Worked science calculations

Percentage mass change

A plant-tissue piece has initial mass 4 g and final mass 3.4 g after soaking. A second piece starts at 7 g and has the same percentage change.

Percentage change = (final mass − initial mass) ÷ initial mass × 100

1. Calculate final mass minus initial mass, including its sign.

   Worked answer

   3.4 − 4 = -0.6 g.

2. Calculate the percentage mass change, including its sign.

   Worked answer

   (3.4 − 4) ÷ 4 × 100 = -15%. The original mass is the denominator.

3. Predict the second piece’s final mass.

   Worked answer

   7 × (1 − 15/100) = 5.95 g.

Surface water was blotted consistently before weighing. Pieces were soaked for the same time and at the same temperature.

Practise percentage mass change with different data · Read the explanation

Magnification and cell size

A cell is 80 µm long. At ×100, its image is measured in millimetres. Later, the same cell is drawn with an image length of 100 mm.

Magnification = image size ÷ actual size. Convert both sizes to the same unit.

1. Convert the cell’s actual length to millimetres.

   Worked answer

   80 ÷ 1,000 = 0.08 mm.

2. Calculate its first image length in millimetres.

   Worked answer

   0.08 × 100 = 8 mm.

3. Calculate the magnification of the later drawing. Leave the unit blank.

   Worked answer

   100 ÷ 0.08 = 1250. Magnification is a ratio with no unit.

Use the stated magnification of the displayed or measured image, not just the microscope setting.

Practise magnification and cell size with different data · Read the explanation

Quadrats and population estimates

Five randomly placed quadrats, each 0.5 m², give counts of 7, 10, 10, 9, 7 plants. The suitable habitat covers 580 m².

Mean count = total count ÷ number of quadrats. Estimate = mean count × habitat area ÷ quadrat area.

1. Calculate the mean number of plants per quadrat.

   Worked answer

   43 ÷ 5 = 8.6 plants per quadrat. Leave the unit field blank for the count.

2. Calculate the estimated number of plants per square metre.

   Worked answer

   8.6 ÷ 0.5 = 17.2 plants per m².

3. Estimate the total number of plants in the habitat.

   Worked answer

   17.2 × 580 = 9976 plants. This is an estimate.

Sampling is representative of the suitable habitat. The estimate describes abundance, not an exact census.

Practise quadrats and population estimates with different data · Read the explanation

Comparing enzyme rates

Amylase reaches the same starch-test endpoint in 50 s in condition A and 100 s in condition B. Samples are tested every 10 s.

Rate comparison value = 1 ÷ endpoint time. This compares the same endpoint under controlled conditions.

1. Calculate the rate comparison value for A as 1 ÷ time.

   Worked answer

   1 ÷ 50 = 0.02 s⁻¹. Keep extra digits until your final answer.

2. Calculate the rate comparison value for B as 1 ÷ time.

   Worked answer

   1 ÷ 100 = 0.01 s⁻¹.

3. Calculate A’s rate comparison value divided by B’s. Leave the unit blank.

   Worked answer

   (1/50) ÷ (1/100) = 100/50 = 2. A has 2 times B’s comparison value.

Starting amounts, pH and endpoint are the same. Reciprocal endpoint time is a rate comparison, not a direct measurement of moles reacted per second.

Practise comparing enzyme rates with different data · Read the explanation

Concentration and dilution

A uniform solution has concentration 10 g/dm³ and final volume 100 cm³. Water is added to make a final volume of 400 cm³, with no solute lost.

Concentration = solute mass ÷ solution volume. 1 dm³ = 1,000 cm³. Dilution conserves solute mass.

1. Convert the original volume to dm³.

   Worked answer

   100 ÷ 1,000 = 0.1 dm³.

2. Calculate the solute mass in the original solution.

   Worked answer

   10 × 0.1 = 1 g.

3. Calculate the concentration after dilution.

   Worked answer

   1 ÷ 0.4 = 2.5 g/dm³.

Practise concentration and dilution with different data · Read the explanation

Moles and reacting masses

Carbon reacts completely in excess oxygen: C + O₂ → CO₂. Use molar masses C = 12 g/mol and CO₂ = 44 g/mol. The starting carbon mass is 4.8 g.

Amount in mol = mass in g ÷ molar mass in g/mol. Use the balanced equation to compare amounts.

1. Calculate the amount of carbon in moles.

   Worked answer

   4.8 ÷ 12 = 0.4 mol C.

2. Calculate the expected mass of carbon dioxide.

   Worked answer

   0.4 mol C gives 0.4 mol CO₂. Mass = 0.4 × 44 = 17.6 g.

3. Calculate the mass of oxygen that combined with the carbon.

   Worked answer

   17.6 − 4.8 = 12.8 g oxygen.

Only the stated product forms, oxygen is in excess and no product is lost. These supplied molar masses are rounded. This is a calculation, not a practical method.

Practise moles and reacting masses with different data · Read the explanation

Rates from gas-volume data

A gas-producing reaction is measured at equal time intervals. Use the table to compare mean rates.

Mean rate = change in volume ÷ elapsed time. Calculate differences between the interval endpoints.

1. Calculate the mean rate from 0 to 20 s.

   Worked answer

   (18 − 0) ÷ 20 = 0.9 cm³/s.

2. Calculate the mean rate from 20 to 60 s.

   Worked answer

   (39 − 18) ÷ (60 − 20) = 0.525 cm³/s.

3. Calculate the mean rate over the entire 80 s.

   Worked answer

   45 ÷ 80 = 0.5625 cm³/s.

Collection conditions stay constant and the apparatus does not leak. Values are an illustrative dataset.

Practise rates from gas-volume data with different data · Read the explanation

Bond-energy calculations

For H₂ + Cl₂ → 2HCl, use these supplied average bond energies: H–H 436, Cl–Cl 240, H–Cl 431, all in kJ/mol of bonds. A later reaction uses 2 mol H₂ with enough chlorine.

Energy change = energy to break bonds − energy released forming bonds, for the equation as written.

1. Calculate the energy needed to break the reactant bonds for 1 mol H₂ reacting.

   Worked answer

   436 + 240 = 676 kJ. Bond breaking requires energy.

2. Calculate the energy released when the product bonds form for 1 mol H₂ reacting.

   Worked answer

   2 × 431 = 862 kJ released. Enter the positive amount released.

3. Calculate the estimated signed energy change for 2 mol H₂ reacting.

   Worked answer

   (676 − 862) × 2 = -372 kJ. The negative sign means energy is transferred to the surroundings.

This is a calculation using average gas-phase bond energies. Values estimate the energy change; this is not a practical method.

Practise bond-energy calculations with different data · Read the explanation

Current, resistance and power

An ideal supply provides 10 V across a resistor carrying 100 mA for 5 minutes. The resistor stays at constant temperature.

R = V/I, P = VI, E = Pt. Convert milliamperes to amperes and minutes to seconds.

1. Convert the current to amperes.

   Worked answer

   100 ÷ 1,000 = 0.1 A.

2. Calculate the resistance.

   Worked answer

   10 ÷ 0.1 = 100 Ω.

3. Calculate the energy transferred during the stated time.

   Worked answer

   P = 10 × 0.1 = 1 W. Time = 300 s. E = 1 × 300 = 300 J.

Potential difference and current stay constant over the interval.

Practise current, resistance and power with different data · Read the explanation

Acceleration and distance

A vehicle moves forwards at 6 m/s, then accelerates uniformly to 13 m/s in 7 s. It does not reverse direction.

a = (v − u)/t. For constant acceleration, distance = (u + v)t ÷ 2.

1. Calculate the change in velocity.

   Worked answer

   13 − 6 = 7 m/s.

2. Calculate the acceleration.

   Worked answer

   (13 − 6) ÷ 7 = 1 m/s².

3. Calculate the distance travelled during this interval.

   Worked answer

   Mean velocity = (6 + 13) ÷ 2 = 9.5 m/s. Distance = 9.5 × 7 = 66.5 m.

Acceleration is constant. Because motion is forwards throughout, distance equals the change in position.

Practise acceleration and distance with different data · Read the explanation

Heating and efficiency

A 0.8 kg sample warms by 8 °C. Its specific heat capacity is 2100 J/(kg °C). Useful heating is 75% of the electrical energy supplied.

Energy transferred = mass × specific heat capacity × temperature rise. Efficiency = useful energy ÷ input energy.

1. Calculate the sample’s useful thermal energy increase.

   Worked answer

   0.8 × 2100 × 8 = 13440 J.

2. Calculate the electrical energy supplied.

   Worked answer

   13440 ÷ 0.75 = 17920 J.

3. Calculate the energy transferred elsewhere.

   Worked answer

   17920 − 13440 = 4480 J.

No change of state occurs. Use the stated constant specific heat capacity and efficiency.

Practise heating and efficiency with different data · Read the explanation

Frequency, wavelength and period

32 complete waves pass a fixed point in 4 s. The wavelength is 50 cm.

Frequency = number of complete waves ÷ time. Wave speed = frequency × wavelength. Period = 1 ÷ frequency.

1. Calculate the frequency.

   Worked answer

   32 ÷ 4 = 8 Hz.

2. Calculate the wave speed in metres per second.

   Worked answer

   λ = 50/100 = 0.5 m. v = 8 × 0.5 = 4 m/s.

3. Calculate the period of one wave.

   Worked answer

   T = 1 ÷ 8 = 0.125 s.

The source frequency and wave speed remain steady during measurement.

Practise frequency, wavelength and period with different data · Read the explanation

Check a calculation

Write the equation, substitute values with matching units and keep extra digits until the final answer. Check the size and sign: a mass loss has a negative percentage change, and dilution reduces concentration when no solute is lost.

Use the units guide when a conversion is unclear. The practice checks allow normal rounding; aim for at least three significant figures where rounding is needed. These activities develop methods across the sciences and do not label a particular examination board or tier.