Combining errors in independent measurements.

            Many experiments depend on several measured quantities, each of which has an associated error. The rules for combining the errors are given below. These would not be examined, but they can be used in your everyday work and will give you a feel for the magnitudes involved and more importantly will train you in the quantitative assessment of error.

            In all the following cases, the two independent measurements A and B with errors DA and DB, are combined to give the quantity X which has the associated error DX.

Errors in sums or differences:

The straightforward combination of individual errors in sums or differences gives an overall error bound that is too pessimistic. It is unlikely that the individual errors will both be in the same direction, though not of course impossible.

If X = A + B or  X = A – B , then: DX =  √{(DA)² + (DB)²}

The expression is the same whether the quantities are added or subtracted. If more than two quantities are involved the expression is simply extended in a similar fashion, that is if X = A + B – C, then:

DX = √{(DA)² + (DB)² + (DC)²}.

The resulting error is always larger than any of the individual errors, but not as large as their sum.

If two values that are very similar are subtracted, the resulting error can be very large. Such procedures are best avoided if at all possible – see example 2.

Example 1: In an experiment to determine the enthalpy of neutralisation of sodium hydroxide by hydrochloric acid, the initial temperature was (19.2 ± 0.2) ^oC, and the final temperature (26.4 ± 0.2) ^oC. What is the temperature rise?

                                              DT = (T₂ – T₁) ± DT

                                                     =  (26.4 – 19.2) ^oC ± DT

                                                      = 7.2 ^oC ± DT.

The error DT is given by: DT  = √{(DT₁)² + (DT₂)²}

                                                = √(0.2 ^oC)² + (0.2 ^oC)²}

                                                = √{(0.08)( ^oC)²} = 0.28 ^oC.

Thus:                                        DT = (7.2 ± 0.28) ^oC.

This result would enable you to calculate the error in the value of DH_{neutralisation}.

Example 2: In an experiment to measure a quantity DQ, two values Q₁ = (90 ± 2) and Q₂ = (82 ± 2) were obtained.

 Q₁ – Q₂  =  (90 – 82) ± DQ;                    DQ  =  √{(DQ₁)²  +  (DQ₂)²}  =  √(2² + 2²)  =  √8  =  2.8.

 Thus  Q₁ – Q₂  =  8 ±  2.8.

 The error in each of the values of Q is about 2%; the error in their difference is 35%!

Errors in products or ratios:

If X =  AB or X = A/B, then  DX/X = √{(DA/A)² + (DB/B)²}

 Notice that unlike the case of sums or differences above we have to use fractional errors when dealing with products or ratios. If this is not done then the units go badly awry – the fractional errors are numbers, not physical quantities. The quantities that are multiplied and divided are physical quantities and will not necessarily (or even usually) have the same units. By using fractional errors the answer will have the correct units, because of X (and its units) being in the denominator of the left-hand side of the expression.

Example: Tin reacts with iodine on heating under reflux in a solvent of perchoroethylene to give an orange solid of formula SnI_x. In an experiment to find x, (3.00 ± 0.01) g of iodine was found to have reacted with (0.70 ± 0.01) g of tin. What is the value of x? (Molar masses: iodine atoms 126.9 g mol ^{– 1} ; tin 118.7 g mol ^{– 1}.)

 In the following 'amount' is used in its technical chemical sense of 'number of moles'.

 Amount of I atoms        = mass of I / molar mass of I

                                    = (3.00 ± 0.01)g/126.9 g mol^-1

 Amount of Sn atoms     =  mass of Sn / molar mass of Sn

                                    =  (0.70 ± 0.01)g / 118.7 g mol ^{– 1}

x =  4.00 ± Dx.

 The molar masses are constants, so it is necessary only to calculate the errors m_E in the mass of element E:

            Dx/x     =  √{(Dm_I / m_I )² + (Dm_Sn / m_Sn )²}

                                    =  √(0.01g / 3.00 g)² + (0.01 g / 0.70 g)²}

                                    = √(1.1 x 10 ^{– 5}  + 2.0 x 10 ^{– 4} )

                                    = √(2.11 x 10 ^{– 4} )         =  0.015.

Therefore:         Dx        =   0.015 x  =  0.015 x 4   =   0.06.

Thus:                x           =   4.00 ±  0.06.

The fractional error in the mass of the tin is some ten times that of the iodine; this makes the error in the weighing of the tin dominant. In fact the fractional error in the mass of the tin (0.014) is nearly the same as the fractional error in the value of x (0.015). If one fractional error is less than a third of another, the smaller one can usually be ignored.

Some final points:

·          Errors should be quoted to 2 s.f. at most. Errors are uncertainty limits and are statistical by their nature.

·          With sums or differences, ignore any error that is less than a third of the largest error.

·          With products or ratios,  ignore any fractional error that is less than one-third of the largest error.

·          Find out what the dominant errors are likely to be in an experiment, and concentrate on reducing those.

·          If differences of two similar quantities are involved take especial care to reduce the error in those quantities as much as possible.

·          If powers of quantities are to be taken, take especial care to reduce the error in those quantities as much as possible.