Difficulty: Intermediate | Prerequisites: Basic thermochemistry concepts, mole calculations, specific heat capacity.
This lab sits within the thermochemistry unit of general chemistry. You are measuring how much heat is absorbed or released when an ionic solid dissolves in water, then converting that measurement into a molar enthalpy of dissolution. The technique (calorimetry) is one you will use again for neutralisation reactions, Hess's law problems, and any lab where energy changes matter. You should already be comfortable converting between grams and moles and using the specific heat equation.
When an ionic solid dissolves in water, it either absorbs or releases heat. You measure the temperature change of the water (the surroundings), use q = mcΔT to calculate the heat involved, flip the sign to get the system's heat, then divide by moles of solute to find the molar enthalpy of dissolution (ΔH_rxn). A positive ΔH_rxn means the dissolution is endothermic (the solution cools down); a negative one means exothermic (the solution warms up).
Calorimetry
The experimental measurement of heat changes during a chemical or physical process, using a device called a calorimeter. In simple terms, this means you are using a thermometer and a known mass of water to figure out how much energy a reaction gave off or absorbed.
Enthalpy of dissolution (ΔH_diss or ΔH_rxn)
The enthalpy change when one mole of a solute dissolves in a solvent (usually water) at constant pressure. Reported in kJ/mol. Think of it as the energy price tag for pulling apart one mole of a solid and spreading its ions through water.
q_surroundings
The heat absorbed or released by the surroundings (in this lab, the water and the calorimeter cup). Calculated as q = mcΔT. In simple terms, this is what the thermometer actually tells you: did the water get warmer or cooler, and by how much energy?
q_system
The heat absorbed or released by the system (the dissolving solute). Equal in magnitude but opposite in sign to q_surroundings. Think of it as the reaction's side of the energy ledger. If the water lost heat, the reaction gained it, and vice versa.
Endothermic process
A process that absorbs heat from its surroundings. ΔH is positive. The solution temperature drops. In simple terms, the dissolving solid sucks heat out of the water, so the water gets colder.
Exothermic process
A process that releases heat to its surroundings. ΔH is negative. The solution temperature rises. In simple terms, the dissolving solid dumps heat into the water, so the water gets warmer.
Specific heat capacity (c)
The amount of energy required to raise the temperature of one gram of a substance by one degree Celsius. For water, c = 4.184 J/(g·°C). Think of it as how stubborn a substance is about changing temperature. Water is quite stubborn, which is why it works well as a calorimetry medium.
Molar mass (M)
The mass of one mole of a substance, in grams per mole (g/mol). Found by summing the atomic masses of every atom in the formula. Think of it as the conversion factor that takes you from grams on the balance to moles in the equation.
Every calculation in this lab follows the same five steps:
Find the total mass of the solution (m). Add the mass of the solid solute to the mass of the water. For example, 2.0221 g of NH₄NO₃ added to 50 g of water gives m = 52.0221 g.
Calculate q_surroundings using q = mcΔT. Multiply the total mass by the specific heat of water (4.184 J/g·°C) by the observed temperature change. The sign of ΔT matters: if the temperature drops, ΔT is negative, and q_surroundings is negative.
Determine q_system. Flip the sign of q_surroundings. Energy is conserved: q_system = −q_surroundings. If the surroundings lost 587 J, the system gained +587 J.
Convert grams of solute to moles. Divide the mass of solute by its molar mass.
Calculate ΔH_rxn in kJ/mol. Divide q_system (in joules) by moles of solute, then divide by 1000 to convert J to kJ.
Mass of solution: 2.0221 g + 50 g = 52.0221 g
q_surroundings: 52.0221 g × 4.184 J/g·°C × (−2.70 °C) = −587 J
q_system: +587 J (opposite sign)
Moles: 2.0221 g ÷ 80.043 g/mol = 0.025263 mol
ΔH_rxn: 587 J ÷ 0.025263 mol × (1 kJ / 1000 J) = +23.2 kJ/mol
The positive sign confirms an endothermic dissolution. The water temperature dropped by 2.70 °C because the solid absorbed heat as it dissolved.
Mass of solution: 2.0319 g + 50 g = 52.0319 g
q_surroundings: 52.0319 g × 4.184 J/g·°C × (−0.79 °C) = −172 J
q_system: +172 J
Moles: 2.0319 g ÷ 25.939 g/mol = 0.078334 mol
ΔH_rxn: 172 J ÷ 0.078334 mol × (1 kJ / 1000 J) = +2.20 kJ/mol
Also endothermic, but a much smaller ΔH_rxn than NH₄NO₃. The temperature barely moved (only −0.79 °C).
Mass of solution: 2.0098 g + 50 g = 52.0098 g
q_surroundings: 52.0098 g × 4.184 J/g·°C × (−0.47 °C) = −102 J
q_system: +102 J
Moles: 2.0098 g ÷ 78.07 g/mol = 0.02574 mol
ΔH_rxn: 102 J ÷ 0.02574 mol × (1 kJ / 1000 J) = +3.96 kJ/mol
Endothermic again. CaF₂ has a very small temperature change and relatively few moles, yielding a modest ΔH_rxn.
Solute | ΔT (°C) | q_system (J) | Moles | ΔH_rxn (kJ/mol) |
|---|---|---|---|---|
NH₄NO₃ | −2.70 | +587 | 0.02526 | +23.2 |
LiF | −0.79 | +172 | 0.07833 | +2.20 |
CaF₂ | −0.47 | +102 | 0.02574 | +3.96 |
All three are endothermic. NH₄NO₃ has by far the largest molar enthalpy of dissolution, which is why ammonium nitrate instant cold packs work: the dissolution absorbs a substantial amount of heat from its surroundings.
Temperature drops → ΔT is negative → q_surroundings is negative → q_system is positive → endothermic (ΔH > 0)
Temperature rises → ΔT is positive → q_surroundings is positive → q_system is negative → exothermic (ΔH < 0)
Heat absorbed or released by the surroundings:
q_{\text{surr}} = m \times c \times \Delta TWhere m = total mass of solution (g), c = specific heat capacity of water (4.184 J/g·°C), ΔT = final temperature minus initial temperature (°C).
Relationship between system and surroundings:
q_{\text{sys}} = -q_{\text{surr}}Moles of solute:
n = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}}Molar enthalpy of dissolution:
\Delta H_{\text{rxn}} = \frac{q_{\text{sys}}}{n} \times \frac{1 \text{ kJ}}{1000 \text{ J}}Note: the lab assumes the specific heat of the dilute solution equals that of pure water and that the calorimeter itself absorbs negligible heat. These are standard simplifying assumptions for a coffee-cup calorimeter.
Instant cold packs use the endothermic dissolution of NH₄NO₃ (or sometimes urea) to cool injuries without ice. When you squeeze the pack, the inner pouch breaks and the solid dissolves in water, pulling heat from the surroundings. The large positive ΔH_diss of ammonium nitrate is exactly what makes the pack cold enough to be useful.
On the exothermic side, dissolving calcium chloride (CaCl₂) in water releases enough heat to melt ice on roads and pavements. The same calorimetry principles from this lab apply to designing industrial processes where heat management during mixing or dissolving is critical, from pharmaceutical manufacturing to concrete curing.
Students often think a negative q_surroundings means the process is exothermic. It does not. A negative q_surroundings means the surroundings lost heat, which means the system gained heat, which means the process is endothermic.
Students sometimes use only the mass of the water (50 g) instead of the total mass of the solution (water + solute). The solute is part of the solution and absorbs or releases heat too. Always add both masses.
Forgetting to flip the sign when going from q_surroundings to q_system is one of the most common errors. The two quantities always have opposite signs.
Students occasionally report ΔH_rxn in J/mol instead of kJ/mol (or vice versa) without converting. Always check your units at the end of the calculation.
⚠️ You will almost certainly be asked to calculate ΔH_rxn from calorimetry data. Know the five-step method cold.
⚠️ Sign convention questions are a favourite. Be ready to explain why q_system and q_surroundings have opposite signs, and what the sign of ΔH tells you about the direction of heat flow.
⚠️ Expect a question comparing two or more solutes. You may need to explain why one has a larger ΔH_rxn than another, connecting it to lattice energy and hydration energy (the Born-Haber approach to dissolution).
⚠️ The assumption that c_solution = c_water (4.184 J/g·°C) is a simplification. An exam may ask you to state or justify this assumption, or to explain how the result would change if the actual specific heat were different.
True or False: If the temperature of the solution decreases during dissolution, the process is exothermic. (False: a temperature decrease means the system absorbed heat from the surroundings, so the process is endothermic.)
Fill in the blank: q_system = ______ q_surroundings. (negative, i.e. q_system = −q_surroundings)
True or False: When calculating the mass for q = mcΔT, you should use only the mass of the water. (False: use the total mass of water plus solute.)
Fill in the blank: To convert ΔH_rxn from J/mol to kJ/mol, divide by ______. (1000)
True or False: A dissolution with ΔH_rxn = +23.2 kJ/mol would cause the solution temperature to rise. (False: positive ΔH means endothermic, so the temperature drops.)
Q: A student dissolves 3.50 g of KNO₃ (molar mass 101.10 g/mol) in 50.0 g of water. The temperature drops by 1.85 °C. Calculate ΔH_rxn for the dissolution of KNO₃.
A: Total mass = 53.50 g. q_surr = 53.50 × 4.184 × (−1.85) = −414 J. q_sys = +414 J. Moles = 3.50 / 101.10 = 0.03462 mol. ΔH_rxn = 414 / 0.03462 / 1000 = +12.0 kJ/mol.
Q: In a dissolution experiment, q_surroundings is calculated as +350 J. Is this dissolution endothermic or exothermic? What is q_system?
A: q_system = −350 J. Because q_system is negative, the system released heat. The dissolution is exothermic.
Q: Explain why the mass used in q = mcΔT includes both the solute and the solvent, not just the solvent.
A: Once the solute dissolves, it becomes part of the solution. The entire solution (solute + solvent) is what changes temperature, so the entire mass must be used in the heat calculation.
Q: Two solutes are dissolved in identical amounts of water. Solute A causes a temperature drop of 3.0 °C; Solute B causes a drop of 0.5 °C. Can you conclude that Solute A has a larger ΔH_rxn per mole? Why or why not?
A: Not necessarily. ΔH_rxn depends on both q_system and the number of moles dissolved. If Solute A has many more moles than Solute B, the per-mole enthalpy could actually be smaller despite the larger temperature change. You need to complete the full calculation.
Q: In the lab data, NH₄NO₃ had a ΔH_rxn of +23.2 kJ/mol while CaF₂ had +3.96 kJ/mol. What does this tell you about the relative strength of lattice forces versus hydration forces in each compound?
A: A large positive ΔH_rxn means the energy needed to break apart the crystal lattice significantly exceeds the energy released when the ions are hydrated. For NH₄NO₃, the mismatch is large. For CaF₂, the lattice energy and hydration energy nearly balance, leaving only a small net energy absorption.
This material connects directly to Hess's law, which uses the additivity of enthalpy changes to calculate ΔH for reactions you cannot measure directly. The ΔH_diss values you calculate here are exactly the kind of data Hess's law problems use as inputs.
Dissolution enthalpy can also be broken into two competing steps using a Born-Haber-style cycle: lattice energy (breaking the solid apart, always endothermic) and hydration energy (surrounding ions with water, always exothermic). Whether dissolution is net endothermic or exothermic depends on which term wins. This connects to ionic bonding, crystal structure, and ion-dipole interactions.
The calorimetry technique itself reappears in acid-base neutralisation experiments and in bomb calorimetry (constant volume rather than constant pressure), which is used to measure the energy content of fuels and food.
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