lecture topic i: problem solving in chemistry significant figures and scientific notation allowed...

LECTURE TOPIC I: Problem Solving in Chemistry

• Significant Figures and Scientific Notation

• Allowed Digits and “Rounding Off”

• Dimensional Analysis and SI Unit Conversions

• Density and Percent Problems

(Kotz & Treichel text: 1.6-1.8)

Handling Numbers in Chemistry: The Necessary Skills:

Correct use of significant figures (“sig figs”)

Handling scientific notation

Rounding off computational values

Knowledge of SI, Metric, and English Units

Use of dimensional analysis as problem solving tool

Correct use of significant figures (“SF’s”)

All measured values are limited in scope by the accuracy of the instrument used.

When correctly expressed, all measured values contain all the digits which can be read directly plus one estimated digit.

The digits in measured values are described as “significant figures” when they are presented in this fashion.

2.35 cm ? sf 2.0 in ? sf

4.689 g ? sf 2000 g ? sf

25.01 oz ? sf .0005 g ? sf

2.00 lb ? sf 200. ft ? sf

TEST YOUR MEMORY, Counting SF’s:

Correct Number of SF’s:

2.35 cm 3 sf 2.0 in 2 sf

4.689 g 4 sf 2000 g 1 sf

25.01 oz 4 sf .0005 g 1 sf

2.00 lb 3 sf 200. ft 3 sf

RECALL THE RULES, COUNT ONLY:

1. All non zero digits: 2.35 cm, 3 SF 4.689 g, 4 SF

2. All Internal zeros: 25.01 oz, 4 SF

3. All ending zeros when the decimal point is expressed: 2.00 lb, 3 SF 2.0 in, 2 SF 200. ft, 3 SF

DO NOT COUNT AS SIG FIGS:

1. Ending zeros, no decimal point 2000 g, 1 SF

2. All beginning zeros .0005 g, 1 SF

Group Work (Test Yourself!)

These “Group Work Exercises” will occur 2-4 times during each two hour lecture session...

Your “table-mates” or nearest neighbors will make up your group... minimum 2, maximum 4, ideal: 3.

Pick a “secretary” to write down your solutions; all members sign name on top right hand corner of 1st page; hand in at end of lecture, front desk...

The group work counts 5 points/lecture session!

I will present the solution key in class following eachactivity, but it will not be in the Topics slides.

The Group Work Solution Keys will be found on-line at groups.yahoo.com along with the homework and resources for each topic....

On-line students: “group work activities” are to be usedas “test yourself” exercises as you proceed through each topic; do the activity and then check out the answers before proceeding...

GROUP WORK 1.1: How many SF’s?

1,000 lb 25.351 g

.000203 mg 1.00 X 103 m

3.040 in 29.00 yd

2500 mi 33.0 gal

420. L .0045 cm

150 mL 12 eggs

SCIENTIFIC NOTATION

Scientific Notation is an alternate method of expressing numerical values in which the original value is multiplied or divided by ten until thereis only ONE digit to the left of the decimal.

The resulting number is multiplied by 10 raised to the appropriate power to restore the worthof the original value.

.0000270 = 2.70 X 10-5 = 2.70 10x10x10x10x10

27,000 = 2.70 X 104 = 2.70 x10x10x10x10

Scientific Notation is used to express very large and very small values, and to facilitate expression of some value to correct number of SF’s.

1,000,000 = 1.00 X 106 (3 SF) or 1 X 106 (1 SF)

1. Recall: 10-1 = .1 100 = 1 101 = 10 102 =100

Equivalent Values: 27.0 = 27.0 x 1 = 27.0 x 100

2. Small Numbers: subtract one from the power of ten for each right move of decimal:

.000027= .000027 x 100 = 2.7 x 10 0 - 5 = 2.7 X 10-5

3. Large Numbers: add one to the power of ten for each left move of decimal:

27,000 = 27,000 x 100 = 2.7 x 10 0 + 4 = 2.7 X 104

The Methodology:

In a nutshell:

Add +1 to power of ten for each LEFT MOVE Add -1 to power of ten for each RIGHT MOVE

.0 0 0 0 2 7 0 = 2.70 X 10 -5

2 7 0 3 . 7 = 2.7037 X 10 +3

2 7 3.5 X 10-5 = 2.735 X 10-5+2

= 2.735 X 10-3

GROUP WORK 1.2

Place into Scientific Notation:

95,000 (4 SF’s) .00593 X 10-4

.0000008090 0.02030 X 10+5

4578.2 346.00 X 10 +4

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 1.3

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:

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

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 = 12 in. 1760 yd = 1 mi

SI/ ENG gate

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

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


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


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 1.4

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

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

solids

liquids

gases

SOLVING DENSITY PROBLEMS

1. Density itself can be calculated from experimentalvalues: Density = mass of object, solution, substance volume occupied (mL, cm3, L)

Volume can be determined in several ways:

a) direct measurement, liquidb) liquid displacement, solidc) measurement of dimensions, calculation

Volume by Displacement:

20.0 mL

28.5 mLFluid displacement: 28.5 mL -20.0 mL

8.5 mL

8.5 mL = 8.5 cm3

A liquid in which the solid does not dissolve issuitable for this technique

Calculation, Volume:

V = l X w X ht (rectangle) V = e3 (cube) V = r2 ht (cylinder) radius = diameter / 2; all dimensions in same units

diameterheight

l

w

ht

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

SOLVING DENSITY PROBLEMS

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

“conversion factors”

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

( D, oil= .92 g / mL) 1 mL oil = .92 g oil 453.6 g = 1 lb

Pathway: mL g lb

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

Density conversion factor

GROUP WORK 1.5: Setup and solve as Conversion Type Problem

A gold coin is 2.75 cm in diameter and .50 cm thick.If the density of gold is 19.3 g / cm3, what is the massof the coin in grams?

Note:

Volume = r2 ht r = d / 2 ht = “thickness”

D, Au = 19.3 g/cm3 19.3 g Au = 1 cm3 Au

Percent Composition by Mass

“Percent composition” is a convenient way to describe a mixture or solution or compound in terms of mass of the part contained in 100 mass units of the whole:

“This solution is 15% salt” means that for every 15 g of salt there is 100 g of solution or: 15 g salt = 100 g solution

“The brass alloy is 15 % tin and 45% copper”

100 g alloy = 15 g tin = 45 g copper

Like Density, Percent Composition is: calculated from experimental values used as a conversion factor.

Calculation of % by mass (g):

% Part = g part X 100% g whole

A brass alloy weighing 79.456 g was found to contain34.29 g of copper. What is the % of Cu in the alloy?

%Cu = 34.29 g Cu X 100% = 43.16% Cu 79.456 g alloy = 43.16 g Cu in 100 g alloy

Percent as a Conversion Factor

What mass in grams of a brass alloy would contain 25.00 g Cu if the alloy is 43.16% Cu?

Question: 25.00 g Cu = ? g brass

Relationship: 43.16 g Cu = 100 g brass

Setup and solve: 25.00 g Cu X 100 g brass = 57.92 g brass 43.16 g Cu

% factor

Other Percent Relationships,Solutions:

Generally, chemists use percent in terms of “mass, g per 100 g of the whole”.

However, for solutions, % is sometimes given in terms of g solute per volume in mL of the solution:

“ 5.00% solution of salt, 5.00 g salt / 100 mL solution”

5.00 g solute salt = 100 mL solution You must carefully note how problem defines % !

If a 5.00 % salt solution is made up to contain 5.00 gof salt for every 100 mL solution, how many mL of this solution would contain 19.0 g of the salt?

Question: 19.0 g salt = ? mL solution

Relationship: 5.00 g salt = 100 mL solution

19.0 g salt X 100 mL soltn = 380. mL solution 5.00 g salt

Density/Percent Solution Problems

Automobile batteries are filled with sulfuric acid, which is a solution of liquid H2SO4 in water.

What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution, if the density of the solutionis 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Analysis of Problem

1.“Describing the Scene”:


3.“Giving the Relationships”:

if the density of the solution is 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

2. “Stating the Question”:

What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution


What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution, if the density of the solutionis 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Question: 500. mL soltn = ? g H2SO4

Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H2SO4 = 100 g soltn

Pathway: mL soltn g soltn g H2SO4

Question: 500. mL soltn = ? g H2SO4


Pathway: mL soltn g soltn g H2SO4

Setup and solve:

500. mL soltn X density factor X % factor = g H2SO4

500. mL soltn X 1.285 g soltn X 38.08 g H2SO4 1 mL soltn 100 g soltn

= 244.664 g H2SO4 = 245 g H2SO4

Group Work 1.6


How many mL of the acid solution would contain 15.00 gH2SO4, if the density of the solution is 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Question: 15.00 g H2SO4 = ? mL soltn


Pathway: g H2SO4 g soltn mL soltn

% D

lecture topic i: problem solving in chemistry significant figures and scientific notation allowed...

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