SIGNIFICANT FIGURES• Significant figures are the numbers in a measurement that represent the

certainty of the measurement, plus one number representing an estimate.

• COUNTING ZEROS AS SIGNIFICANT FIGURES• Leading zeros are never significant figures.• Buried zeros are always

significant figures.• Trailing zeros are generally

significant figures.

SIGNIFICANT FIGURES (continued)• The answer obtained by multiplication or division must contain

the same number of significant figures (SF) as the quantity with the fewest number of significant figures used in the calculation.

SF 2 SF 2SF 4

194625.19 5.4 325.4

SF 2 SF 2SF 4

96.0169.0 5.4 325.4

SIGNIFICANT FIGURES (continued)• The answer obtained by addition or subtraction must contain

the same number of places to the right of the decimal (prd) as the quantity in the calculation with the fewest number of places to the right of the decimal.

prd 1 prd 1prd 3

8.10825.10 5.5 325.5

prd 1 prd 1prd 3

2.0175.0 5.5 325.5

ROUNDING RULES FOR NUMBERS• If the first of the nonsignificant figures to be dropped from an

answer is 5 or greater, all the nonsignificant figures are dropped and the last remaining significant figure is increased by one.

• If the first of the nonsignificant figures to be dropped from an answer is less than 5, all nonsignificant figures are dropped and the last remaining significant figure is left unchanged.

Round 10.825 to 1place to the right of the decimal.⇒10.8

Round −0.175 to 1 place to the right of the decimal.⇒ −0.2

EXACT NUMBERS• Exact numbers are numbers that have no uncertainty (they

do not affect significant figures).• A number used as part of a defined relationship between

quantities is an exact number (e.g. 100 cm = 1 m).• A counting number obtained by counting individual objects

is an exact number (e.g. 1 dozen eggs = 12 eggs).• A reduced simple fraction is an exact number (e.g. 5/9 in

equation to convert ºF to ºC).

USING UNITS IN CALCULATIONS• The factor-unit method for solving numerical problems is a

four-step systematic approach to problem solving.

• Step 1: Write down the known or given quantity. Include both the numerical value and units of the quantity.

• Step 2: Leave some working space and set the known quantity equal to the units of the unknown quantity.

• Step 3: Multiply the known quantity by one or more factors, such that the units of the factor cancel the units of the known quantity and generate the units of the unknown quantity.

• Step 4: After you generate the desired units of the unknown quantity, do the necessary arithmetic to produce the final numerical answer.

SOURCES OF FACTORS• The factors used in the factor-unit method are fractions

derived from fixed relationships between quantities. These relationships can be definitions or experimentally measured quantities.

• An example of a definition that provides factors is the relationship between meters and centimeters: 1m = 100cm. This relationship yields two factors:

and

m 1

cm 100

cm 100

m 1

FACTOR UNIT METHOD EXAMPLES

• A length of rope is measured to be 1834 cm. How many meters is this?

• Solution: Write down the known quantity (1834 cm). Set the known quantity equal to the units of the unknown quantity (meters). Use the relationship between cm and m to write a factor (100 cm = 1 m), such that the units of the factor cancel the units of the known quantity (cm) and generate the units of the unknown quantity (m). Do the arithmetic to produce the final numerical answer.

m 34.18cm 100

m 1cm 1834

m cm 1834

PERCENTAGE • The word percentage means per one hundred. It is the

number of items in a group of 100 such items.

• PERCENTAGE CALCULATIONS• Percentages are calculated using the equation:

• In this equation, part represents the number of specific items included in the total number of items.

100whole

part%

EXAMPLE PERCENTAGE CALCULATION

• A student counts the money she has left until pay day and finds she has $36.48. Before payday, she has to pay an outstanding bill of $15.67. What percentage of her money must be used to pay the bill?

• Solution: Her total amount of money is $36.48, and the part is what she has to pay or $15.67. The percentage of her total is calculated as follows:

%96.4210048.36

67.15100

whole

part%

DENSITY• Density is the ratio of the mass of a sample of matter divided

by the volume of the same sample.

or

volume

massdensity

v

md

DENSITY CALCULATION EXAMPLE• A 20.00 mL sample of liquid is put into an empty beaker that

had a mass of 31.447 g. The beaker and contained liquid were weighed and had a mass of 55.891 g. Calculate the density of the liquid in g/mL.

• Solution: The mass of the liquid is the difference between the mass of the beaker with contained liquid, and the mass of the empty beaker or 55.891g -31.447 g = 24.444 g. The density of the liquid is calculated as follows:

mL

g222.1

mL 20.00

g 444.24

v

md