04. Sig Figs in Calcs tutorial - Foothill
[Pages:6]04. Sig Figs in Calcs tutorial.doc
Introduction
In this tutorial you will learn the procedural steps necessary to determine the correct number of significant figures to keep in a calculation. When measured values are used in a calculation, the uncertainty in the measured value(s) will necessarily lead to some uncertainty in the results of a calculation. To find the uncertainty (error) in the result of a calculation you will use a set of rules that that are organized by the mathematical operation performed on the measured quantity. We will only consider three types of operations: 1) multiplication/division/power, 2) addition/subtraction and 3) logarithmic/exponential functions.
It is to your benefit to learn these rules and apply them regularly in the laboratory setting. Each and every laboratory instructor will be looking for your ability to handle significant figures in calculations. You will need to practice until the analysis becomes routine for you. Contrary to what you may have heard the rules are easy to follow and apply. Students get into trouble when they forget to APLY THE SIGNIFICANT FIGURE RULES FOR EVERY MATHEMATICAL OPRATION IN A SEQUENTIAL MANNER. Students often get lazy and just "guess" at how many significant figures the result should have instead of taking the time to apply the rules. Most students guess wrong. Don't be one of them!
The three rules
Several examples will be given for each rule.
1. Multiplication, Division or Power Functions ? round to the fewest significant figures
The result of the calculation will contain the same number of significant figures as there are in the measurement with the fewest significant figures. Find the error digit by counting significant figures from left to right in the result of the calculation. The last digit retained is the error (uncertain) digit.
Examples: In each example the number of significant figures in the result is indicated in parenthesis.
1. Let the mass of a penny be 3.1533 g with a volume of 0.44 mL. The density would
be: d = mass = 3.1533g = 7.2 g (2). Only two significant figures because the volume has
volume 0.44mL
mL
the fewest (2) significant figures.
2. Let the radius of a sphere be 2.11 cm. The volume would then be:
V = 4 !r3 = 4 ! (2.11cm)3 = 39.3cm3 (3). Three significant figures are kept since the radius
3
3
had 3 significant figures and cubing a number is the same as multiplying 3 times.
3. The measurements of a room are 8.3'x17.2'x13.7'. The volume of the room in cubit feet would be: 8.3' x17.2 ' x13.7 ' = 1956 ft 3(2) ! 2.0x103 ft 3 . Only two significant figures in the
result since the smallest dimension (8.3') had only two significant figures. This must be writen in scientific notation to drop the trailing zeros that are not significant.
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2. Addition or Subtraction ? round to the largest absolute error This is the rule most students forget to apply correctly. It will require you to assign a precision (absolute error) to every measurement used in the calculation. There is no easier way! For addition or subtraction the result is rounded to the digit that has the same magnitude as the greatest absolute error found in any one of the measurements that were added and/or subtracted. This means that the error digit of the result matches the magnitude of the greatest absolute error found in any one of the measurements. To apply this rule you must assign absolute errors to each measurement, compare them and retain the largest error. You then round the result to magnitude of this error. Practice! The last digit retained is the error (uncertain) digit.
Remember, when assigning absolute error, the absolute error is ?1 unit in the error digit.
1. Solve: 72.33 g ? 32.127 g = 40.203 g a. Assign absolute error to each number: 72.33 (?0.01) 32.127 (?0.001) b. The largest absolute error is retained (?0.01) c. Round the result to the second decimal place, the magnitude of the largest error. 40.20 g
2. Solve: 1345 mL ? 423.2 mL = 921.8 mL a. Assign absolute error to each number: 1345 (?1) 423.2 (?0.1) b. The largest absolute error is retained (?1) c. Round the result to the ones place, the magnitude of the largest error. 922 mL
3. When adding numbers, you may find the number of significant figures will increase. Converting 25.8 ?C (3 sig figs) to Kelvin gives: 25.8+273.15 = 299.0 K (4 sig figs)
4. Likewise, when subtracting numbers you may find the number of significant figures decreases 34.46 ml ? 27.88 mL (4 sig figs each) = 6.58 mL (3 sig figs).
5. When adding/subtracting numbers written in scientific notation, it is necessary to convert the numbers to the same power of ten before adding/subtracting to correctly compare absolute errors.
Solve: 8.63x10?3 g + 9.62x10?2 g = 0.10483 g a. Rewrite all numbers in the same power of ten. Usually the largest power present. In this case 10-2. 8.63x10?3g 0.863x10?2g
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b. Assign absolute errors to each number, RETAIN THE POWER OF TEN! 0.863x10?2g (?0.001x10-2) 9.62x10?2 (?0.01 x10-2)
c. The largest absolute error is retained (?0.01 x10-2)
d. Write the result in the same power of ten: 0.10483 g 10.483x10-2 g
e. Round the result to the second decimal place (x10-2), the magnitude of the largest error. 10.48x10-2 g
f. Rewrite in standard scientific notation or in decimal form: 10.48x10-2 g 1.048x10-1 g 0.1048 g
3. Logarithms and Exponentials:
1. Logarithms: When you take a logarithm of a number, the result is viewed as having two parts: the number to the right of the decimal that is called the mantissa and the number to the left of the decimal that is called the characteristic. The characteristic simply gives the power of 10 in the original number and is not considered when counting significant figures. Significant figures are only counted in the mantissa as follows: the mantissa of the logarithm will contain the same number of significant figures as the original number. On your calculator you have two log functions, base 10 is designated log(x) and base e (natural log) is designated ln(x). These rules apply to either function
a. log(3.000) = 0.4771
When you take the log of 3.000 the characteristic is zero and the mantissa (.4771) is given to four significant figures.
b. log(3.0) = 0.48
Here, the mantissa has two significant figures.
c. log(3) = 0.5
Here, the mantissa has only one significant figure.
d. log(2.78x106) = 6.444
The characteristic is 6 and the mantissa is given to three significant figures.
e. log(1.2301x10?4) = -3.91006 The characteristic is -3 and the mantissa is given to five significant figures.
2. Antilogarithms: Remember that the characteristic is not considered a significant figure. When taking an antilogarithm, retain the same number of significant figures in the result as there are in the mantissa of the original number. This means all your numbers should be written in decimal form to count the number of digits in the mantissa. The antilog functions on your calculator are 10x or ex.
a. 10(5.89) = 7.8x105
The mantissa .89 contains two significant figures, therefore the result of the antilog contains two significant figures.
b. 10(-0.0123) = 1.029
The mantissa .0123 contains four significant figures, (The leading zero after the decimal counts when taking
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an antilog.) therefore the result of the antilog contains four significant figures.
Problems with More than One Step
When doing multiple step calculations, YOU MUST determine the correct number of significant figures at each step, there are no short cuts! Carry extra digits through until the end and then round the final answer to the correct number of significant figures. It is helpful to underline the last digit that should be retained in each step when applying each significant figure rule. Keeping at least 1 extra significant figure in each intermediate answer in a multi-step calculation prevents the accumulation of rounding errors. Underline the error digit after each step to help you!
1. Example: Find the density of a metal object given the following data:
a. Mass of beaker: 36.215 g
b. Mass of beaker plus metal: 125.69 g
c. Water level in 100-mL graduated cylinder: 50.5 mL
d. Water level after submerging metal: 89.0 mL
i. Mass of metal: 125.69 g - 36.215 g = 89.475 g
ii. Volume of water: 89.0 mL ? 50.5 mL = 38.5 mL
iii. Density of metal: 89.475g = 2.32 g
38.5mL
mL
2. Example: Find the mass of carbon dioxide produced at 25 ?C, 1.044 atm pressure in the
following combustion reaction: C2H5OH(l) + 3 O2(g) 3 H2O(l) + 2 CO2(g). Given 22.74 L of air is consumed that is 20.2 % oxygen by volume in the combustion.
a. Volume of oxygen consumed: 22.74 L air * 0.202 L O2/1 L air = 4.593 L O2
b.
Moles of oxygen consumed = PV/RT =
1.044atm * 4.593L 0.08206L * atm
= 0.1961molO2
* 298K
mol * K
c.
Moles of CO2 produced: 0.1961 mol O2 *
2molCO2 3molO2
= 0.1307molCO2
d. Grams of CO2 produced: 0.1307 mol CO2 * 44.01g/mol = 5.75 g CO2
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Self-Test
Complete the following calculations. Round the final result to the correct number of significant digits. Check your answers by reviewing the next page.
1. It takes 10.5 s for a sprinter to run 100.00 m. Calculate the average speed of the sprinter in meters per second and mi/hr.
2. The mass of an empty 10-mL graduated cylinder is 25.442 g. After adding 8.5 mL of liquid the mass increases to 32.402 g. Calculate the density of the liquid in g/mL and kg/L.
3. The radius of an iodine atom is 140 pm. Find the volume of one iodine atom in pm3 and cm3.
4. pH is defined as ?log[H+] where [H+] is the molarity (M) of hydrogen ions in an aqueous solution. If the molarity of hydrogen ions is 1.32x10-3 M, find the pH.
5. A solution has a pH of 10.72. Find the concentration of hydrogen ions in solution.
6. A Cu/Al alloy contains 95.6% copper by mass. How many milligrams of aluminum are in a 2.7332 g sample of this alloy?
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Answers to Self-Test
1. It takes 10.5 s for a sprinter to run 100.00 m. Calculate the average speed of the sprinter in
meters per second and mi/hr.
100.00m = 9.52 m
10.5s
s
Convert
to
km/hr:
9.524
m s
!"#
1km 1000m
$%&
!"#
1mi 1.6093km
$%&
!"#
3600s 1hr
$%&
=
21.3 mi hr
(The extra digit (4) was kept to avoid round-off error in the m/s to mi/hr conversion.)
2. The mass of an empty 10-mL graduated cylinder is 25.442 g. After adding 8.5 mL of liquid
the mass increases to 32.402 g. Calculate the density of the liquid in g/mL and kg/L.
32.402g ! 25.442g = 6.960g = 0.81 g (2 sig figs)
8.5mL
8.5mL
mL
Convert
to
kg/L:
0.819
g ! 1kg $ ! 1mL $ mL "# 1000g %& "# 10'3 L %&
=
kg 0.82
L
3. The radius of an iodine atom is 140 pm. Find the volume of one iodine atom in pm3 and cm3.
V = 4 !r3 = 4 ! (140 pm)3 = 1.15x107 pm3 (3 sig figs based on the radius)
3
3
Convert to cm3: (The extra digit (...49) was kept to avoid round-off error.)
1.149 x10 7
pm 3
" 10!12 cm % 3 #$ 10!2 pm &'
=
1.149 x10 7
pm
3
" #$
10!30 cm 1 pm 3
3
% &'
=
1.15x10!23 cm3
4. pH is defined as ?log[H+] where [H+] is the molarity (M) of hydrogen ions in an aqueous solution. If the molarity of hydrogen ions is 1.32x10-3 M, find the pH. pH = -log[H+] = -log(1.32x10-3) = -(-2.879) = 2.879.
(3 digits in the mantissa since there were 3 sig figs in the molarity.)
5. A solution has a pH of 10.72. Find the concentration of hydrogen ions in solution. Rearranging and using the properties of logarithms: log[H+] = -pH. 10log[H + ] = [ H + ] = 10! pH = 10!10.72 = 1.9x10!11 M
(2 sig figs in the molarity since the mantissa of the pH has only 2 decimal places.)
6. A Cu/Al alloy contains 95.6% copper by mass. How many milligrams of aluminum are in a
1.0332 g sample of this alloy?
Al% in the alloy: (100%-95.6%) = 4.4%.
(The 100% is exact, so 1 decimal place for Al%.)
1.0332gAlloy
! "#
4.4 gAl 100gAlloy
$ %&
! 1mg $ "# 10'3 g %&
=
45mgAl
(2 sig figs since mass % Al is 2 sig figs.)
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