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Chapter 1 Section 3: Measurements and Calculations in Chemistry 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. Chapter 1 Section 3: Measurements and Calculations in Chemistry. Accuracy and Precision.

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Chapter 1 Section 3: Measurements and Calculations in Chemistry

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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.


Chapter 1 section 3 measurements and calculations in chemistry

Chapter 1Section 3: Measurements and Calculations in Chemistry


Accuracy and precision

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.


Accuracy and precision1

Accuracy and Precision

BOTH PRECISE AND ACCURATE

NEITHER PRECISE NOR ACCURATE

PRECISE BUT NOT ACCURATE


Significant figures sig figs

Significant Figures (sig figs)

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


Rules for determining sig figs

Rules for Determining Sig Figs

  • All non-zero digits are significant.

    123= 3 sig figs

    8,476 = 4 sig figs

    934,526 = 6 sig figs


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


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


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


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


How many sig figs

How Many Sig Figs?

  • 25 cm

  • 305 cm

  • 0.00123 in

  • 400 g

  • 400. m

  • 0.94600 mL


How many sig figs1

How Many Sig Figs?

  • 25 cm = 2

  • 305 cm = 3

  • 0.00123 in = 3

  • 400 g = 1

  • 400. m = 3

  • 0.94600 mL = 5


Answers

Answers

a.) 5,490,000 m

b.) 0.0134793 mL

c.) 31,950 cm2

d.) 192.67 m2

e.) 790 cm

f.) 389,278,000 J


Calculations with sig figs addition and subtraction

Calculations with Sig Figs:Addition and Subtraction

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

  • Complete the calculation.

  • Determine the number of digits to the right of the decimal point in each measurement used in the calculations.

  • 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


Example

Example

3.95 g + 2.87 g + 213.6 g = 220.42 g

3.95 g has 2 numbers after the decimal

2.87 g has 2 numbers after the decimal point

213.6 g has 1 number after the decimal point

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

Answer = 220.4 g


Example1

Example

Solve: 4.999 mL – 1.2 mL

4.999 mL – 1.2 mL = 3.799 mL

4.999 mL has 3 numbers after the decimal

1.2 mL has 1 number after the decimal

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

Answer = 3.8 mL


Practice

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


Answers don t forget units

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


Calculations with sig figs multiplication and division

Calculations with Sig FigsMultiplication and Division

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

  • Complete the calculation.

  • Determine the number of sig figs in each measurement used in the calculations.

  • Round answer to the same number of sig figs as the least accurate measurement.

Example: 651 cm x 75 cm


Example2

Example

Solve: 651 cm x 75 cm

651 cm x 75 cm = 48825 cm2

651 cm has 3 sig figs

75 cm has 2 sig figs

Therefore, the answer must have 2 sig figs.

Answer = 49000 cm2


Practice1

Practice

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

1.) 7.835 kg / 2.5 L

2.) 14.75 L / 1.20 s

3.) 360 cm x 51 cm x 9.07 cm

4.) 5.18 m x 0.77 m x 10.22 m

5.) 34.95 g / 11.169 cm3


Answers1

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/L

2.) 14.75 L / 1.20 s = 12.34 s

3.) 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


Practice2

Practice

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

2.) (246.83 g/26) – 1.349 g


Answer

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


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


Chapter 1 section 3 measurements and calculations in chemistry

Answer

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 : 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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