I. Units of Measurement (p. 33 - 39) CH. 2 - MEASUREMENT I. Units of Measurement (p. 33 - 39)
A. Number vs. Quantity Quantity - number + unit UNITS MATTER!!
B. SI Units Quantity Symbol Base Unit Abbrev. Length l meter m Mass m kilogram kg Time t second s Temp T kelvin K Amount n mole mol
B. SI Units Prefix Symbol Factor mega- M 106 kilo- k 103 BASE UNIT --- 100 deci- d 10-1 centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12
M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L Combination of base units. Volume (m3 or cm3) length length length 1 cm3 = 1 mL 1 dm3 = 1 L D = M V Density (kg/m3 or g/cm3) mass per volume
D. Density Mass (g) Volume (cm3)
Problem-Solving Steps 1. Analyze 2. Plan 3. Compute 4. Evaluate
D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3) An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass. GIVEN: V = 825 cm3 D = 13.6 g/cm3 M = ? WORK: M = DV M = (13.6 g/cm3)(825cm3) M = 11,200 g
D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? GIVEN: D = 0.87 g/mL V = ? M = 25 g WORK: V = M D V = 25 g 0.87 g/mL V = 29 mL
II. Using Measurements (p. 44 - 57) CH. 2 - MEASUREMENT II. Using Measurements (p. 44 - 57)
A. Accuracy vs. Precision Accuracy - how close a measurement is to the accepted value Precision - how close a series of measurements are to each other ACCURATE = CORRECT PRECISE = CONSISTENT
B. Percent Error Indicates accuracy of a measurement your value accepted value
B. Percent Error % error = 2.9 % A student determines the density of a substance to be 1.40 g/mL. Find the % error if the accepted value of the density is 1.36 g/mL. % error = 2.9 %
C. Significant Figures Indicate precision of a measurement. Recording Sig Figs Sig figs in a measurement include the known digits plus a final estimated digit 2.35 cm
C. Significant Figures Counting Sig Figs (Table 2-5, p.47) Count all numbers EXCEPT: Leading zeros -- 0.0025 Trailing zeros without a decimal point -- 2,500
Counting Sig Fig Examples C. Significant Figures Counting Sig Fig Examples 1. 23.50 1. 23.50 4 sig figs 2. 402 2. 402 3 sig figs 3. 5,280 3. 5,280 3 sig figs 4. 0.080 4. 0.080 2 sig figs
C. Significant Figures (13.91g/cm3)(23.3cm3) = 324.103g 324 g Calculating with Sig Figs Multiply/Divide - The # with the fewest sig figs determines the # of sig figs in the answer. (13.91g/cm3)(23.3cm3) = 324.103g 4 SF 3 SF 3 SF 324 g
C. Significant Figures 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL Calculating with Sig Figs (con’t) Add/Subtract - The # with the lowest decimal value determines the place of the last sig fig in the answer. 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL 7.85 mL 224 g + 130 g 354 g 224 g + 130 g 354 g 7.9 mL 350 g
C. Significant Figures Calculating with Sig Figs (con’t) Exact Numbers do not limit the # of sig figs in the answer. Counting numbers: 12 students Exact conversions: 1 m = 100 cm “1” in any conversion: 1 in = 2.54 cm
C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL) 4 SF 2 SF = 2.390625 g/mL 2.4 g/mL 2 SF 6. 18.9 g - 0.84 g 18.1 g 18.06 g
D. Scientific Notation 65,000 kg 6.5 × 104 kg Converting into Sci. Notation: Move decimal until there’s 1 digit to its left. Places moved = exponent. Large # (>1) positive exponent Small # (<1) negative exponent Only include sig figs.
D. Scientific Notation Practice Problems 7. 2,400,000 g 8. 0.00256 kg 9. 7 10-5 km 10. 6.2 104 mm 2.4 106 g 2.56 10-3 kg 0.00007 km 62,000 mm
D. Scientific Notation Calculating with Sci. Notation (5.44 × 107 g) ÷ (8.1 × 104 mol) = Type on your calculator: EXP EE EXP EE ENTER EXE 5.44 7 8.1 ÷ 4 = 671.6049383 = 670 g/mol = 6.7 × 102 g/mol
Chemistry Binder Organization
III. Unit Conversions (p. 40 - 42) CH. 2 - MEASUREMENT III. Unit Conversions (p. 40 - 42)
A. SI Prefix Conversions Symbol Factor mega- M 106 kilo- k 103 BASE UNIT --- 100 deci- d 10-1 move left move right centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12
M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L Combination of base units. Volume (m3 or cm3) length length length 1 cm3 = 1 mL 1 dm3 = 1 L D = M V Density (kg/m3 or g/cm3) mass per volume
D. Density Mass (g) Volume (cm3)
Problem-Solving Steps 1. Analyze 2. Plan 3. Compute 4. Evaluate
D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3) An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass. GIVEN: V = 825 cm3 D = 13.6 g/cm3 M = ? WORK: M = DV M = (13.6 g/cm3)(825cm3) M = 11,200 g
D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? GIVEN: D = 0.87 g/mL V = ? M = 25 g WORK: V = M D V = 25 g 0.87 g/mL V = 29 mL
Homework P. 54 Practice problems 1 & 2
Conversion Factors Problems Dimensional Analysis Conversion Factors Problems
B. Dimensional Analysis A tool often used in science for converting units within a measurement system Conversion Factor A numerical factor by which a quantity expressed in one system of units may be converted to another system
B. Dimensional Analysis The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out
B. Dimensional Analysis Steps: 1. Identify starting & ending units. 2. Line up conversion factors so units cancel. 3. Multiply all top numbers & divide by each bottom number. 4. Check units & answer.
B. Dimensional Analysis Lining up conversion factors: = 1 1 in = 2.54 cm 2.54 cm 2.54 cm 1 = 1 in = 2.54 cm 1 in 1 in
B. Dimensional Analysis How many milliliters are in 1.00 quart of milk? qt mL 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL
B. Dimensional Analysis You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3. lb cm3 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g = 35 cm3
B. Dimensional Analysis How many liters of water would fill a container that measures 75.0 in3? in3 L 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3 = 1.23 L
B. Dimensional Analysis 5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? cm in 8.0 cm 1 in 2.54 cm = 3.2 in
B. Dimensional Analysis 6) Taft football needs 550 cm for a 1st down. How many yards is this? cm yd 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft = 6.0 yd
B. Dimensional Analysis 7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? cm pieces 1.3 m 100 cm 1 m 1 piece 1.5 cm = 86 pieces