^-1)
$n$ = Amount of solute, in moles (mol)
$V$ = Volume of solution, in litres (l)

Concentration measures how much solute is dissolved in a given volume of solution, in moles per litre (mol l^-1 or mol/l).

Volumes given in cm³ must be converted to litres first by dividing by 1,000 (e.g. 20 cm³ = 0.020 l). Rearranged: $n = c V$ or $V = \frac{n}{c}$ .

Worked example - Concentration of chloride ions

Calculate the concentration, in mol l^-1, of chloride ions in a 20 cm³ sample containing 0.01 mol of Cl^- ions.

Step 1 - Convert volume to litres

20 cm³ = 0.020 l

Step 2 - Calculate concentration

$c = \frac{0.01}{0.020} = 0.5 {mol l}^{- 1}$

note 4 of 10

Mass from moles by a balanced equation

A balanced equation shows the ratio of moles reacting and forming, which allows a mass of one substance to be converted into a mass of another.
Every mass-to-mass calculation follows three steps:
1. Mass to moles, using the first substance's GFM ( $n = \frac{m}{GFM}$ ).
2. Moles to moles, using the mole ratio from the balanced equation.
3. Moles to mass, using the second substance's GFM ( $m = n \times GFM$ ).

Worked example - Mass of lithium chloride formed

1.0 g of lithium carbonate reacted with excess dilute hydrochloric acid. Calculate the mass, in grams, of lithium chloride, LiCl, formed.

Li₂CO₃(s) + 2HCl(aq) → 2LiCl(aq) + CO₂(g) + H₂O(l)

Step 1 - Moles of lithium carbonate

GFM of Li₂CO₃ = (2 × 7) + 12 + (3 × 16) = 74 g mol^-1

$n = \frac{1.0}{74} = 0.0135 mol$

Step 2 - Apply the mole ratio

The equation shows 1 mol of Li₂CO₃ → 2 mol of LiCl:

$n (LiCl) = 2 \times 0.0135 = 0.0270 mol$

Step 3 - Convert to mass

GFM of LiCl = 7 + 35.5 = 42.5 g mol^-1

$\begin{matrix} m & = n \times GFM \\ = 0.0270 \times 42.5 \\ = 1.15 g \end{matrix}$

note 5 of 10

Formula for average rate

$Average rate = \frac{Δ quantity}{Δ t}$

Average rate measures how much a quantity changes over a given time interval.

The unit depends on what was measured (e.g. cm³ s^-1, cm³ min^-1, or g min^-1).

Worked example - Rate of hydrogen production

A student measured the volume of hydrogen produced when calcium reacts with water.

Time (min)	0	1	2	3	4	5
Volume of gas produced (cm³)	0	32	40	46	48	48

Calculate the average rate of reaction, in cm³ min^-1, between 1 and 4 minutes.

Step 1 - Read values at the chosen times

At 1 minute: 32 cm³
At 4 minutes: 48 cm³

Step 2 - Calculate average rate

$\begin{matrix} Average rate & = \frac{48 - 32}{4 - 1} \\ = 5.3 {cm}^{3} {min}^{- 1} \end{matrix}$

note 6 of 10

Formula for percentage composition

$% composition = \frac{A_{r} \times number of atoms}{GFM of compound} \times 100$

Where:

$A_{r}$ = Relative atomic mass of the element
Number of atoms is how many atoms of that element the formula shows
GFM is the gram formula mass of the compound, in g mol^-1

The percentage composition of an element is its share of a compound's total mass.

Worked example - Phosphorus in phosphoric acid

Calculate the percentage by mass of phosphorus in phosphoric acid, H₃PO₄.

Step 1 - Find the GFM

GFM of H₃PO₄ = (3 × 1) + 31 + (4 × 16) = 98 g mol^-1

Step 2 - Calculate percentage composition

$\begin{matrix} Percentage composition & = \frac{31}{98} \times 100 \\ = 31.6 % \end{matrix}$

note 7 of 10

Reading and predicting percentage yield

Percentage yield is the amount of product actually made, as a percentage of the maximum that could be made.
At National 5, yield data is usually provided in a table or graph to analyse rather than calculate from masses.

Worked example table - Yield of ammonia

Temperature (°C)	100	200	400	500	600	700
Percentage yield of ammonia (%)	97	87	46	28	17	10

Typical question types

Reading a value - Read directly from the table (e.g. at 200 °C, yield is 87%).
Describing the trend - State what happens to one variable as the other changes (e.g. as temperature increases, the percentage yield decreases).
Predicting a value - Follow the pattern beyond the given values (e.g. at 800 °C, predicting around 3%).
Plotting data - Place the independent variable (condition) on the x-axis and percentage yield on the y-axis, drawing a smooth curve of best fit.

note 8 of 10

Formula for energy transferred

$E_{h} = c m Δ T$

Where:

$E_{h}$ = Energy transferred, in kilojoules (kJ)
$c$ = Specific heat capacity of water, 4.18 kJ kg^-1 °C^-1 (from the data booklet)
$m$ = Mass of water, in kilograms (kg); water has a density of 1 g cm^-3, so 100 cm³ of water is 0.100 kg
$Δ T$ = Temperature change, in degrees Celsius (°C)

When a reaction heats or cools water, the energy transferred is calculated from the temperature change.

Worked example - Energy absorbed by heated water

A student burned propanol to heat 100 cm³ of water, recording a mean temperature rise of 21 °C. Calculate the energy, in kJ, absorbed by the water.

Step 1 - Convert the mass of water to kilograms

100 cm³ = 0.100 kg

Step 2 - Calculate energy

$E_{h} = 4.18 \times 0.100 \times 21 = 8.78 kJ$

note 9 of 10

Units and significant figures

Units in calculation answers

Units are usually not required unless specifically asked for in the question.
If you write a unit, it must be correct - an incorrect unit loses a mark (penalised at most once per paper).
If unsure of a unit that wasn't requested, leave it off.

Rounding and significant figures

Round answers sensibly based on the data given in the question.
Avoid writing out full recurring calculator displays (e.g. write 31.6% rather than 31.63265306%).
Intermediate rounding errors can be avoided by keeping unrounded numbers in the calculator until the final step.

note 10 of 10

Common exam questions

Question 1

Calculate the number of moles in 4.6 g of sodium, Na (Ar = 23).

$n = \frac{m}{GFM} = \frac{4.6}{23} = 0.2$ mol.

Question 2

A solution contains 0.02 mol of solute dissolved in 500 cm³ of solution. Calculate its concentration in mol l^-1.

Convert the volume to litres: 500 cm³ is 0.500 l. $c = \frac{n}{V} = \frac{0.02}{0.500} = 0.04$ mol l^-1.

Question 3

Magnesium reacts with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Calculate the mass of MgCl₂ formed from 2.4 g of magnesium (A_r of Mg = 24.5, GFM of MgCl₂ = 95.5).

$n (Mg) = \frac{2.4}{24.5} = 0.098$ mol. The ratio of Mg to MgCl₂ is 1:1, so $n ({MgCl}_{2}) = 0.098$ mol. $m = n \times GFM = 0.098 \times 95.5 = 9.36$ g.

Question 4

The volume of gas collected in a measuring cylinder inverted over water rises from 12 cm³ to 36 cm³ in the first 40 seconds of a reaction. Calculate the average rate of reaction over this time.

$Average rate = \frac{36 - 12}{40} = 0.6$ cm³ s^-1.

Question 5

Calculate the percentage by mass of nitrogen in ammonia, NH₃ (A_r of N = 14, GFM of NH₃ = 17).

$\frac{14}{17} \times 100 = 82.4 %$ .

Question 6

A table shows the percentage yield of a reaction at three pressures: 100 atmospheres, 46%; 200 atmospheres, 60%; 300 atmospheres, 68%. Describe how the percentage yield changes as the pressure increases.

As the pressure increases, the percentage yield increases, from 46% to 68%.

Question 7

A student mixes two solutions in an insulated cup and records a temperature rise of 6.5 °C in 50 cm³ of solution. Calculate the energy released, using the specific heat capacity of water.

Convert the volume to a mass in kilograms: 50 cm³ is 0.0500 kg. $E_{h} = c m Δ T = 4.18 \times 0.0500 \times 6.5 = 1.4$ kJ.

Question 8

A student finishes a calculation with "= 24" for a question that did not ask for a unit. Should they add one?

It is not required, so leaving it off costs nothing. Adding a unit is only worth doing if it is correct, because getting it wrong when it was not needed can still cost the one unit mark available across the whole paper.

Question 9

Why does National 5 Chemistry have no significant-figures rule, and what should you do instead?

No tolerance is stated anywhere in the papers or the specification, so instead of aiming for a fixed number of figures, round your answer to a sensible number that matches the precision of the data you were given, rather than copying a long calculator display.

key terms

TermDefinition

Gram formula massThe mass of one mole of a substance, found by adding the relative atomic masses of every atom in its formula.

ConcentrationA measure of how much of a substance is present in a given volume, usually how much solute is dissolved in a given volume of solution.

Average rateThe change in a measured quantity, such as a volume of gas or a mass, divided by the time interval over which it changed.

Percentage by massThe percentage of a compound's total mass that comes from a particular element. It is also called percentage composition.

Percentage yieldThe amount of product actually made, given as a percentage of the maximum amount that could be made (the theoretical yield).

end of notes

Key points to remember

9.4 - Plotting Graphs & Bar ChartsNext up

contents

Chemistry

9.3 - Processing Data & Calculations

12 min

note 1 of 10

What a calculation question looks like

Most numerical questions on the National 5 Chemistry paper give you data and ask you to work an answer out, often in more than one step.

A 20 cm³ water sample gave a precipitate of silver chloride with a mass of 1.435 g.

(a) Calculate the number of moles of silver chloride, AgCl, present in this sample.

(b) Using your answer to part (a), calculate the concentration, in mol l^-1, of chloride ions in this sample.

How multi-step questions are marked

Each step earns its own credit - Part (a) is marked on its own even though part (b) depends on it.
A wrong value carried forward still earns credit, as long as it's used correctly from then on. A slip in part (a) doesn't cost you part (b) if you divide correctly by the volume there.

The data booklet

Carries the relative atomic masses needed to find a formula mass.
Gives the relationships used across calculations, including $E_{h} = c m Δ T$ , $n = c V$ , and average rate.
You don't need to memorise relative atomic masses or formulas - you need to know which one to choose and how to substitute into it.

note 2 of 10

Formula for amount in moles

$n = \frac{m}{GFM}$

Where:

$n$ = Amount of substance, in moles (mol)
$m$ = Mass, in grams (g)
$GFM$ = Gram formula mass, found by adding the relative atomic masses of every atom in the formula (from the data booklet)

Turning a mass into an amount in moles uses the gram formula mass (GFM).

Worked example - Moles of silver chloride

A 20 cm³ water sample gave a precipitate of silver chloride, AgCl, with a mass of 1.435 g. Calculate the number of moles of AgCl present.

Step 1 - Find the GFM

GFM of AgCl = 108 + 35.5 = 143.5 g mol^-1

Step 2 - Calculate moles

$n = \frac{1.435}{143.5} = 0.01 mol$

note 3 of 10

Formula for concentration

$c = \frac{n}{V}$

Where:

$c$ = Concentration, in moles per litre (mol l^-1)
$n$ = Amount of solute, in moles (mol)
$V$ = Volume of solution, in litres (l)

Concentration measures how much solute is dissolved in a given volume of solution, in moles per litre (mol l^-1 or mol/l).

Volumes given in cm³ must be converted to litres first by dividing by 1,000 (e.g. 20 cm³ = 0.020 l). Rearranged: $n = c V$ or $V = \frac{n}{c}$ .

Worked example - Concentration of chloride ions

Calculate the concentration, in mol l^-1, of chloride ions in a 20 cm³ sample containing 0.01 mol of Cl^- ions.

Step 1 - Convert volume to litres

20 cm³ = 0.020 l

Step 2 - Calculate concentration

$c = \frac{0.01}{0.020} = 0.5 {mol l}^{- 1}$

note 4 of 10

Mass from moles by a balanced equation

A balanced equation shows the ratio of moles reacting and forming, which allows a mass of one substance to be converted into a mass of another.
Every mass-to-mass calculation follows three steps:
1. Mass to moles, using the first substance's GFM ( $n = \frac{m}{GFM}$ ).
2. Moles to moles, using the mole ratio from the balanced equation.
3. Moles to mass, using the second substance's GFM ( $m = n \times GFM$ ).

Worked example - Mass of lithium chloride formed

1.0 g of lithium carbonate reacted with excess dilute hydrochloric acid. Calculate the mass, in grams, of lithium chloride, LiCl, formed.

Li₂CO₃(s) + 2HCl(aq) → 2LiCl(aq) + CO₂(g) + H₂O(l)

Step 1 - Moles of lithium carbonate

GFM of Li₂CO₃ = (2 × 7) + 12 + (3 × 16) = 74 g mol^-1

$n = \frac{1.0}{74} = 0.0135 mol$

Step 2 - Apply the mole ratio

The equation shows 1 mol of Li₂CO₃ → 2 mol of LiCl:

$n (LiCl) = 2 \times 0.0135 = 0.0270 mol$

Step 3 - Convert to mass

GFM of LiCl = 7 + 35.5 = 42.5 g mol^-1

$\begin{matrix} m & = n \times GFM \\ = 0.0270 \times 42.5 \\ = 1.15 g \end{matrix}$

note 5 of 10

Formula for average rate

$Average rate = \frac{Δ quantity}{Δ t}$

Average rate measures how much a quantity changes over a given time interval.

The unit depends on what was measured (e.g. cm³ s^-1, cm³ min^-1, or g min^-1).

Worked example - Rate of hydrogen production

A student measured the volume of hydrogen produced when calcium reacts with water.

Time (min)	0	1	2	3	4	5
Volume of gas produced (cm³)	0	32	40	46	48	48

Calculate the average rate of reaction, in cm³ min^-1, between 1 and 4 minutes.

Step 1 - Read values at the chosen times

At 1 minute: 32 cm³
At 4 minutes: 48 cm³

Step 2 - Calculate average rate

$\begin{matrix} Average rate & = \frac{48 - 32}{4 - 1} \\ = 5.3 {cm}^{3} {min}^{- 1} \end{matrix}$

note 6 of 10

Formula for percentage composition

$% composition = \frac{A_{r} \times number of atoms}{GFM of compound} \times 100$

Where:

$A_{r}$ = Relative atomic mass of the element
Number of atoms is how many atoms of that element the formula shows
GFM is the gram formula mass of the compound, in g mol^-1

The percentage composition of an element is its share of a compound's total mass.

Worked example - Phosphorus in phosphoric acid

Calculate the percentage by mass of phosphorus in phosphoric acid, H₃PO₄.

Step 1 - Find the GFM

GFM of H₃PO₄ = (3 × 1) + 31 + (4 × 16) = 98 g mol^-1

Step 2 - Calculate percentage composition

$\begin{matrix} Percentage composition & = \frac{31}{98} \times 100 \\ = 31.6 % \end{matrix}$

note 7 of 10

Reading and predicting percentage yield

Percentage yield is the amount of product actually made, as a percentage of the maximum that could be made.
At National 5, yield data is usually provided in a table or graph to analyse rather than calculate from masses.

Worked example table - Yield of ammonia

Temperature (°C)	100	200	400	500	600	700
Percentage yield of ammonia (%)	97	87	46	28	17	10

Typical question types

Reading a value - Read directly from the table (e.g. at 200 °C, yield is 87%).
Describing the trend - State what happens to one variable as the other changes (e.g. as temperature increases, the percentage yield decreases).
Predicting a value - Follow the pattern beyond the given values (e.g. at 800 °C, predicting around 3%).
Plotting data - Place the independent variable (condition) on the x-axis and percentage yield on the y-axis, drawing a smooth curve of best fit.

note 8 of 10

Formula for energy transferred

$E_{h} = c m Δ T$

Where:

$E_{h}$ = Energy transferred, in kilojoules (kJ)
$c$ = Specific heat capacity of water, 4.18 kJ kg^-1 °C^-1 (from the data booklet)
$m$ = Mass of water, in kilograms (kg); water has a density of 1 g cm^-3, so 100 cm³ of water is 0.100 kg
$Δ T$ = Temperature change, in degrees Celsius (°C)

When a reaction heats or cools water, the energy transferred is calculated from the temperature change.

Worked example - Energy absorbed by heated water

A student burned propanol to heat 100 cm³ of water, recording a mean temperature rise of 21 °C. Calculate the energy, in kJ, absorbed by the water.

Step 1 - Convert the mass of water to kilograms

100 cm³ = 0.100 kg

Step 2 - Calculate energy

$E_{h} = 4.18 \times 0.100 \times 21 = 8.78 kJ$

note 9 of 10

Units and significant figures

Units in calculation answers

Units are usually not required unless specifically asked for in the question.
If you write a unit, it must be correct - an incorrect unit loses a mark (penalised at most once per paper).
If unsure of a unit that wasn't requested, leave it off.

Rounding and significant figures

Round answers sensibly based on the data given in the question.
Avoid writing out full recurring calculator displays (e.g. write 31.6% rather than 31.63265306%).
Intermediate rounding errors can be avoided by keeping unrounded numbers in the calculator until the final step.

note 10 of 10

Common exam questions

Question 1

Calculate the number of moles in 4.6 g of sodium, Na (Ar = 23).

$n = \frac{m}{GFM} = \frac{4.6}{23} = 0.2$ mol.

Question 2

A solution contains 0.02 mol of solute dissolved in 500 cm³ of solution. Calculate its concentration in mol l^-1.

Convert the volume to litres: 500 cm³ is 0.500 l. $c = \frac{n}{V} = \frac{0.02}{0.500} = 0.04$ mol l^-1.

Question 3

Magnesium reacts with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Calculate the mass of MgCl₂ formed from 2.4 g of magnesium (A_r of Mg = 24.5, GFM of MgCl₂ = 95.5).

$n (Mg) = \frac{2.4}{24.5} = 0.098$ mol. The ratio of Mg to MgCl₂ is 1:1, so $n ({MgCl}_{2}) = 0.098$ mol. $m = n \times GFM = 0.098 \times 95.5 = 9.36$ g.

Question 4

The volume of gas collected in a measuring cylinder inverted over water rises from 12 cm³ to 36 cm³ in the first 40 seconds of a reaction. Calculate the average rate of reaction over this time.

$Average rate = \frac{36 - 12}{40} = 0.6$ cm³ s^-1.

Question 5

Calculate the percentage by mass of nitrogen in ammonia, NH₃ (A_r of N = 14, GFM of NH₃ = 17).

$\frac{14}{17} \times 100 = 82.4 %$ .

Question 6

As the pressure increases, the percentage yield increases, from 46% to 68%.

Question 7

A student mixes two solutions in an insulated cup and records a temperature rise of 6.5 °C in 50 cm³ of solution. Calculate the energy released, using the specific heat capacity of water.

Convert the volume to a mass in kilograms: 50 cm³ is 0.0500 kg. $E_{h} = c m Δ T = 4.18 \times 0.0500 \times 6.5 = 1.4$ kJ.

Question 8

A student finishes a calculation with "= 24" for a question that did not ask for a unit. Should they add one?

Question 9

Why does National 5 Chemistry have no significant-figures rule, and what should you do instead?

key terms

TermDefinition

Gram formula massThe mass of one mole of a substance, found by adding the relative atomic masses of every atom in its formula.

ConcentrationA measure of how much of a substance is present in a given volume, usually how much solute is dissolved in a given volume of solution.

Average rateThe change in a measured quantity, such as a volume of gas or a mass, divided by the time interval over which it changed.

Percentage by massThe percentage of a compound's total mass that comes from a particular element. It is also called percentage composition.

Percentage yieldThe amount of product actually made, given as a percentage of the maximum amount that could be made (the theoretical yield).

end of notes

Key points to remember

9.4 - Plotting Graphs & Bar ChartsNext up