Question

*A standard sample containing 10.3% acetone yielded the following results: 10.2 9.9 10.3% acetone On analysis, another standard containing 0.40% acetone gave the following: 0.38 0.34 0.35% acetone a. Compare the precision of the two analyses in relative and absolute deviations from the mean. b. Compare the errors associated with the two analyses in absolute and relative terms.

Answer 1

Given data set for 10.3% acetone yield:

x1 = 10.2, x2= 9.9, x3= 10.3

We know the formula to mean as: x= (x1+x2+x3)/3

Substituting theses values,

Mean(x) = (10.2 + 9.9 +10.3)/3

So, Mean= 10.13

Now, calculating the standard deviation:

S= $\sqrt{}$ (x- x1)2+(x- x2)2+(x- x3)2/3

Substituting the given,

S = $\sqrt{}$ (10.13-10.20)2 + (10.13- 9.9)2 + (10.13 - 10.30)2 /3

S= $\sqrt{}$ 0.09/3 = 0.17

Now, Relative deviation is calculated as: (100*S)/x

Substituting the given vaues, Relative deviation = (100*0.17)/10.13

Relative deviation= 1.7%

Now, Absolute deviation= (|10.13-10.20| + |10.13- 9.9| + |10.13 - 10.30|)/3

= 0.5/3 = 0.2%

Data set for 0.40% acetone yield:

x1 = 0.38, x2= 0.34, x3= 0.35

We know the formula to mean as: x= (x1+x2+x3)/3

Substituting theses values,

Mean(x) = (0.38+0.34+0.35)/3

So, Mean= 0.36

Now, calculating the standard deviation:

S= $\sqrt{}$ (x- x1)2+(x- x2)2+(x- x3)2/3

Substituting the given,

S = $\sqrt{}$ (0.36-0.38)2 + (0.36- 0.34)2 + (0.36 -0.35)2 /3

S= 0.02

Now, Relative deviation is calculated as: (100*S)/x

Substituting the given vaues, Relative deviation = (100*0.02)/0.36

Relative deviation= 5.55%

Now, Absolute deviation= (|0.36-0.38| + |0.36- 0.34| + |0.36 -0.35|)/3

= 0.02%

(a) Now, we can say that while talking about relative deviation, 10.3% acetone is more precise.

In case of absolute deviation, 0.40% acetone is more precise.