Significant Significant FiguresFigures

Rule #1Rule #1

All non-zero digit numbers All non-zero digit numbers AREARE significantsignificant

Ex: 8.526 = 4 SD’sEx: 8.526 = 4 SD’s

Ex: 2345.785 = 7 SD’sEx: 2345.785 = 7 SD’s

Rule #2Rule #2 Zeros between non-zero digit Zeros between non-zero digit

numbers numbers AREARE significant significant

Ex: 90,204 = 5 SD’sEx: 90,204 = 5 SD’s

Ex: 70.08 = 4 SD’sEx: 70.08 = 4 SD’s

Rule #3Rule #3

All zeros to the left of the first All zeros to the left of the first non-zero number are non-zero number are NOTNOT significant.significant.

Ex: 0.0024 = 2 SD’sEx: 0.0024 = 2 SD’s

Ex: 0.1 = 1 SDEx: 0.1 = 1 SD

Rule #4Rule #4 When a number ends in zeros that When a number ends in zeros that

are to the right of a decimal point, are to the right of a decimal point, they they AREARE significant. significant.

Ex: 0.200 = 3 SD’sEx: 0.200 = 3 SD’s

Ex:3.0 = 2 SD’sEx:3.0 = 2 SD’s

Ex: 0.050 = 2 SD’s (use Rule #3 also)Ex: 0.050 = 2 SD’s (use Rule #3 also)

Rule #5Rule #5 When a number ends in zeros that When a number ends in zeros that

are not to the right of a decimal are not to the right of a decimal point, the numbers are not point, the numbers are not necessarily significant.necessarily significant.

Ex: 9,000,000,000 = 1 SDEx: 9,000,000,000 = 1 SD

Ex: 50,100 = 3 SD’sEx: 50,100 = 3 SD’s

Ex: 2,500 = 2 SD’sEx: 2,500 = 2 SD’s

Practice ProblemsPractice Problems 5.432 =5.432 = 0.189 =0.189 = 2,873.0 =2,873.0 = 0.000235 =0.000235 = 2,500 =2,500 = 1.04 X 101.04 X 101414 = = 0.002300 =0.002300 = 7.500 X 107.500 X 1088 = = 1.000 X 101.000 X 1033 = = 48.625403 =48.625403 =

4 SD’s3 SD’s

5 SD’s3 SD’s

2 SD’s3 SD’s

4 SD’s4 SD’s4 SD’s8 SD’s

Sig Fig Quick Quiz Sig Fig Quick Quiz 8.89 =8.89 = 0.16 =0.16 = 2,500.0 =2,500.0 = 0.00067345 =0.00067345 = 36,700 =36,700 = 6.093 X 106.093 X 101919 = = 0.0060 =0.0060 = 7.9800 X 107.9800 X 102424 = = 148.000 X 10148.000 X 1033 = = 408.6025403 =408.6025403 =

3 SD’s2 SD’s

5 SD’s5 SD’s

3 SD’s4 SD’s

2 SD’s5 SD’s6 SD’s

10 SD’s

Sig Figs in MathSig Figs in Math Multiplying and DividingMultiplying and Dividing

12.257m12.257m

X 1.162mX 1.162m

14.2426234m14.2426234m22

(5 SD)

(4 SD)

Which number is smaller 5 or 4?

4 is smaller than 5 therefore you only want 4 SD in your answer!

14.24m2 (4 SD’s)

Sig Figs in MathSig Figs in Math

Adding and SubtractingAdding and Subtracting

3.95 g3.95 g

2.879 g2.879 g

+ 213.6 g+ 213.6 g

220.429 g220.429 g

Look AFTER the DECIMAL ONLY

(2 SDAD)

(3 SDAD)

(1SDAD)

Same as multiplication and division. Choose the smallest number for the answer.

220.4 g (1 SDAD)

Sig Fig Math ProblemsSig Fig Math Problems 0.1273 mL – 0.000008 mL=0.1273 mL – 0.000008 mL=

(12.4 cm X 7.943 cm) + 0.0064 cm(12.4 cm X 7.943 cm) + 0.0064 cm22

(246.83 g / 26 L) – 1.349 g/L=(246.83 g / 26 L) – 1.349 g/L=

(4 SDAD) (1 SDAD)0.127292 =0.1 mL

(Only want 1 SDAD)

(3 SD) (4 SD)

98.4932 cm2 =Only want 3 SD

98.5 cm2+ 0.0064 cm2 =(1 SDAD) (2 SDAD)

98.5064 =(Only want 1 SDAD)

98.5 cm2

(5 SD) (2 SD)

9.4934615 =Only want 2 SD

9.5 g/L- 1.349 g/L =8.151 g/L =(1 SDAD) (3 SDAD) (Only want 1 SDAD)

8.2 g/L

Sig Fig Math Practice ProblemsSig Fig Math Practice Problems

0.8102m X 3.44m=0.8102m X 3.44m=

94.20g / 3.16722mL=94.20g / 3.16722mL=

32.897g + 14.2g=32.897g + 14.2g=

34.09L – 1.230L= 34.09L – 1.230L=

2.79m2

(4 SD) (3 SD)

(4 SD) (6 SD)29.74g/mL

(3 SDAD) (1 SDAD)47.1g

(2 SDAD) (3 SDAD)32.86L

Sig Fig Sample Problem A (Pg. 59)Sig Fig Sample Problem A (Pg. 59) A student heats 23.62 g of a solid and observes that A student heats 23.62 g of a solid and observes that

its temperature increases from 21.6 its temperature increases from 21.6 °C to 36.79 °C. °C to 36.79 °C. Calculate the temperature increase per gram of Calculate the temperature increase per gram of solid.solid.

1.1. Gather InformationGather InformationMass of Solid = 23.62 gMass of Solid = 23.62 gInitial Temp = 21.6 °C Initial Temp = 21.6 °C Final Temp = 36.79 °C Final Temp = 36.79 °C

2.2. Plan WorkPlan WorkA. Calculate temp. increaseA. Calculate temp. increase

(Final temp – Initial temp) = temp increase(Final temp – Initial temp) = temp increase

B. Per gram of solidB. Per gram of solid temp increasetemp increase = temp increase per g of solid = temp increase per g of solid gramgram

3.3. CalculateCalculateA. (36.79 °C – 21.6 °C ) = 15.19 °C =A. (36.79 °C – 21.6 °C ) = 15.19 °C =

B. B. 15.2 °C15.2 °C = 0.6435224 = = 0.6435224 = 23.62 g 23.62 g

15.2 °C

0.644 °C/g

Sig Fig Problem PracticeSig Fig Problem Practice

Do Pg 59 Practice Problems 1 & 2Do Pg 59 Practice Problems 1 & 2

Practice Problem #1 (Pg. Practice Problem #1 (Pg. 59)59) Perform the following calculations, and express the answers Perform the following calculations, and express the answers

with the correct number of significant figures.with the correct number of significant figures.

A.A. 0.1273 mL – 0.000008 mL = 0.1273 mL – 0.000008 mL =

B.B. (12.4 cm X 7.943 cm) + 0.0064 cm(12.4 cm X 7.943 cm) + 0.0064 cm22 = =

C.C. (246.83 g / 26) – 1.349 g =(246.83 g / 26) – 1.349 g =

0.127292 = 0.1 mL

98.4932 cm2 = 98.5 cm2 + 0.0064 cm2 = 98.5064 =98.5 cm2

9.4934615 g = 9.5 g – 1.349 g = 8.151 g = 8.2 g

4 SDAD 1 SDAD

3 SD 4 SD

1 SDAD 2 SDAD

5 SD 2 SD

1 SDAD 3 SDAD

Practice Problem #2 (Pg. 59)Practice Problem #2 (Pg. 59) A student measures the mass of a beaker filled with corn oil to A student measures the mass of a beaker filled with corn oil to

be 215.6 g. The mass of the beaker is 110.4 g. Calculate the be 215.6 g. The mass of the beaker is 110.4 g. Calculate the density of the corn oil if its volume is 114 cmdensity of the corn oil if its volume is 114 cm33..

1.1. Gather InfoGather Info

A. Mass of beaker + oil = 215.6 gA. Mass of beaker + oil = 215.6 g

B. Mass of empty beaker = 110.4 gB. Mass of empty beaker = 110.4 g

C. Volume of corn oil = 114 cmC. Volume of corn oil = 114 cm33

2.2. Plan WorkPlan Work

Mass of oil = A – B = XMass of oil = A – B = X

Density of oil = Density of oil = X__ X__

CC

3.3. CalculateCalculate

Mass of oil = 215.6 g – 110.4 g =Mass of oil = 215.6 g – 110.4 g =

Density of oil = Density of oil = 105.2 g__ 105.2 g__ = =

114 cm114 cm33

0.923 g/cm34 SD

3 SD

105.2 g1 SDAD 1 SDAD