Simple rules of thumb for calculations
Level 0 (green)- this is basic material that you have probably encountered already, although the approach may be slightly different. No prior knowledge is assumed.
Since a good deal of quantitative chemistry uses measured quantities in calculations and measured quantities are only defined to a finite precision it is necessary to understand how to handle errors in a calculation.
This topic comes in two parts, we start with some very simple rules of thumb for simple calculations. This is mainly intended for quick estimation. It is generally safe in the sense that it will not normally underestimate the error in the solution, but it is not actually the optimal way of following errors through, which will be found in the tutorial on Propagation of Errors.
The rule of thumb is add the absolute errors.
For example if you subtract two quantities, A and B with estimated errors eA and eB, the result will be A – B with an estimated absolute error of eA + eB.
The rule of thumb is add the relative errors.
For example if you divide two quantities, A and B with estimated errors eA and eB, the relative errors will be rA = eA / A and rB = eB / B. The result will be A / B with an estimated relative error of rA + rB.
So that the absolute error will be (A / B)x( rA + rB)
The reasons for these rules are fairly obvious: they represent worst case scenarios where the errors reinforce each other. Of course in reality it is just as likely that the errors will partially cancel, which is why a more sophisticated method is necessary.
Example - calculate the density of water
The density of water is measured by taking a known volume from a burette and weighing it. The volume was 5.00 cm3 with an estimated error of 0.05 cm3 (1%) and the mass was 4.98 g with an estimated error of 0.15 g (3%). The density is given by d = m/V = 0.996 g cm-3 with a relative error of 1% + 3% = 4%, corresponding to an absolute error of 0.04 g cm-3.
Example - measure the enthalpy of solution of NH4Cl.
This is a typical undergraduate first year practical. The strategy is to work out the heat capacity of the apparatus using a reaction whose enthalpy change is known (NaOH + HCl), and then to dissolve a known amount of NH4Cl in the same apparatus and use the temperature change and the known heat capacity to work out the enthalpy of solution.
The calculation proceeds as follows:
- 50.00 ± 0.05 mL of HCl whose concentration is 1.00 ± 0.01 M was mixed with 50.00 ± 0.05 mL of NaOH whose concentration is 1.00 ± 0.01 M.
- The amount of each reactant is calculated from n = cV, and so relative errors are needed. The relative error in each volume is 0.1% and the relative error in each concentration is 1%. Hence the amount of the limiting reagent is 50.0 mmol with a relative error of 1.1%.
- The solutions are mixed and the temperature changes by DT = 6.8 ± 0.1 K (1.5%). The enthalpy of reaction is known to be -59.6 kJ mol-1 with an estimated error of 0.05 kJ mol-1 (0.1%).
- The heat capacity of the system is therefore -q / DT = -n DH / DT = 438 J K-1 with estimated error of 1.1% (n) + 0.1% (DH) + 1.5% (DT) = 2.7%, corresponding to 438 J K-1 ± 12 J K-1.
- Now 5.25 g ± 0.01 g (0.2%) of NH4Cl is weighed out, corresponding to 0.09815 mol with a relative error of 0.2%. (The error of the relative formula mass is too small to bother about).
- The calorimeter is filled with 100.0 mL ± 0.1 mL of water (0.1%). The calorimeter is the same as before, but the volume of water may not be exactly the same. The difference in the volumes will be 0.0 mL ± 0.2 mL. Given the density (0.998234 g / mL) and specific heat capacity of water (4.1819 J K-1 g-1), in both of which the error is too small to affect anything, this translates to an additional error in the heat capacity of the system of 0.8 J K-1. Hence the heat capacity of the system is 438 J K-1 ± 13 J K-1 (3.0%).
- The temperature change is -4.1 K ± 0.1 K (2.5%), and so the heat change is q = –c DT = +1.80 kJ with a relative error of 3% (from c) + 2.5% (from DT) = 5.5%.
- Finally we use the amount of reactant to work out the molar enthalpy change: DH = q/n = 18.3 kJ mol-1, with a relative error of 5.5% (from q) + 0.2% (from n) = 5.7%, i.e. DH = 18.3 kJ mol-1 ± 1.0 kJ mol-1.