Question

Question: Determine the % relative error for the following quantities:

(a) 2.51 - 1.67

(b) (0.504 ± 0.002) ÷ (1251 ± 3 L/mole)

If there is no uncertainty given for a specific value, assume an absolute error of “1” in the last digit. For example, if there is no uncertainty provided for the value of “0.024”, you will assume the absolute error to be ±0.001 or 0.024±0.001 (i.e. an uncertainty of “1” in the last significant digit)

Please show formula(s) and work. Express answer with the correct number of sig figs.

Answer 1

So if x ± Δx be an experimental value, absolute error is Δx and relative error is (Δx/x) and % relative error is [(Δx/x) * 100]

In case of addition and subtraction of experimental values with errors in each individual values, the error is always added up in the final value.

For addition; x ± Δx = a ± Δa + b ± Δb we get, x= a+b and ± Δx = ± Δa ± Δb

For Subtraction; x ± Δx = (a ± Δa) - (b ± Δb) we get, x= a-b but ± Δx = ± Δa ± Δb

Thus in both cases the net error just adds up.

In case of multiplication and division of experimental values with errors in each individual values, the error propagation is a bit different. It can be calculated as follows

In case of multiplication we have (x ± Δx) = (a ± Δa) * (b ± Δb) or, x(1 ± Δx/x) = a(1 ± Δa/a) * b(1 ± Δb/b)

we get, x=ab and (1 ± Δx/x) = (1 ± Δa/a) * (1 ± Δb/b) = 1 ± Δa/a ± Δb/b ± (Δa/a*Δb/b) = 1 ± Δa/a ± Δb/b neglecting (Δa/a*Δb/b) as it is multiplication of two very small values giving even lesser value.

(1 ± Δx/x )= 1 ± Δa/a ± Δb/b or, ± Δx/x = ± Δa/a ± Δb/b or, Δx = x(± Δa/a ± Δb/b)

For division also this formula holds good.

So getting to question.

(a.) 2.51 - 1.67 = (2.51 ± 0.01) - (1.67 ± 0.01) errors assumed as per instruction

So (2.51 ± 0.01) - (1.67 ± 0.01) = (2.51 - 1.67) ± 0.01 ± 0.01 = 0.84‬ ± 0.02

Absolute value = 0.84

Absolute error = ± 0.02

Working formula = > x ± Δx = (a ± Δa) - (b ± Δb) we get, x= a-b and ± Δx = ± Δa ± Δb

% Relative error = [(0.02/0.84) * 100] = 2.38%

(b.) (0.504 ± 0.002) ÷ (1251 ± 3 )

Absolute value = 0.504 ÷ 1251 = 0.0004029

Absolute error = ± 0.0004 ( 0.002/0.504 + 3/1251) = ± 0.0000025

(0.504 ± 0.002) ÷ (1251 ± 3 ) = 0.0004029 + 0.0000025

Working formula = > (x ± Δx) = (a ± Δa) ÷ (b ± Δb) , we get x = a÷b and Δx = x(± Δa/a ± Δb/b)

% Relative error = [(0.0000025/0.0004029) * 100] = 0.62%