Thermometric, Rate, Gravimetric and Gas-Volume Experiments

AS · 14 min

When the Paper 3 quantitative question is not a titration, it is one of four other experiment types named in the syllabus: a thermometric experiment (measuring a temperature change to find an enthalpy change), a rates experiment (timing how long something takes to happen), a gravimetric experiment (heating a solid and recording the mass change), or a gas-volume experiment (collecting a gas over water). This note covers the method, the results table, the graph and the calculation for each, with the evaluation points examiners look for. The treatment of uncertainties, significant figures and graph-drawing rules common to all of them is in Measurement, uncertainty and presenting results.

Thermometric experiments

The principle

When a reaction happens in solution in an insulated container, the energy released or absorbed changes the temperature of the solution. Assuming all the energy goes into (or comes from) the water:

Key result

q = mc\Delta T

$q$ = energy transferred, in J
$m$ = mass of the solution in g (taken as the volume in $\text{cm}^3$ , since the density is about $1.00\ \text{g cm}^{-3}$ )
$c$ = specific heat capacity of the solution, taken as that of water, $4.18\ \text{J g}^{-1}\ \text{K}^{-1}$
$\Delta T$ = temperature change in K (the same number as in °C)

Then $\Delta H = -\dfrac{q}{n}$ , where $n$ is the moles of the limiting reactant (or of water formed, for neutralisation). The sign is negative if the temperature rose (exothermic).

Watch out

The mass in $q = mc\Delta T$ is the mass of the solution that changes temperature, not the mass of a solid reactant added to it. Convert $q$ from J to kJ before dividing by moles to get $\text{kJ mol}^{-1}$ , and always include the sign.

Method

Measuring an enthalpy change in solution

Use a measuring cylinder or pipette to put a known volume of one solution in a polystyrene cup (placed in a beaker for stability), with a lid.
Record the temperature every 30 s or every minute for a few minutes, to establish the starting temperature.
At a recorded time, add the second reagent (solution or a weighed solid, in excess if it is not the limiting reagent), stir and continue recording the temperature regularly for several minutes after the maximum (or minimum).
Plot temperature against time and extrapolate both lines back to the time of mixing to find $\Delta T$ , correcting for heat lost while the reaction was happening.

The cooling correction

The temperature rise is never instantaneous, so the cup is losing heat to the surroundings while it warms up. The highest temperature recorded is therefore lower than it would have been with no heat loss. Extrapolating the cooling line back to the moment of mixing estimates the temperature the solution would have reached instantly.

(0, 21) -- (3.5, 21) (3.5, 31.2) -- (10, 29.575) (3.5, 21) -- (3.5, 31.2) (0, 21) (1, 21) (2, 21) (3, 21) (4, 28.6) (4.5, 30.6) (5, 30.8) (6, 30.6) (7, 30.3) (8, 30.1) (9, 29.8) (10, 29.6)

In this graph zinc powder is added to copper(II) sulfate solution at 3.5 minutes. The cooling line extrapolated back to 3.5 minutes gives 31.2 °C, so $\Delta T = 31.2 - 21.0 = 10.2\ \text{K}$ , larger than the highest reading (30.8 °C) minus 21.0.

Using Hess's law with two experiments

Some enthalpy changes cannot be measured directly, for example the hydration of anhydrous copper(II) sulfate ( $\ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)}$ ), because the reaction is slow and cannot be controlled. Instead, two enthalpy changes of solution are measured and combined:

$\Delta H_1$ : $\ce{CuSO4(s) + aq -> CuSO4(aq)}$
$\Delta H_2$ : $\ce{CuSO4.5H2O(s) + aq -> CuSO4(aq)}$
By Hess's law: $\Delta H_{\text{hydration}} = \Delta H_1 - \Delta H_2$ .

The same idea is used for the decomposition of a carbonate (react both the carbonate and the oxide with acid) and in many Paper 3 questions.

Rate experiments

The thiosulfate–acid ("disappearing cross") experiment

\ce{S2O3^2-(aq) + 2H+(aq) -> S(s) + SO2(g) + H2O(l)}

The sulfur forms as a fine pale yellow precipitate that makes the mixture gradually opaque. A cross drawn on paper under the flask is viewed from above; the time taken for the cross to disappear is recorded.

Method

Effect of concentration on rate

Measure different volumes of sodium thiosulfate solution (e.g. 50, 40, 30, 20, 10 $\text{cm}^3$ ) into a conical flask and make each up to the same total volume (50 $\text{cm}^3$ ) with distilled water, so the concentration of thiosulfate is proportional to its volume.
Place the flask on a paper marked with a cross.
Add a fixed volume of acid (e.g. 10 $\text{cm}^3$ of $1.0\ \text{mol dm}^{-3}$ HCl), start the stopwatch and swirl once.
Stop the clock when the cross can no longer be seen from above. Record the time.
Repeat for each concentration, keeping the temperature, total volume, acid volume and concentration, the flask and the cross the same.

Because the reaction is stopped at the same point each time (the same amount of sulfur), the rate is proportional to $\dfrac{1}{t}$ .

volume of $\ce{Na2S2O3}$ / cm³	volume of water / cm³	time / s	$\dfrac{1}{t}$ / s⁻¹
50.0	0.0	20	0.050
40.0	10.0	25	0.040
30.0	20.0	33	0.030
20.0	30.0	50	0.020
10.0	40.0	100	0.010

y = 0.001 x (10, 0.010) (20, 0.020) (30, 0.0303) (40, 0.040) (50, 0.050)

A graph of $\dfrac{1}{t}$ against volume of thiosulfate (proportional to concentration) is a straight line through the origin: the rate is directly proportional to the concentration of thiosulfate.

The same apparatus is used for the effect of temperature: warm the thiosulfate to different temperatures in a water bath, measure the temperature at the start (and end) of each run, and keep concentrations constant.

Other rate methods include measuring the volume of gas given off at intervals (gas syringe), the loss in mass as a gas escapes (balance), or the time for a colour to appear (an iodine "clock" with starch).

Gravimetric experiments

Water of crystallisation

The syllabus example is heating a hydrated salt in a crucible to drive off the water of crystallisation, then calculating $x$ in the formula $\ce{MgSO4.xH2O}$ or $\ce{CuSO4.xH2O}$ .

Method

Finding water of crystallisation

Weigh an empty crucible (and lid).
Add the hydrated salt and reweigh.
Heat on a pipe-clay triangle, gently at first then more strongly, with the lid ajar (to let water vapour escape but limit spitting).
Cool (in a desiccator if available) and weigh.
Heat to constant mass: reheat, cool and reweigh until two consecutive masses agree (for example within 0.01 g). This shows all the water has been removed.
Calculate the mass of anhydrous salt and the mass of water lost, convert both to moles, and find the ratio.

The method is the same for thermal decomposition, for example of a Group 2 carbonate, where the mass lost is carbon dioxide.

Gas-volume experiments

A gas produced in a reaction can be collected over water in an inverted measuring cylinder or burette, or in a gas syringe. The syllabus example is finding the composition of a solid from the volume of $\ce{CO2}$ produced when a carbonate reacts with acid.

Method

Collecting a gas over water

Fill a measuring cylinder (or burette) completely with water and invert it in a trough of water, with no air bubble.
Weigh the solid in a small tube, then put it with excess acid in a conical flask fitted with a bung and delivery tube leading under the cylinder. (To avoid losing gas, the solid is often kept separate, for example in a small tube inside the flask, and the flask tipped to mix after the bung is in place.)
When no more gas is produced, read the volume of gas at eye level.
Use $n = \dfrac{V}{24.0}$ (room conditions, $V$ in $\text{dm}^3$ ) and the equation to calculate the amount of reactant.

Limitations: carbon dioxide is slightly soluble in water, so some is lost when it is collected over water (a gas syringe avoids this); gas can escape before the bung is inserted; the measuring cylinder has a fairly large uncertainty; and the volume depends on temperature and pressure.

Worked examples

Enthalpy change of neutralisation

$25.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ hydrochloric acid is mixed with $25.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ sodium hydroxide in a polystyrene cup. The temperature rises by $13.5\ \text{K}$ . Calculate the enthalpy change of neutralisation. (Assume $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ and density $1.00\ \text{g cm}^{-3}$ .)

Solution

$m = 25.0 + 25.0 = 50.0\ \text{g}$

$q = 50.0 \times 4.18 \times 13.5 = 2822\ \text{J} = 2.822\ \text{kJ}$

$n(\ce{H2O}) = n(\ce{HCl}) = \dfrac{2.00 \times 25.0}{1000} = 0.0500\ \text{mol}$

\Delta H = -\frac{2.822}{0.0500} = -56.4\ \text{kJ mol}^{-1}

Negative because the temperature rose (exothermic).

Displacement reaction with a cooling correction

Excess zinc powder is added to $50.0\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ copper(II) sulfate. From the graph above, $\Delta T = 10.2\ \text{K}$ . Calculate $\Delta H$ for $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$ , and explain why the extrapolated value is used.

Solution

$q = 50.0 \times 4.18 \times 10.2 = 2132\ \text{J}$

$n(\ce{Cu^2+}) = \dfrac{0.200 \times 50.0}{1000} = 0.0100\ \text{mol}$ (limiting, since zinc is in excess)

\Delta H = -\frac{2.132}{0.0100} = -213\ \text{kJ mol}^{-1}

The reaction takes time to finish, and heat is lost to the surroundings during that time, so the highest temperature recorded is lower than the true maximum. Extrapolating the cooling curve back to the time of mixing compensates for this heat loss.

Water of crystallisation

Results: mass of crucible $15.20\ \text{g}$ ; crucible + hydrated magnesium sulfate $17.66\ \text{g}$ ; crucible + residue after heating to constant mass $16.40\ \text{g}$ . Find $x$ in $\ce{MgSO4.xH2O}$ . ( $M(\ce{MgSO4}) = 120.4$ , $M(\ce{H2O}) = 18.0$ )

Solution

Mass of anhydrous $\ce{MgSO4} = 16.40 - 15.20 = 1.20\ \text{g}$ ; $n = \dfrac{1.20}{120.4} = 9.967 \times 10^{-3}\ \text{mol}$

Mass of water $= 17.66 - 16.40 = 1.26\ \text{g}$ ; $n = \dfrac{1.26}{18.0} = 0.0700\ \text{mol}$

x = \frac{0.0700}{9.967 \times 10^{-3}} = 7.02 \approx 7

The formula is $\ce{MgSO4.7H2O}$ .

Exam-style: Hess's law from two experiments

Experiment 1: $3.99\ \text{g}$ of anhydrous copper(II) sulfate is added to $50.0\ \text{cm}^3$ of water; the temperature rises by $7.6\ \text{K}$ .

Experiment 2: $6.24\ \text{g}$ of hydrated copper(II) sulfate, $\ce{CuSO4.5H2O}$ , is added to $50.0\ \text{cm}^3$ of water; the temperature falls by $1.3\ \text{K}$ .

Calculate the enthalpy change of hydration of anhydrous copper(II) sulfate. ( $M$ : $\ce{CuSO4}$ 159.6, $\ce{CuSO4.5H2O}$ 249.6; $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ )

Solution

Experiment 1: $n = \dfrac{3.99}{159.6} = 0.02500\ \text{mol}$ ; $q_1 = 50.0 \times 4.18 \times 7.6 = 1588\ \text{J}$

$\Delta H_1 = -\dfrac{1.588}{0.02500} = -63.5\ \text{kJ mol}^{-1}$ (exothermic)

Experiment 2: $n = \dfrac{6.24}{249.6} = 0.02500\ \text{mol}$ ; $q_2 = 50.0 \times 4.18 \times 1.3 = 272\ \text{J}$

$\Delta H_2 = +\dfrac{0.272}{0.02500} = +10.9\ \text{kJ mol}^{-1}$ (endothermic: the temperature fell)

Hess cycle: $\ce{CuSO4(s)} \to \ce{CuSO4.5H2O(s)} \to \ce{CuSO4(aq)}$ has the same overall change as $\ce{CuSO4(s)} \to \ce{CuSO4(aq)}$ :

\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2 = -63.5 - (+10.9) = -74.4\ \text{kJ mol}^{-1}

Exam-hard: composition of a carbonate mixture

$1.00\ \text{g}$ of a mixture of magnesium carbonate and calcium carbonate reacts with excess hydrochloric acid, producing $250\ \text{cm}^3$ of carbon dioxide at room conditions, collected in a gas syringe.

(a) Calculate the percentage by mass of magnesium carbonate in the mixture.

(b) Explain why a gas syringe is better than collecting over water here, and identify the measurement with the largest effect on the result.

( $M$ : $\ce{MgCO3}$ 84.3, $\ce{CaCO3}$ 100.1)

Solution

(a) $n(\ce{CO2}) = \dfrac{0.250}{24.0} = 0.01042\ \text{mol}$

Each carbonate gives one $\ce{CO2}$ : $\ce{MCO3 + 2HCl -> MCl2 + H2O + CO2}$ .

Let the mass of $\ce{MgCO3}$ be $m$ g, so the mass of $\ce{CaCO3}$ is $(1.00 - m)$ g:

\frac{m}{84.3} + \frac{1.00 - m}{100.1} = 0.01042

$0.011862m + 0.009990 - 0.009990m = 0.01042$

$0.001872m = 0.000427$ , so $m = 0.228\ \text{g}$

Percentage of $\ce{MgCO3}$ $= 22.8\%$ .

(b) Carbon dioxide is slightly soluble in water, so some would dissolve if collected over water, giving a volume that is too small. A gas syringe avoids contact with water.

The result is very sensitive to the gas volume: the answer comes from the small difference between the two molar masses, so a 1% error in volume changes the calculated composition by much more than 1%. The gas volume measurement is the most significant source of error.

Exam tip

In thermometric questions, show $q = mc\Delta T$ with numbers, convert J to kJ, divide by moles of the limiting reagent, and give a sign.
Typical evaluation: heat loss to the surroundings (improve with a lid, insulation, or by using the extrapolation method), heat capacity of the cup ignored, assuming the solution has the density and specific heat capacity of water, incomplete reaction of a solid.
In rate questions, state the variable that is changed, the one that is measured, and at least two that are controlled. Say that $\dfrac{1}{t}$ is used as a measure of rate.
The disappearing-cross end-point is a judgement: this gives a random error, which the syllabus accepts as an observer error. "Human error" alone is not accepted.
In gravimetric questions, the improvement examiners want is almost always heat to constant mass.

Practical skills

Summary of key controlled variables and improvements:

experiment	control	main source of error	improvement
enthalpy of solution or reaction	volume of liquid, mass of solid, starting temperature, cup and lid	heat loss to surroundings	lid, insulation, extrapolate cooling curve, use a thermometer reading to 0.1 °C
thiosulfate rates	total volume, acid volume and concentration, temperature, same cross and flask	judging when the cross disappears	use a light sensor / colorimeter; repeat and average
water of crystallisation	same balance throughout	incomplete dehydration; decomposition of the anhydrous salt if overheated	heat to constant mass; heat gently at first
gas volume	mass of solid, excess acid, temperature	gas lost before bung inserted; $\ce{CO2}$ dissolving in water	gas syringe; keep reagents separate until sealed

Summary

Thermometric: $q = mc\Delta T$ (mass of solution, $c = 4.18$ ), $\Delta H = -q/n$ ; extrapolate the cooling curve to the time of mixing; combine two measured values with Hess's law when needed.
Rates: time for a fixed amount of change; rate $\propto 1/t$ ; vary one factor, control the others.
Gravimetric: heat to constant mass; mass lost = water or gas; convert to moles and find the ratio.
Gas volume: collect over water or in a gas syringe; $n = V / 24.0$ ; beware solubility of $\ce{CO2}$ and gas lost before sealing.
Always state the main source of error and a realistic improvement.

Practice

Question

State why the mass used in $q = mc\Delta T$ for a neutralisation is the total volume of the two solutions in $\text{cm}^3$ .
$50.0\ \text{cm}^3$ of water in a polystyrene cup rises in temperature from $20.5\ ^\circ\text{C}$ to $27.0\ ^\circ\text{C}$ when $0.0100\ \text{mol}$ of a solid dissolves. Calculate the enthalpy change of solution.
Explain why a polystyrene cup is used rather than a glass beaker.
In the thiosulfate experiment, why is water added so that the total volume is always $50\ \text{cm}^3$ ?
A student heats $2.50\ \text{g}$ of hydrated copper(II) sulfate, $\ce{CuSO4.xH2O}$ , to constant mass. The residue weighs $1.60\ \text{g}$ . Calculate $x$ . ( $M$ : $\ce{CuSO4}$ 159.6, $\ce{H2O}$ 18.0)
The student in question 5 stops after one heating. Explain the effect on the value of $x$ .
$0.200\ \text{g}$ of impure calcium carbonate gives $44.0\ \text{cm}^3$ of carbon dioxide at room conditions with excess acid. Calculate the percentage purity. ( $M(\ce{CaCO3}) = 100.1$ )
In the thiosulfate experiment, the time taken with $20.0\ \text{cm}^3$ of thiosulfate is 52 s. Predict the time with $40.0\ \text{cm}^3$ (same total volume), and explain.
Suggest how you would use two thermometric experiments to find the enthalpy change for $\ce{MgCO3(s) -> MgO(s) + CO2(g)}$ , which cannot be measured directly.
In a neutralisation experiment, the temperature rise is $6.8\ \text{K}$ and the thermometer is calibrated at $1\ ^\circ\text{C}$ intervals. Calculate the maximum percentage error in $\Delta T$ , and suggest how it could be halved without changing the thermometer.

Answers

The energy released heats the whole mixed solution. Its density is assumed to be $1.00\ \text{g cm}^{-3}$ , so the mass in g equals the total volume in $\text{cm}^3$ .
$q = 50.0 \times 4.18 \times 6.5 = 1359\ \text{J} = 1.359\ \text{kJ}$ . $\Delta H = -1.359 / 0.0100 = -136\ \text{kJ mol}^{-1}$ (exothermic, temperature rose).
Polystyrene is a good thermal insulator and has a very low heat capacity, so less heat is lost to (or absorbed by) the container and the measured temperature change is closer to the true value.
So that the concentration of thiosulfate is directly proportional to the volume of thiosulfate used, and the depth of liquid (which affects how the cross is seen) is the same in every run.
Mass of water $= 2.50 - 1.60 = 0.90\ \text{g}$ , $n = 0.0500\ \text{mol}$ . $n(\ce{CuSO4}) = 1.60 / 159.6 = 0.01003\ \text{mol}$ . $x = 0.0500 / 0.01003 = 4.99 \approx 5$ .
Some water may remain, so the mass of residue is too high and the mass of water lost too low. The calculated value of $x$ would be too small.
$n(\ce{CO2}) = 0.0440 / 24.0 = 1.833 \times 10^{-3}\ \text{mol} = n(\ce{CaCO3})$ . Mass $= 0.1835\ \text{g}$ . Purity $= 0.1835 / 0.200 \times 100 = 91.8\%$ .
About 26 s. Doubling the volume (at constant total volume) doubles the thiosulfate concentration; rate is proportional to concentration (as the straight line through the origin shows), so the rate doubles and the time halves.
Measure the enthalpy change for $\ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l) + CO2(g)}$ ( $\Delta H_1$ ) and for $\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}$ ( $\Delta H_2$ ), each by adding a weighed solid to excess acid in a polystyrene cup. Then by Hess's law $\Delta H = \Delta H_1 - \Delta H_2$ .
Each reading has uncertainty $\pm 0.5\ ^\circ\text{C}$ ; $\Delta T$ uses two readings, so $\pm 1.0\ \text{K}$ . Percentage error $= 1.0 / 6.8 \times 100 = 14.7\%$ . Doubling the quantities of both reagents (concentrations, keeping the volumes) would roughly double $\Delta T$ , halving the percentage error.

Thermometric experiments

The principle

Method

The cooling correction

Using Hess's law with two experiments

Rate experiments

The thiosulfate–acid ("disappearing cross") experiment

Gravimetric experiments

Water of crystallisation

Gas-volume experiments

Worked examples

Practice

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