04. Sig Figs in Calcs tutorial

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 �C 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:

parenthesis.

In each example the number of significant figures in the result is indicated in

1. Let the mass of a penny be 3.1533 g with a volume of 0.44 mL. The density would

mass

3.1533g

g

be: d =

(2). Only two significant figures because the volume has

=

= 7.2

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:

4

4

V = ! r 3 = ! (2.11cm)3 = 39.3cm 3 (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) ! 2.0x10 3 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 �C 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 �C 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 �C 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 �C 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�C3 g + 9.62x10�C2 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�C3g �� 0.863x10�C2g

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b. Assign absolute errors to each number, RETAIN THE POWER OF TEN!

0.863x10�C2g (��0.001x10-2)

9.62x10�C2 (��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�C4) = -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.

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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 �C 50.5 mL = 38.5 mL

iii. Density of metal:

89.475g

g

= 2.32

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

= 0.1961molO2

0.08206L * atm

* 298K

mol * K

c. Moles of CO2 produced: 0.1961 mol O2 *

2molCO2

= 0.1307molCO2

3molO2

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 �Clog[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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