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DIMENSIONAL ANALYSIS

DIMENSIONAL ANALYSIS. A.K.A FACTOR-LABELING USING UNITS TO SOLVE A MATH PROBLEM. TYPES. USE A FORMULA OR CONVERSION FACTORS SHOW A RELATIONSHIP BETWEEN DIFFERENT UNITS OF MEASUREMENT EX: 12 in = 1 ft 1 in = 2.54 cm 100 cm = 1 m 1 kg = 2.2 lbs. HOW TO USE.

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DIMENSIONAL ANALYSIS

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  1. DIMENSIONAL ANALYSIS A.K.A FACTOR-LABELING USING UNITS TO SOLVE A MATH PROBLEM

  2. TYPES • USE A FORMULA OR • CONVERSION FACTORS • SHOW A RELATIONSHIP BETWEEN DIFFERENT UNITS OF MEASUREMENT • EX: 12 in = 1 ft 1 in = 2.54 cm 100 cm = 1 m 1 kg = 2.2 lbs

  3. HOW TO USE • IDENTIFY THE GIVEN AND THE UNKNOWN. • WRITE DOWN THE APPROPRIATE CONVERSION FACTOR(S) TO USE. (OR PLUG NUMBERS INTO THE FORMULA)

  4. TO SOLVE (WHEN USING A FORMULA) • IDENTIFY THE GIVENS AND UNKNOWN. • SUBSTITUTE THE GIVENS INTO THE FORMULA • REARRANGE THE FORMULA TO SOLVE FOR THE UNKNOWN

  5. EXAMPLE • THE MASS OF AN OBJECT IS 5.87g AND THE VOLUME IS 7.98mL. WHAT IS THE DENSITY OF THE OBJECT?

  6. IDENTIFY GIVEN & UNKNOWN D = ? m = 5.87 g D = m/v v = 7.98 mL SUBSTITUTE THE NUMBERS D = 5.87g = 0.736 g/mL 7.98mL

  7. TO SOLVE (WHEN USING CONVERSION FACTORS) • PUT YOUR GIVEN OVER 1. (THIS TURNS IT INTO A FRACTION) • MULTIPLY THE GIVEN BY THE APPROPRIATE CONVERSION FACTOR(S). THE UNITSTELLYOU WHICH CONVERSION FACTOR TO USE & HOW TO SET IT UP.

  8. CANCEL UNITS. IF THE UNITS CANCEL THEN YOU CAN DO THE MATH. • SIG. FIGS.COME FROM THEGIVEN.

  9. EXAMPLE • A WOMAN NEEDS 76.0 m OF FABRIC. HOW MANY YARDS IS THIS? • GIVEN: 76.0 m • UNKNOWN: ? YARDS • CONVERSION FACTOR(S): 0.3048 m = 1 ft 3 ft = 1 yd

  10. UNKNOWN UNIT • 76.0 m x 1 ft x 1 yd = 1 0.3048 m 3 ft CALC. ANS: 83.11461067 yds ROUNDED ANS: 83.1 yds

  11. PRACTICE PROBLEMSCOPY AND WORK 1) The density of an object is 7.71 g/mL. Its mass is 51.12g. What volume does the object occupy?

  12. 2) A box has a mass of 4.5 kg and a volume of 6.4L. Calculate the density in g/cm3. Use dimensional analysis to convert your metric units.

  13. ANSWERS • 1) D = 7.71g/mL m = 51.12g v = ? D = m v solve for this

  14. WORK: v = m D v = 51.12g 6.63 mL 7.71g/mL =

  15. PROBLEM 2 REQUIRES 3 STEPS: • 1ST CONVERT kg TO g • 2ND CONVERT L TO cm3 • USE DENSITY EQUATION

  16. 2) GIVEN: 4.5kg UNK: g C.F.: 1kg = 1000g 4.5kg X 1000g = 4500g 1 1kg

  17. GIVEN: 6.4L UNK: mL C.F.:1L = 1000 mL 1mL = 1cm3 6.4L X 1000mL X 1 cm3= 1 1L 1mL 6400cm3

  18. WORK: D = m/v D= 4500g = 6400cm3 0.70g/cm3

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