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Significant Figures

Significant Figures. Rule #1. All non-zero digit numbers ARE significant Ex: 8.526 = 4 SD’s Ex: 2345.785 = 7 SD’s. Rule #2. Zeros between non-zero digit numbers ARE significant Ex: 90,204 = 5 SD’s Ex: 70.08 = 4 SD’s. Rule #3.

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Significant Figures

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  1. Significant Figures

  2. Rule #1 • All non-zero digit numbers ARE significant Ex: 8.526 = 4 SD’s Ex: 2345.785 = 7 SD’s

  3. Rule #2 • Zeros between non-zero digit numbers ARE significant Ex: 90,204 = 5 SD’s Ex: 70.08 = 4 SD’s

  4. Rule #3 • All zeros to the left of the first non-zero number are NOT significant. Ex: 0.0024 = 2 SD’s Ex: 0.1 = 1 SD

  5. Rule #4 • When a number ends in zeros that are to the right of a decimal point, they ARE significant. Ex: 0.200 = 3 SD’s Ex:3.0 = 2 SD’s Ex: 0.050 = 2 SD’s (use Rule #3 also)

  6. Rule #5 • When a number ends in zeros that are not to the right of a decimal point, the numbers are not necessarily significant. Ex: 9,000,000,000 = 1 SD Ex: 50,100 = 3 SD’s Ex: 2,500 = 2 SD’s

  7. Practice Problems 4 SD’s 3 SD’s 5 SD’s • 5.432 = • 0.189 = • 2,873.0 = • 0.000235 = • 2,500 = • 1.04 X 1014 = • 0.002300 = • 7.500 X 108 = • 1.000 X 103 = • 48.625403 = 3 SD’s 2 SD’s 3 SD’s 4 SD’s 4 SD’s 4 SD’s 8 SD’s

  8. Sig Fig Quick Quiz 3 SD’s 2 SD’s 5 SD’s • 8.89 = • 0.16 = • 2,500.0 = • 0.00067345 = • 36,700 = • 6.093 X 1019 = • 0.0060 = • 7.9800 X 1024 = • 148.000 X 103 = • 408.6025403 = 5 SD’s 3 SD’s 4 SD’s 2 SD’s 5 SD’s 6 SD’s 10 SD’s

  9. Sig Figs in Math • Multiplying and Dividing 12.257m X 1.162m 14.2426234m2 Which number is smaller 5 or 4? 4 is smaller than 5 therefore you only want 4 SD in your answer! (5 SD) (4 SD) 14.24m2 (4 SD’s)

  10. Sig Figs in Math Look AFTER the DECIMAL ONLY • Adding and Subtracting 3.95 g 2.879 g + 213.6 g 220.429 g Same as multiplication and division. Choose the smallest number for the answer. (2 SDAD) (3 SDAD) (1SDAD) 220.4 g (1 SDAD)

  11. Sig Fig Math Problems 0.1 mL 0.127292 = (Only want 1 SDAD) (4 SDAD) (1 SDAD) • 0.1273 mL – 0.000008 mL= • (12.4 cm X 7.943 cm) + 0.0064 cm2 • (246.83 g / 26 L) – 1.349 g/L= (3 SD) (4 SD) 98.5064 = 98.5 cm2 + 0.0064 cm2 = 98.4932 cm2 = 98.5 cm2 (2 SDAD) Only want 3 SD (1 SDAD) (Only want 1 SDAD) (5 SD) (2 SD) 9.5 g/L - 1.349 g/L = 8.151 g/L = 8.2 g/L 9.4934615 = Only want 2 SD (1 SDAD) (3 SDAD) (Only want 1 SDAD)

  12. Sig Fig Math Practice Problems 2.79m2 (4 SD) (3 SD) • 0.8102m X 3.44m= • 94.20g / 3.16722mL= • 32.897g + 14.2g= • 34.09L – 1.230L= 29.74g/mL (6 SD) (4 SD) 47.1g (1 SDAD) (3 SDAD) 32.86L (2 SDAD) (3 SDAD)

  13. Sig Fig Sample Problem A (Pg. 59) • A student heats 23.62 g of a solid and observes that its temperature increases from 21.6 °C to 36.79 °C. Calculate the temperature increase per gram of solid. • Gather Information Mass of Solid = 23.62 g Initial Temp = 21.6 °C Final Temp = 36.79 °C • Plan Work A. Calculate temp. increase (Final temp – Initial temp) = temp increase B. Per gram of solid temp increase = temp increase per g of solid gram • Calculate A. (36.79 °C – 21.6 °C ) = 15.19 °C = B. 15.2 °C = 0.6435224 = 23.62 g 15.2 °C 0.644 °C/g

  14. Sig Fig Problem Practice • Do Pg 59 Practice Problems 1 & 2

  15. Practice Problem #1 (Pg. 59) 0.127292 = 0.1 mL • Perform the following calculations, and express the answers with the correct number of significant figures. • 0.1273 mL – 0.000008 mL = • (12.4 cm X 7.943 cm) + 0.0064 cm2 = • (246.83 g / 26) – 1.349 g = 1 SDAD 4 SDAD 3 SD 4 SD 98.4932 cm2 = 98.5 cm2 + 0.0064 cm2 = 98.5064 = 1 SDAD 2 SDAD 98.5 cm2 5 SD 2 SD 9.4934615 g = 9.5 g – 1.349 g = 8.151 g = 8.2 g 1 SDAD 3 SDAD

  16. Practice Problem #2 (Pg. 59) • 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 density of the corn oil if its volume is 114 cm3. • Gather Info A. Mass of beaker + oil = 215.6 g B. Mass of empty beaker = 110.4 g C. Volume of corn oil = 114 cm3 • Plan Work Mass of oil = A – B = X Density of oil = X__ C • Calculate Mass of oil = 215.6 g – 110.4 g = Density of oil = 105.2 g__ = 114 cm3 1 SDAD 1 SDAD 105.2 g 4 SD 0.923 g/cm3 3 SD

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