Identifying Unknown Salts with Calorimetry
Written by Charisa
Abstract
The purpose of this lab is to identify the unknown salts using the change in temperature to calculate heat released. With the given “q” values for all three identities of the unknowns, one can compare the calculated heat of the reaction to the given value to identify the salt. The unknowns chosen were unknown “C” for trials 1 and 2, and unknown “A” for trials 3 and 4. The initial temperature was taken before the salt was added, and after the salt had dissolved for 10 minutes in 30 second intervals.
Procedure
Bluedoor. Endothermic Enthalpy – Calorimetry. Bluedoor Labs. 29 Oct. 2013. www.bluedoorlabs.com/online_manual
Data Unknown “C”
Trial 1– mass = 5.061g Trial 2–Mass – 5.033g
Trial 1 Temp Change = -0.4 Degrees C Trial 2 Temp Change = -0.6 Degrees C ____________________________________________________________________________________ Unknown “A”
Trial 3–mass = 5.023g Trial 4–mass = 5.016
Trial 3 temp Change = 7.9 Degrees C Trial 4 Temp Change = 7.9 Degrees C
Calculations
To calculate the heat released by reaction: Unknown A
q = m*C*(Change in temp) where:
m = average mass of salt in g + mass of water
C = specific heat of water (4.186 J/g*°C)
(Change in temp) = average change in temperature
q = 85.047g* 4.186*(-.5) q = -178 J
Unknown C
q = 85.020g* (4.186 J/g*°C)*(7.9) q = 364J
To calculate Heat of Solution
Based on the calculated q values and the chart of Heat of Solution, it can be determined that Unknown A is LiCl
Based on the appearance, and left over values in the chart of the heat of solution, it can be determined that unknown C is NaCl
Editor’s note (October 21, 2026): The calculations above mix up the two unknowns and misstate what the signs mean. The student’s measurements and salt identifications are left unchanged; here is how to read them.
- The labels are swapped. The line labeled “Unknown A” actually uses Unknown C’s data: 85.047 g is Unknown C’s average salt mass (5.061 g and 5.033 g average to 5.047 g) plus 80 g of water (the post doesn’t state the water mass; 80 g is what remains after subtracting the salt, and it fits both calculation lines), and −0.5 °C is the average of Unknown C’s two temperature changes (−0.4 and −0.6 °C). The line labeled “Unknown C” uses Unknown A’s data (average salt mass 5.0195 g, so about 85.020 g with the water, and a change of +7.9 °C).
- The Unknown A heat is off by a lot. 85.020 g × 4.186 J/g·°C × 7.9 °C ≈ 2,810 J, not 364 J.
- A temperature drop means heat was absorbed, not released. q = m × c × ΔT gives the heat gained or lost by the water. The dissolving salt has the opposite sign: qreaction = −qsolution (OpenStax Chemistry 2e, 5.2). So Unknown C’s −178 J for the water means its dissolution absorbed about 178 J (endothermic). Unknown A’s temperature rise means its dissolution released about 2,810 J (exothermic, qreaction ≈ −2,810 J).
- Bond breaking does not release energy. The Conclusion below says the salt “released energy from the bonds breaking.” Pulling the ions of a solid apart always takes energy. Heat is given off when the separated ions are surrounded by water molecules (solvation). Whether a salt warms or cools the water depends on which of the two is larger (OpenStax Chemistry 2e, 11.1).
With those fixes, the identifications agree with published heats of solution. LiCl dissolves exothermically (about −37.0 kJ/mol), which fits Unknown A’s large temperature rise. NaCl dissolves slightly endothermically (about +3.9 kJ/mol), which fits Unknown C’s small drop (LibreTexts, Enthalpies of Solution, Table 9.5.1).
Conclusion
In conclusion, this experiment was a success in showing the different amounts of heat released based on the type of salt added to the water. As the salt crystals dissolved in the water, it released energy from the bonds breaking, which contributed to the temperature raising based on what type of salt was added. Table salt only changed the temperature of the water slightly, while the LiCl caused the temperature of the water to jump, then slowly decrease. The overall temperature change was 7.9 degrees, versus a cooldown of .4 degrees for the NaCl. As far as errors are concerned, the measurements may have been taken before the salt was completely dissolved, causing the temperature to change. To combat this error, one might need to employ a different method of stirring to make sure everything is dissolved.
How coffee-cup calorimetry identifies a salt
A calorimeter measures the heat a process gives off or takes in by watching the temperature of its surroundings, which here is the water in a polystyrene “coffee cup.” The heat gained or lost by the solution is
qsolution = m × c × ΔT
where m is the mass of the solution, c is its specific heat (taken to be that of water, about 4.184 J/g·°C), and ΔT is the final temperature minus the initial temperature. If no heat escapes, the heat of the dissolving process is equal and opposite: qreaction = −qsolution.
- Temperature rises: the process is exothermic. It releases heat into the water, and qreaction is negative.
- Temperature falls: the process is endothermic. It draws heat from the water, and qreaction is positive. Instant cold packs work this way: ammonium nitrate absorbs heat as it dissolves.
Coffee-cup calorimeters exchange some heat with the room, so they give less accurate values than sealed research calorimeters. Even so, the sign and rough size of the temperature change are usually enough to tell apart salts whose heats of solution differ as much as LiCl and NaCl. Source: OpenStax Chemistry 2e, 5.2 Calorimetry. The same heat-balance idea is used to find a metal’s specific heat in our Specific Heat of Metals lab experiment.