1. Chemistry Units and Unit Conversions Section 2.1

2. Conversion Factor A fractional expression that relates or connects two different units –Example: 1 min = 60 s 1 min/60 s OR 60 s/1 min

5. Multiples of Base Units Page 17 and 18 of text

6. Conversions To go from one unit to another, you need a conversion factor This allows you to cancel out units until you have the unit you are interested in

7. Method Step 1: Identify the unknown amount and its unit Step 2: Identify the initial amount and its unit Step 3: Identify the conversion factor. Multiply the initial amount by the conversion factor so that one of the units cancels the unit of the initial amount out Step 4: Do the math.

8. Example If 0.200mL of gold has a mass of 3.86g, what is the mass of 5.00mL of gold? Solution: Unknown amount and unit: # of grams Initial amount and unit: 5.00mL Conversion Statement: 0.200mL/3.86g # of g = 5.00mL * 3.86g/0.200mL = 96.5g

9. Unit converting to: m Initial amount: 535 cm Conversion factor: 1 m / 100 cm Solution: # of m = 535 cm *(1 m)/(100 cm) = 5.35 m Convert 535 cm into m

10. Example 6: If a car can go 190 km in 2 h, how far can the car go in 10 h? Dissect the sentence to determine the conversion statement, what you are trying to determine and the initial amount. 1) Km/hour from Km/2 hour 2) Initial amount: in 10 h. 3) Conversion statement: If a car can go 190 km in 2h. 4) Math We are looking for km; thus, # km = 10 h * (190 km)/(2 h) = 850 km.. Notice that the h cancels out.

11. Example 7: If a 4.25 mL of gold has a mass of 5.40 g, what is the mass of 5.00 mL of gold? 1) grams (g) 2) Initial amount: 5.00 mL. 3) Conversion factor statement: 4.25 mL of gold has a mass of 5.40 g Conversion factor: 5.40 g / 4.25 mL gold. 4) 5.00 mL Au * (5.40 g Au)/(4.25 mL Au) = 6.35 g Au

12. Homework Homework: - Q. 1 a-j Page 11 - Q. 2 a-e, h,j Page 14