Calorimetry answer 1

In an experiment to measure the enthalpy of solution of sodium carbonate-10-water, 8.324 g of Na₂CO₃.10H₂O was dissolved in 100 cm³ of distilled water in a plastic cup. There was a temperature change from 18.1 ºC to 13.9 ºC. Calculate the enthalpy change of solution. What errors occur in this experiment?

Specific heat capacity of water = 4.18 J g^-1 K^-1

We have only one chemical dissolving in water here, so there is no concern as to which chemical is in excess.

Na₂CO₃.10H₂O = 286

\ number of moles of Na₂CO₃.10H₂O = 8.324/286 = 0.0291 mol

Mass of water = 100 g

Temperature change = 13.9 - 18.1 = - 4.2 ºC

\ heat absorbed = 4.18 × 100 × 4.2 = 1756 J = 1.76 kJ

\ enthalpy of solution = + 1.76/0.0291 = + 60 kJ mol^-1

A couple of points to note here. The positive sign, indicating the endothermic reaction, must be included with your answer. The answer has been given to two sig. figs. to match with the two sig. figs. of the temperature change measurement.

There are a couple of measurement errors here. The measuring cylinder should be accurate to around 1 cm³, giving a percentage error of 1/100 × 100 = ± 1%. The temperature change measurement is accurate to 0.1 ºC, giving an error of 0.1/4.2 × 100 = ± 2.4%. If we were being precise, we could also mention that we have given the molar mass to the nearest whole number, actually Na₂CO₃.10H₂O = 286.1

Another error is heat gain. Be careful here, it is very easy with calorimetry experiments to automatically talk about heat loss (as most reactions are exothermic), but this reaction is endothermic. The temperature falls below that of the surroundings, so heat is gained by the reaction mixture, and so the temperature fall is less than that expected.

We made the usual assumption that the solid, the cup and the thermometer had no heat capacity. In other words, all the heat was lost from the water. In this case we did not take account of the water present in the hydrated salt. 0.0291 moles of Na₂CO₃.10H₂O contains 0.291 moles of water, with a mass of 0.291 × 18 = 5.2 grams. This is really an avoidable error as we could have used 105.2 g of water in our calculation. This gives a final answer of + 64 kJ mol^-1 (a reasonable difference).

The sodium carbonate may not be entirely pure, meaning less than the weighed amount was actually used. If your background reading is very good, you may be aware that sodium carbonate decahydrate is efflorescent. This will affect the result as not all our carbonate is likely to be fully hydrated.

We should have checked for complete solution, so incomplete reaction should not be an error here.

If you're still uncertain about error analysis check out the link.