Scientific Notation - Answers
EXPERIMENTAL ERRORS
Some instruments are capable of a higher degree of accuracy than others, e.g., a micrometer
measurement should be more accurate than a meter stick measurement.
Definitions:
Accuracy is a statement of how close a measurement is to the true value. Accuracy is sometimes
quantified using the term ‘percent recovery’. For example, if after adding 10.0 grams of pure
standard to a beaker and running it through an analytical procedure, we measure the quantity
recovered to be 10.4 grams, we would report 104% recovery.
10.4
 measured value 
% recovery  
 100%  104% recovery
  100% 
10.0
 true value 
More commonly error is reported.
Error is a statement of the magnitude of the difference between a true/accepted value and a
measured value. Both ‘absolute’ error and ‘relative’ error are used.
Absolute Error: The absolute error is simply the actual error, i.e., the difference between the
measured and the true value. In the previous example, the absolute error was + 0.4 grams. Note
that error is always assigned a negative or positive value. If the measured value is greater than the
true value, then the error is positive
absolute error = (measured value – true value) = (10.4 – 10.0) = +0.4 grams
Relative Error: The relative error is often more meaningful than the absolute error. The relative
error can be reported in different ways such as % error and standard error (in statistics).
 0.4
 absolute error 
percent error  
 100%   4.0% error
  100% 
10.0
 true value 
Problem 1: Consider a case with a recovery of 49.0 grams on 50.0 grams of standard.
Calculate and compare the:

% recovery
(ans. = 98.0% recovery)

absolute error
(ans. = -1.0 g)

relative error (%)
(ans. = -2.0%)
Despite the larger absolute error in the second case (-1.0 g), it is more accurate than the first case
because the magnitude of its relative error is less (only 2.0 %).
Sources of Error:
Some of the more common causes of error include the following:

interpreting instrument scales incorrectly (inexcusable and not uncommon)

recording data incorrectly

misinterpreting the units of measurement

not ‘zeroing’ the apparatus

using improper procedures

wrong angle of vision when reading analog scales (e.g., meter needles), this is called ‘parallax
error’

incorrect mathematical calculations/conversions
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Scientific Notation - Answers
All the above are ‘human’ errors and can be eliminated. In SPC lingo these are described as
‘bias’, ‘assignable causes’ or ‘special’ causes of error/variation.
Errors can be reduced or eliminated by taking some precautions:

Take 2 or more readings of all measurements, preferably checked by another student.

Carefully follow an approved procedure.

Examine equipment. Is it in good working order (no leaks, properly zeroed)? Check the
instrument zero between readings and re-zero if necessary to correct for ‘drift’.

Compare experimental values with theoretical and ask yourself if the results are reasonable.

Graph your data to assess its consistency and variation. Look for patterns/trends in variation
suggesting a special cause of error. Apply line/curve fitting and determine the correlation
coefficient of the data to the regression equation.

Be conscientious.
Built-in Instrument Error:
Certain error is unavoidable due to instrument limitations such as aging, friction, ambient
temperature change, instrument scales with poor precision, etc. Such causes of errors in SPC
lingo are termed ‘common causes’ and are random in their direction. Instrumentation lingo calls
this ‘built-in-error’.
For instruments manufactured by reliable instrument companies the built-in-error is never greater
than one-half of the smallest unit marked on the instrument scale. The actual error may be less but
it should never be more.
Problem 2:
The smallest division on a meter stick is 1 mm and therefore the built-in-error or accuracy of the
meter stick is  0.5 mm.
If the meter stick is used to measure a length of 28.63 cm, calculate the % error built-in to this
measurement.
( 0.17%, 286.3 mm  0.5 mm)
If the meter stick is used to measure a length of 10.0 mm, calculate the % error built-in to this
measurement. ( 5%, 10 mm  0.5 mm)
Some manufacturers state the accuracy of their instruments. A voltmeter may be marked to state
its accuracy as 2% of its full-scale reading. If the full-scale reading is 250 V, the meter is actually
accurate to  5 V.
absolute error 
2%
 250 V  5V
100%
Problem 3:
Calculate the % error in voltmeter readings taken on the above instrument for:
a) a reading of 115 V
(ans. =  4.3%)
b) a reading of 10 V
(ans. =  50%!)
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Scientific Notation - Answers
Combining Errors in Calculations:
When several measurements are taken and combined in formulas, the resultant error will be a
combination of individual errors. Although errors may cancel, we must calculate the maximum
possible error assuming that errors are additive.
First convert absolute errors to % errors
The maximum possible error is determined by adding the % errors together
If a reading is raised to a power in a calculation, then the % error for that part is the power times
the % error. (Recall that applying an exponent is simply repeated multiplication).
Example:
An ammeter reads 2.54  0.05 A. An ohmmeter reads 90.6  0.5 . Calculate the absolute and
relative error of the voltage and power of the circuit.
V = IR = (2.54)  (90.6) = 230V
Relative error in voltage:
 0.05 0.5 


  100%  0.0197  0.00552   100%  230 V  2.52%
 2.54 90.6 
Absolute error in voltage:
 2.52% 

  230 V  230  5.80V
 100% 
P = I2R = (2.54)2  (90.6) = 585 W
Relative error in power:
  0.05   0.5  
 2

   100 %  0.0394  0.00519   100 %  585 W  4.49%
  2.54   90.6  
Absolute error in power:
 4.49% 
585 watts  
   26.3 W
 100% 
Summary for Combining Errors:
When adding or subtracting measurements, the total absolute error is the sum of individual
absolute errors.
When multiplying or dividing measurements, the total relative error is the sum of individual
relative errors.
When a measurement is raised to a power, the total relative error is equal to the relative error of
the value multiplied by its exponent (in the formula).
Problem 4:
In testing an electrical circuit, a resistance of 98.3 is measured by an ohmmeter graduated from
0 to 500  with an accuracy of 0.25%, full scale. A current of 3.15 A was measured with an
ammeter that had an accuracy of  0.02 A. Calculate the power of the circuit and determine the
absolute and relative error of the answer. (ans. = 975 W  2.5% or  25 W)
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Scientific Notation - Answers
Problem 5:
The quantity of heat absorbed by a substance is calculated using the formula Q = mCpT, where
Q = quantity of heat (Btu), m = mass of material (lb), Cp = heat capacity (Btulb-1F°-1), and T =
temperature difference (F°).
The heat capacity of the material is 1.032  0.002 Btulb-1F°-1.
An empty beaker was weighed at 0.328 lb. and then material was added giving a total weight of
2.639 lb. The balance used had an accuracy of  0.003 lb.
The thermometer used had an accuracy of  0.5 F°. The initial temp. was 35.6 °F and the final
temp. was 72.3 °F.
a) Determine the % error in the weighing (ans. =  0.26%)
b) Determine the % error in the temperature difference (ans. =  2.7%)
c) Calculate the quantity of heat, Q, transferred (ans. = 87.5 Btu)
d) Determine the absolute and relative error of Q (ans. =  2.8 Btu and  3.2%)
Problem 6:
The smallest division on a barometer scale is 1 mmHg. A student reads the atmospheric pressure
on the barometer as 751.2 mmHg.
a) State the absolute error of the pressure (ans. =  0.5 mmHg)
b) State the upper and lower values of the pressure based on this reading (751.7  750.7)
c) State the % error of the pressure reading (ans. =  0.07%)
Problem 7:
A student determines the density of a liquid organic unknown compound by transferring the
contents of a 10 mL class A volumetric pipet into a preweighed sample bottle on a top loading
balance. The mass of liquid was determined by difference.
The tolerance of the pipet is stated as 0.02 mL
The tolerance of the top loader balance is 0.005 g for each weighing
The mass of the empty sample bottle was 30.10 g
The mass of bottle + liquid was 39.20 g
Calculate:
a)
b)
c)
d)
e)
f)
the density of the liquid, showing the correct number of significant figures (0.910 g/mL)
the % error of the volume dispensed from the pipet. Assume no human error. ( 0.2 %)
the % error of the mass of the liquid weighed. (ans. =  0.11%)
the % error of the calculated density (ans. =  0.31%)
the absolute error of the calculated density (ans. =  0.0028 g/mL)
the range of possible densities, showing the correct number of sig. figs. (0.907  0.913)
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Scientific Notation - Answers
Problem 8:
One way to determine the MW of a volatile liquid is to vaporize a weighed sample and measure its
volume at a known temperature and pressure. The MW can be calculated using the ideal gas law
in the form shown below.
PV 
m
RT
M
P = pressure measured by a Hg barometer (115.3  0.05 kPa)
V = volume measured in a gas burette (50.1  0.1mL)
m = mass of sample weighed on analytical balance (0.1041  0.0001 g)
R = 8.3143435  0.0042% Jmol-1K-1
T = temperature of the sample (299.2  0.1 K)
a) determine the MW of the unknown (ans. = 44.8 g/mol)
b) determine the % error in the calculated MW (ans. =  0.38%)
c) determine the absolute error in the calculated MW. (ans. =  0.1689 g)
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