1. 1 Dimensional Analysis DHS Chemistry

2. 2 Note: From this point on, unless told otherwise, it is expected that all answers will be reported using the sig fig rules. All numbers greater than 999 and less than 0.01 are to be reported in scientific notation. For multiplication and division, the measurement with the lowest number of sig figs will determine how many sig figs are in the answer. (Usually go by the sig figs in the given)

3. 3 Dimensional Analysis Dimensional analysis is a powerful problem- solving method that is based on the fact that any number or expression can be multiplied by one without changing its value. It is the most important skill you will learn for the rest of your science career.

4. 4 Conversion factors = relationships To use dimensional analysis you will use conversion factors and fraction cancellation. Anytime you can say two things are equal to each other (or the same as each other), you can make 2 conversion factors out of it --- each conversion factor is equal to 1.

5. 5 Setting Up Ratios 100 centimeters = 1 meter can be written as 100 cm OR 1 m___ 1 m 100 cm 8 slices = 1 pizza can be written as 1 pizza OR 8 slices___ 8 slices 1 pizza 5 g/mL can be written as 5 g OR 1 mL 1 mL 5 g $1.99 per pound can be written as $1.99 OR 1 lb 1 lb $1.99

6. 6 The Format # x # x # x # = 1 # # # OR # # # = 1 # # Note: Unless you are using 1 as a spot filler, all numbers should ALWAYS include units

7. 7 Understanding order of operations & typing in calculator keys correctly Solve. 10 13 6 = 2 4 Method 1: 10 X 13 X 6 = 780 then divide 97. 5 2 x 4 = 8 Method 2: 10 X 13 X 6 / 2 / 4 = 97.5 130 780 390 97.5 Method 3: 10 X 13 / 2 X 6 / 4 = 97.5 130 65 390 97.5

8. 8 Try It 152 3 1.2 4 = 1 4 5 14 5 0.30857

9. 9 Canceling out units Just as numbers are multiplied and divided, units are too. Cancel out all units that are located in both the numerator and denominator to reveal the left over units. cm in feet yard = cm in feet yard

10. 10 Try It week day hr min sec = 1 week day hr min sec

11. 11 Putting the numbers & units together 3 weeks 7 day 24 hr 60 min 60 sec = 1 1 week 1 day 1 hr 1 min =1814400 sec Rules?

12. 12 Using Dimensional Analysis 1)Start by writing GWR (given, want, relationships) 2)Identify the given. 3)Determine the units you want. 4)Choose the relationships that will allow you to convert from your given to the want.

13. 13 Using Dimensional Analysis Set-up your problem: 5)Start with the given as a clean fraction. 6)Write your multiplication symbol and place your relationships carefully so everything above and below each other is equal to each other, and make sure your units cancel out diagonally. 7)Any units that did not cancel out are part of the units in your answer.

14. 14 I. One-Step Conversions

15. 15 282.3 pizzas I. One- Step Conversions EX 1: Determine the number of slices in 282.3 pizzas Step 1) what are you given? Step 2) What unit do you want? Step 3) List your relationships (remember from pg 1 of the notes) Step 4) Set up your problem 8 slices slices 1 pizza =2258.4 1 282.3 pizzas Number of slices 1 pizza = 8 slices =2258 slices (SF)

16. 16 Dozen of eggs 1 dozen EX 2: Determine the number of dozen eggs in 3.8 x 10 3 eggs. Step 1) what are you given? Step 2) What unit(s) do you want? Step 3) List your relationships Step 4) Set up your problem 3.8 X 10 3 eggs 12 eggs =316.67 1 3.8 X 10 3 eggs Dozen Eggs 12 eggs = 1 dozen =320 dozen of eggs (SF)

17. 17 Practice 1.Determine the number of feet in 2821 inches. wantgiven 2821 in1 foot feet 12 in = 235.1 1 Relationship 1 foot = 12 inches Start with what’s given

18. 18 Practice 2. Determine the number of g in 0.03455 kg wantgiven 0.03455 kg1000 g g 1 kg = 34.55 1 Relationship 1000g = 1kg Start with what’s given

19. 19 want given 37 mi 1.6093 kmkm 1 mi = 59.5 1 Relationships 1 mi = 1609.3 meter 1 mi = 1.6093 km Start with what’s given 99 mi 1.6093 kmkm 1 mi = 159 1

20. 20 Solving the same problem, but in a different way

21. 21 1 km want given 37 mi 1609.3 mkm 1 mi = 59.5 1 Relationships 1 mi = 1609.3 meter 1 km = 1000 m Start with what’s given 1000 m

22. 22 For every problem, start with: G iven: W ant: R elationships:

23. 23 II. Multi-Step Conversions

24. 24 II. Multi-step conversions Of course, most problems are not simple 1- step conversions. Most problems will require multiple conversion factors. Note: in this class you MUST go through the root unit for any metric conversion. (root units: meter, gram, liter …) Tip: Cancel out all units until you get what you’re looking for.

25. 25 Ex. 1 Convert 0.115 km to cm G iven: W ant: R elationships: 0.115 km ____ cm 100 cm = 1 m 1 km = 1000 m 0.115 km1000 m cm 1 km = 11500 same as 1.15 x 10 4 cm 1 Start with what’s given 1 m 100 cm

26. 26 Ex. 2 Convert 323 mL to cups G iven: W ant: R elationships: 323 mL ____ cups 1000 mL = 1 L 1 L = 1.06 quarts 1 quart = 4 cups 1.06 qt. 323 mL1 L cups 1000 mL 1.36952= 1.37 1 Start with what’s given 1 L 1 qt. 4 cups

27. 27 Practice 1: Convert 7.005 ft to mm G iven: W ant: R elationships: 7.005 ft ____ mm 1 foot = 12 inches 1 inch = 2.54 cm 100cm = 1 m 1m = 1000mm 2.54 cm 7.005 ft 12 in mm 1 ft 2135 1 Start with what’s given 1 in100 cm 1 m 1000 mm 1 m

28. 28 Practice 2: calculate the number of seconds in 2 years G iven: W ant: R elationships: 2 years ____ seconds 1 year = 365.25 days 24hrs = 1 day 60 min = 1 hr 1 min = 60 sec 24 hrs 2 years 365.25 days seconds 1 year 63115200 = 6 X 10 7 1 Start with what’s given 1 day1 hr 60 min 60 sec 1 min

29. 29 What does that price mean? $1.99 = 1lb

30. 30 Solving Word Problems using Dimensional Analysis

31. 31 If you are given multiple numbers in a problem, only one number will be your starting point. The other numbers are relationships that you will use in your problem. If you are given multiple numbers and one of them involves a “/” (e.g. m/s), then always use the “/” as a relationship and start with the other number.

32. 32 If it helps, change any combination unit into a relationship ex. 0.05mL/s 0.05mL = 1 s

33. 33 G iven: W ant: R elationships: 0.05mL/s ____ Liters 1 day = 24 hr 1 hr = 60 min EX: A faucet is dripping at a rate of 0.05 mL/s. How many liters of water will be lost in 1 day? 1 day 0.05mL = 1 sec 1000mL = 1 L 60 sec = 1 min

34. 34 1 min Relationship 0.05 mL60 min EX: A faucet is dripping at a rate of 0.05 mL/s. How many liters of water will be lost in 1 day? wantgiven 1 day24 hr L 1 day = 4.32 1 Relationships 0.05 mL = 1 s60 min = 1 hr 1000mL = 1 L24 hr = 1 day 60s = 1 min Start with what’s given 1 hr 60 s 1 s 1 L 1000 mL = 4 L (SF)

35. 35 Finally, note dimensional analysis can be used anytime you can say something is equal to something else. It does not have to involve standard conversion factors.

36. 36 G iven: W ant: R elationships: ____ feet 24 frames = 1 sec 1 hr = 60 min 1 in = 2.54cm EX 1: Motion pictures are shown at a speed of 24 frames, or individual pictures, each second. If a standard frame is 1.9 cm long, how long will the strand of film be (in ft) for a 2 ½ hr movie? 2.5 hours 1 frame = 1.9cm 1 min = 60 sec 1 ft = 12 in

37. 37 12 in EX 1: Motion pictures are shown at a speed of 24 frames, or individual pictures, each second. If a standard frame is 1.9 cm long, how long will the strand of film be (in feet) for a 2 ½ hour movie? 1 s Relationship 1.9 cm 60 s wantrelationship 2.5 hr 60 min ft 1 hr = 13464.6 1 Relationships 24 frames = 1 s1 frame = 1.9cm 1 hr = 60 min60 s = 1 min 1in = 2.54 cm1 ft = 12 in Start with what’s given 1 min 24 frames 1 frame 1in 2.54 cm want given 1 ft

38. 38 G iven: W ant: R elationships: ____ kg 0.05 g = 1 mL 1000g = 1kg 1L = 1000mL EX 2: The density of an unknown liquid is 0.5 g/mL. What is the mass (in kg) of 2.00 L? 2.00 L

39. 39 1 L EX 2: The density of an unknown liquid is 0.5 g/mL. What is the mass (in kg) of 2.00 L? 1 2.00L 1000g 1 kg =1.00 1000mL Relationship given want Relationships 0.5 g = 1mL 1000g = 1 kg 1000mL = 1 L 0.5g 1mL kg

40. 40 G iven: W ant: R elationships: 0.0899 g = 1 L ____ Liters 1 kg = 1000 g EX 3: What is the volume of 5 kg of hydrogen gas at room conditions? The density of hydrogen at room conditions is 0.0899 g/L. 5 kg

41. 41 1 kg EX 3: What is the volume of 5 kg of hydrogen gas at room conditions? The density of hydrogen at room conditions is 0.0899 g/L. 1 5 kg =55600 1000g Relationship givenwant Relationships 0.0899 g = 1L 1000g = 1 kg 0.0899g 1L L

42. 42 Practice Box Answers 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? 275.88 trips 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? 216000 kg 3. The density of aluminum is 2.7 g/cm3. What volume would a 5.0 kg block of aluminum have? 1851.85 cm 3 4.Oxygen has a density of 1.43 g/L at 0oC and 1 atm. What would be the mass of 5.0 x 105 L of oxygen at those conditions? 715000 g

43. 43 G iven: W ant: R elationships: ____ trips 60 mi = 1 gal 1 trip = 3.5km 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? 10 gallons 1000m = 1km 1 mile = 1609.3 m

44. 44 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? 1000m Relationship 1 trip1609.3 m want relationship 10gal 60 mi trips 1 gal = 275.88 1 Relationships 60 miles = 1 gal1 trip = 3.5 km 1 mi = 1609.3m1000m = 1km Start with what’s given 1 mi 1 km 3.5 km given = 300 trips (SF)

45. 45 G iven: W ant: R elationships: ____ kg 60 s = 1 min 60 min = 1 hr 24 hrs = 1 day 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? 1 day 1 sec = 2500g 1kg = 1000g

46. 46 1 kg 1000g 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? 1 min Relationship 2500g 60 min want 1 day 24 hr kg 1 day = 216000 1 Relationships 1 s = 2500g 60s = 1 min 1000g = 1kg60 min = 1 hr 24 hr = 1 day Start with what’s given 1 hr 60 s 1 s given kg= 2 x 10 5

47. 47 G iven: W ant: R elationships: ____ cm 3 2.7 g = 1 cm 3 1kg = 1000g 3. The density of aluminum is 2.7 g/cm 3. What volume would a 5.0 kg block of aluminum have? 5.0kg

48. 48 1 kg 3. The density of aluminum is 2.7 g/cm 3. What volume would a 5.0 kg block of aluminum have? 1 5.0kg =1851.85 1000g Relationship given want Relationships 2.7 g = 1 cm 3 1000g = 1 kg 1 cm 3 2.7g cm 3 =1900 cm 3 (SF)

49. 49 G iven: W ant: R elationships: 1.43g/L ____ g 4. Oxygen has a density of 1.43 g/L at 0 o C and 1 atm. What would be the mass of 5.0 x 10 5 L of oxygen at those conditions? 5.0 x 10 5 L

50. 50 1 L 4. Oxygen has a density of 1.43 g/L at 0 o C and 1 atm. What would be the mass of 5.0 x 10 5 L of oxygen at those conditions? 1 5.0 X 10 5 L =715000 1.43 g Relationship givenwant Relationships 1.43 g = 1L g =72000 g (SF)= 7.2 x 10 4 g

51. 51 Tips for studying Try doing the review without using the conversion sheet Quiz yourself with the conversions Do extra practice problems (your textbook may have some)

52. 52 More Practice

53. 53 Practice: Convert 5.2 L to cups G iven: W ant: R elationships: 5.2 L ____ cups 1 L = 1.06 quarts 1 quart = 4 cups 1.06 qt. 5.2 L cups 1 = 22.0 Start with what’s given 1 L 1 qt. 4 cups

54. 54 356 g A block of metal measuring 2.5 cm x 7.2 cm x 6.7 cm has a mass of 356 g. What volume of water will 1500 grams of this metal displace? 1 1500g =508 120.6 cm 3 Relationship given want Relationships 120.6 cm 3 = 356g cm 3 =510 cm 3 (SF)