lecture 2 chm 151 ©slg sf's in calculations si, metric, english units dimensional analysis...

LECTURE 2 CHM 151 ©slg

• SF's IN CALCULATIONS• SI, METRIC, ENGLISH UNITS• DIMENSIONAL ANALYSIS• UNIT CONVERSIONS• DENSITY

TOPICS:

SF's IN CALCULATIONS

When doing calculations involving measuredvalues (always the case in science!), you must limit the number of digits in your results to reflectthe degree of uncertainty introduced by these values .

You must be familiar with the rules for number ofSF’s or digits allowed and also with the rules forrounding values down to the allowed number of digits.

Calculations Involving SF’s

Multiplication and Division:

The final answer in a computation involving theseoperations should have no more SF’s than the value in the original problem with the least numberof SF’s.

Addition and Subtraction:

The sum of these operations is allowed no moredigits after the decimal than the original value withthe least number of digits after the decimal.

Rounding off Answers to Correct # SF’s

If first digit to be dropped is <5, drop it and all following digits, leaving rest of number unchanged. Round off 23.45231 to 4 SF’s:

23.45231 = 23.45

Less than five

If the first digit to be dropped is >5, drop it and allfollowing digits, but increase the last “retained digit” by one:

Round off 23.45678 to 4 SF’s:

23.45678 = 23.46

>5

If the first digit to be dropped is exactly five,no non zero digits following, “even up” theresulting rounded-off value:

Increase the last retained digit to make it even if it is odd only.

Round off to 4 SF’s:

23.45500 = 23.46 23.44500 = 23.44 Note:

23.455001 = 23.46 23.445001 = 23.45

SF’s in Calculations, Samples:

1.30 in. X .20 in. X 2960. in. = 769.60 in.3 = 7.7 X 10 2 in.3

3 SF 2 SF 4 SF 2 SF allowed

1.30 in. Since one value has .20 in. no digits after decimal,+ 2960. in. none are allowed!

2961.50 in. = 2962 in.

Group Work

24.569 g - .0055 g = ?

32.35 cm X 21.9 cm X 0.76 cm = ?

54.01 lb + .6489 lb + 1,312.0 lb =?

Calculate and round off to correct # SF’s:

Group Work #1, Answers:

24.569 g - .0056 g = ?

24.569 g - .0055 g 24.5635 g = 24.564 g

“ No more digits after decimal than value withleast number of digits after decimal”

32.35 cm X 21.9 cm X 0.76 cm = ?

32.35 cm X 21.9 cm X 0.76 cm = “538.4334” cm3

allowed SF’s: 2 538.4334 = 5.384334 X 102

= 5.4 X 102 cm3

You may also use 540 cm3 which is also 2 SF’s

54.01 lb + .6789 lb + 1,312.0 lb =?

54.01 lb .6489 lb+ 1,312.0 lb 1,366.6589 lb

= 1,366.7 lb

METRIC AND SI UNITS

Prefixes you should know:

M, 1,000,000 (106) X basic unit “mega”k, 1,000 (103) X basic unit “kilo”

d, 1/10 (10-1) X basic unit “deci”c, 1/100 (10-2) X basic unit “centi”m, 1/1000 (10-3) X basic unit “milli”

, 1/1,000,000 (10-6) X basic unit “micro”n, 1/1,000,000,000 (10-9) X basic unit “nano”p, 1/1,000,000,000,000 (10-12) X basic unit “pico”

Common SI/Metric/ English Conversions (Know!)

1000 g = 1 kg 1000 mg = 1 g (1 g = 10-3 kg ) (1 mg = 10-3 g)

1 lb = 453.6 g

16 oz avoir = 1 lb

SI / English gateway

1. MASS

Unit Conversion Using Dimensional Analysis

Three Steps Involved:

a) Recognizing and stating the question

b) Recognizing and stating relationships

c) Multiplication of Initial value by appropriate conversion factors to get desired value

Dimensional Analysis and Mass ConversionProblems

A typical goal weight for a 5’ 6” female might be 135 pounds. What is this weight in kilograms (kg) ?

1. State the Question: (Use following format): Original value, old unit = ? Value, new unit

135 lb = ? kg

135 lb = ? kgEng / SI conversion

2. State the relationship(s) between the old unit and the new unit:

1 lb = 453.6 g103 g = 1 kg Eng / SI mass

gateway

Problem pathway: lb g kg

135 lb = ? kg

Problem pathway: lb g kg

3. Multiply the old value, unit by the appropriateconversion factor(s) to arrive at the new value, unit:

1 lb = 453.6 g 1 lb = 1 = 453.6 g 453.6 g 1 lb

103 g = 1 kg 103 g = 1 = 1 kg 1 kg 103g

135 lb = ? kg

Problem pathway: lb g kg

3. Setup and Solve:

Old value, unit X factor 1 X factor 2 = new value, unit

135 lb X 453.6 g X 1 kg = 61.236 kg 1 lb 103 g

Calculator answerNot done yet!

135 lb X 453.6 g X 1 kg = 61.236 kg 1 lb 103 g

3 SF 4 SF

Exact, SF’s

Exact

Exact

All defined conversions within a system are exact;The “1” in all conversion relationships is exact

# SF’s allowed, 3

61.236 kg = 61.2 kg (Answer!)

1 m = 100 cm = 103 mm 1000 m = 1 km(1 cm = 10-2 m) (1 m = 10-3 km) (1 mm = 10-3 m) 1 m = 109 nm = 1012 pm

2. LENGTH:

2.540 cm = 1 inch

1 yd = 3 ft = 36 in. 1760 yd = 1 mi

SI/ ENG gate

Orange light has a wavelength of 625 nm. What wouldthis length be in centimeters?

Dimensional Analysis and Length ConversionProblems

1. State the question: 625 nm = ? cm

2. State the needed relationships: 109 nm = 1 m 1 m = 102 cm

Pathway: nm m cm

3. Setup and solve:

625 nm X 1 m X 102 cm = 625 X 102 - 9 cm 109 nm 1 m

= 625 X 10-7 cm

= 6.25 X 10-5 cm

A football field is 100. yd long. What length is this inmeters? (1 m = 39.37 in)

Group Work

In water, H2O, the bond length between each H and O is94 pm. What is this length in millimeters?

A football field is 100. yd long. What length is this inmeters? (1 m = 39.37 in)

1. State question: 100. yd = ? m

2. State relationships: 1 yd = 3 ft 1 ft = 12 in 39.37 in = 1 m

Pathway: yd ft in m

Pathway: yd ft in m

3. Setup and Solve:

100. yd X 3 ft X 12 in X 1 m = 91.44018 m 1 yd 1 ft 39.37 in

= 91.4 m

In water, H2O, the bond length between each H and O is94 pm. What is this length in millimeters?

1. State the question: 94 pm = ? mm

2. State the needed relationships: 1012 pm = 1 m 1 m = 103 mm

Pathway: pm m mm

3. Setup and solve:

94 pm X 1 m X 103 mm = 94 X 103 - 12 mm 1012 pm 1 m

= 94 X 10-9 mm

= 9.4 X 10-8 mm

Pathway: pm m mm

3. VOLUME:

1 L = 1000 mL = 1000 cm3

10 cm

10 cm

10 cm

1 qt = .9463 L= 946.3 mL

SI / ENG GATE

1 qt = 2 pt 1 gal = 4 qt 1pt = 16 fl oz

Dimensional Analysis and Volume ConversionProblems

What is the volume in cm3 occupied by one half gallon( .50 gal) of Sunkist Orange Juice? What is this value inin3?

1. State First Question: .50 gal = ? cm3

2. Relationships: 1 gal = 4 qt 1 qt = 946.3 mL 1 mL = 1 cm3

Pathway: gal qt mL cm3

Pathway: gal qt mL cm3

3. Setup and Solve:

.50 gal X 4 qt X 946.3 mL X 1 cm3 = 1892.6 cm3

1 gal 1 qt 1 mL

= 1.8926 X 103 cm3

= 1.9 X 103 cm3

What is the volume in cm3 occupied by one half gallon( .50 gal) of Sunkist Orange Juice? What is this value inin3? ( .50 gal = 1.9 X 103 cm3)

1. State the Question: 1.9 X 103 cm3 = ? in3

2. State the relationships: 2.540 cm = 1 in

( 2.540 cm)3 = ( 1 in)3

16.39 cm3 = 1 in3

Pathway: cm3 in3

Pathway: cm3 in3

3. Setup and solve:

1.9 X 103 cm3 X 1 in3 = 115.9 in3

16.39 cm3

= 1.159 X 102 in3

= 1.2 X 102 in3

Calculator: enter 1.9, then “E” button, then “3”

Dimensional Analysis and Density Conversions

“Density” is a handy conversion factor which relates the mass of any object or solution or gas to the amountof volume it occupies.

It is a physical property of matter, and can be obtainedfor all elements, most compounds and common solutions in a reference handbook (or on line).

Denser matter “feels heavier” and accounts for theproperties of “floating” and “sinking”

Common Density Values

Magnesium: 1.74 g/cm3

Aluminum: 2.70 g/cm3

Silver: 10.5 g/cm3

Gold: 19.3 g/cm3

Dry air: 1.2 g/L at 25 oC, 1 atm pressure

Water: .917 g/mL at 0 oC 1.00 g/ mL at 4.0 oC .997 g/mL at 25 oCSea Water: 1.025 g/mL at 15 oC Antifreeze: 1.1135 g/mL at 20oC

SOLVING DENSITY PROBLEMS

1. When the Density factor itself is desired:

Density = mass of object, etc (g) volume occupied (mL, cm3, L)

2. In all other cases, where conversion of mass to volume or volume to mass is desired, density is used as a“conversion factor”:

D, Al = 2.70 g/cm3 2.70 g Al = 1 cm3 Al

2.70 g Al = 1 = 1 cm3 Al 1 cm3 Al 2.70 g Al

Sample, Obtaining Density Value

A “cup” is a volume used by cooks in US. One cup is equivalent to 237 mL. If one cup of olive oil has a mass of 205 g, what is the density of the oil in g/mL and inoz avoir / in3?

1. State the question: D, olive oil = ? g/ mL = ? oz avoir / in3

2. State formula: Density = mass, g volume, mL

3. Substitute values and solve:

D = mass = 205 g = .86497 g = .865 g volume 237 mL mL mL

Second question: What is this density factor expressedas oz avoir/ in3?

1. State Question: .865 g = ? oz avoir mL in3

2. State relationships:

453.6 g = 1 lb 1 lb = 16 oz avoir

1 mL = 1 cm3

2.540 cm = 1 in ( 2.540 cm)3 = (1 in)3

16.39 cm3 = 1 in3

.865 g = ? oz avoir mL in3

Pathway: g lb oz avoir; mL cm3 in3

Pathway: g lb oz avoir; mL cm3 in3

3. Setup and Solve:

.865 g X 1 lb X 16 oz avoir X 1 mL X 16.39 cm3 1 mL 453.6 g 1 lb 1 cm3 1 in3

= .50008 oz avoir in3

= .500 oz avoir in3

.865 g = ? oz avoir mL in3

Density as a Conversion Factor:

Peanut oil has a density of .92 g/mL. If the recipe callsfor 1 cup of oil (1 cup = 237 mL), what mass of oil, inlb, are you going to use?

State question: 237 mL oil =? Lb oil

State relationship: 1 mL oil = .92 g oil 453.6 g = 1 lb

Pathway: mL g lb

SETUP AND SOLVE AS GROUP WORK!

Solution:

237 mL oil =? Lb oil

Pathway: mL g lb

237 ml oil X .92 g oil X 1 lb = .48068 lb oil 1 mL oil 453.6 g

= .48 lb oil