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Chapter 1 Section 3: Measurements and Calculations in Chemistry

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Objectives: (1)To determine the number of significant figures in a measurement. (2) To perform calculations using significant figures. Chapter 1 Section 3: Measurements and Calculations in Chemistry. Accuracy and Precision. - PowerPoint PPT Presentation
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Objectives: (1)To determine the number of significant figures in a measurement. (2) To perform calculations using significant figures.
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Page 1: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Objectives: (1)To determine the number of

significant figures in a measurement.

(2) To perform calculations using significant figures.

Page 2: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Chapter 1

Section 3: Measurements and Calculations in Chemistry

Page 3: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Accuracy and Precision

• Accuracy is a description of how close a measurement is to the true value of the quantity measured.

• Precision is the exactness of a measurement.

Page 4: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Accuracy and Precision

NEITHER PRECISE NOR ACCURATE BOTH PRECISE AND ACCURATE

PRECISE BUT NOT ACCURATE

Page 5: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Significant Figures (sig figs)

• Sig Figs tell us how good the data we are using are.

Page 6: Chapter 1 Section 3: Measurements and Calculations in Chemistry
Page 7: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Rules for Determining Sig Figs

1. All non-zero digits are significant.

123 = 3 sig figs

8,476 = 4 sig figs

934,526 = 6 sig figs

Page 8: Chapter 1 Section 3: Measurements and Calculations in Chemistry

2. All zeros, between non-zero digits, are significant.

101 = 3 sig figs

2,000,503 = 7 sig figs

3,009 = 4 sig figs

Page 9: Chapter 1 Section 3: Measurements and Calculations in Chemistry

3. Reading left to right, all ending zeros, without a decimal point, are not significant.

6,000 = 1 sig fig

120 = 2 sig figs

407,000 = 3 sig figs

Page 10: Chapter 1 Section 3: Measurements and Calculations in Chemistry

4. Reading left to right, all ending zeros, with a decimal point, are significant.

400. = 3 sig figs

122.000 = 6 sig figs

5,020.0 = 5 sig figs

Page 11: Chapter 1 Section 3: Measurements and Calculations in Chemistry

5. Reading left to right, all starting zeros after a decimal point are not significant.

0.0012 = 2 sig figs

0.0609 = 3 sig figs

0.000000003 = 1 sig fig

Page 12: Chapter 1 Section 3: Measurements and Calculations in Chemistry

How Many Sig Figs?

a. 25 cm b. 305 cm c. 0.00123 in d. 400 g e. 400. m f. 0.94600 mL

Page 13: Chapter 1 Section 3: Measurements and Calculations in Chemistry

How Many Sig Figs?

a. 25 cm = 2b. 305 cm = 3c. 0.00123 in = 3d. 400 g = 1e. 400. m = 3f. 0.94600 mL = 5

Page 14: Chapter 1 Section 3: Measurements and Calculations in Chemistry
Page 15: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Answers

a.) 5,490,000 mb.) 0.0134793 mLc.) 31,950 cm2

d.) 192.67 m2

e.) 790 cmf.) 389,278,000 J

Page 16: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Calculations with Sig Figs:Addition and Subtraction

Your answer can only be as accurate as the least accurate measurement used in the calculation.

1. Complete the calculation.2. Determine the number of digits to the right of the

decimal point in each measurement used in the calculations.

3. Round answer so that it has that same number of digits to the right of the decimal as the measurement with the smallest number of digits to the right of the decimal.

Example: 3.95 g + 2.87 g + 213.6 g

Page 17: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Example

3.95 g + 2.87 g + 213.6 g = 220.42 g

3.95 g has 2 numbers after the decimal2.87 g has 2 numbers after the decimal point213.6 g has 1 number after the decimal point

Therefore, the answer must have 1 number after the decimal point.

Answer = 220.4 g

Page 18: Chapter 1 Section 3: Measurements and Calculations in Chemistry

ExampleSolve: 4.999 mL – 1.2 mL

4.999 mL – 1.2 mL = 3.799 mL

4.999 mL has 3 numbers after the decimal1.2 mL has 1 number after the decimal

Therefore, the answer must have 1 number after the decimal.

Answer = 3.8 mL

Page 19: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Practice

1.) 0.1273 mL – 0.008 mL

2.) 1.222 cm + 3.55 cm + 1.0 cm

3.) 3.2 g – 1.4 g + 0.33 g

4.) 34.09 L – 1.230 L

5.) 32.89 g + 14.21 g

Page 20: Chapter 1 Section 3: Measurements and Calculations in Chemistry

AnswersDON’T FORGET UNITS!

1.) 0.1273 mL – 0.008 mL = 0.1193 mL 0.119 mL

2.) 1.222 cm + 3.55 cm + 1.0 cm = 5.772 cm 5.8 cm

3.) 3.2 g – 1.4 g + 0.33 g = 2.13 g 2.1 g

4.) 34.09 L – 1.230 L = 32.860 L 32.86 L

5.) 32.89 g + 14.21 g = 47.1 g 47.10 g

Page 21: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Calculations with Sig FigsMultiplication and Division

Your answer can only be as accurate as the least accurate measurement used in the calculation.

1.Complete the calculation.2.Determine the number of sig figs in each

measurement used in the calculations.3.Round answer to the same number of sig figs as the

least accurate measurement.

Example: 651 cm x 75 cm

Page 22: Chapter 1 Section 3: Measurements and Calculations in Chemistry

ExampleSolve: 651 cm x 75 cm

651 cm x 75 cm = 48825 cm2

651 cm has 3 sig figs75 cm has 2 sig figs

Therefore, the answer must have 2 sig figs.

Answer = 49000 cm2

Page 23: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Practice

Perform the following calculations and express the answer in the correct units and number of sig figs.

1.) 7.835 kg / 2.5 L2.) 14.75 L / 1.20 s3.) 360 cm x 51 cm x 9.07 cm4.) 5.18 m x 0.77 m x 10.22 m 5.) 34.95 g / 11.169 cm3

Page 24: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Answers

Perform the following calculations and express the answer in the correct units and number of sig figs.

1.) 7.835 kg / 2.5 L = 3.1 kg/L2.) 14.75 L / 1.20 s = 12.34 s3.) 360 cm x 51 cm x 9.07 cm = 170,000 cm3

4.) 5.18 m x 0.77 m x 10.22 m = 41 m3

5.) 34.95 g / 11.169 cm3 = 3.129 g/cm3

Page 25: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Practice

1.) (12.4 cm x 7.943 cm) + 0.0064 cm2

2.) (246.83 g/26) – 1.349 g

Page 26: Chapter 1 Section 3: Measurements and Calculations in Chemistry

Answer

1.) (12.4 cm x 7.943 cm) + 0.0064 cm2

First multiply and second add. 12.4 cm x 7.943 cm = 98.4932 98.5 cm2

98.5 cm2 + 0.0064 cm2 = 98.50064 cm2 98.5 cm2

2.) (246.83 g/26) – 1.349 g First divide and second subtract. 246.83 g/26 = 9.4934615 g 9.5 g 9.5 g – 1.349 g = 8.151 g 8.2 g

Page 27: Chapter 1 Section 3: Measurements and Calculations in Chemistry

A rectangle measures 87.59 cm by 35.1 mm. Express its area with the proper number of sig figs in the specified unit:

in cm2

in mm2

in m2

Page 28: Chapter 1 Section 3: Measurements and Calculations in Chemistry

AnswerA rectangle measures 87.59 cm by 35.1 mm. Express

its area with the proper number of sig figs in the specified unit:

in cm2 : 87.59 cm x 3.51 cm = 307 cm2

in mm2 : 875.8 mm x 35.1 mm = 30700 mm2

in m2 : 0.8759 m x 0.0351 m = 0.0307 m2


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