Transcript Problem Solving in Chemistry

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.
TEST YOUR MEMORY, Counting SF’s:
2.35 cm
? sf
2.0 in
4.689 g
? sf
2000 g ? sf
25.01 oz
? sf
.0005 g ? sf
2.00 lb
? sf
200. ft ? sf
? sf
Correct Number of SF’s:
2.35 cm
3 sf
2.0 in
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
2 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 each
activity, 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 used
as “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
3
.000203 mg
1.00 X 10 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 there
is only ONE digit to the left of the decimal.
The resulting number is multiplied by 10 raised
to the appropriate power to restore the worth
of the original value.
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.
.0000270 = 2.70 X 10-5 =
2.70
10x10x10x10x10
27,000 = 2.70 X 104 = 2.70 x10x10x10x10
1,000,000 = 1.00 X 106 (3 SF) or 1 X 106 (1 SF)
The Methodology:
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
In a nutshell:
Add +1 to power of ten for each LEFT MOVE
Add -1 to power of ten for each RIGHT MOVE
+1
-1
.0 0 0 0 2 7 0 = 2.70 X 10
-5
2 7 0 3 . 7 = 2.7037 X 10
-5
+3
-5+2
2 7 3.5 X 10 = 2.735 X 10
-3
= 2.735 X 10
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 measured
values (always the case in science!), you must
limit the number of digits in your results to reflect
the degree of uncertainty introduced by these
values .
You must be familiar with the rules for number of
SF’s or digits allowed and also with the rules for
rounding values down to the allowed number of digits.
Calculations Involving SF’s
Multiplication and Division:
The final answer in a computation involving these
operations should have no more SF’s than the
value in the original problem with the least number
of SF’s.
Addition and Subtraction:
The sum of these operations is allowed no more
digits after the decimal than the original value with
the 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 all
following 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” the
resulting 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
1.30 in.
.20 in.
+ 2960.
in.
4 SF
2 SF allowed
Since one value has
no digits after decimal,
none are allowed!
2961.50 in. = 2962 in.
Group Work 1.3
Calculate and round off to correct # SF’s:
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 =?
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
c, 1/100
m, 1/1000
,
n,
p,
(10-1) X basic unit “deci”
(10-2) X basic unit “centi”
(10-3) X basic unit “milli”
1/1,000,000
(10-6) X basic unit “micro”
1/1,000,000,000
(10-9) X basic unit “nano”
1/1,000,000,000,000 (10-12) X basic unit “pico”
Common SI/Metric/ English
Conversions (Know!)
1. MASS
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
2. LENGTH:
1 m = 100 cm = 103 mm
(1 cm = 10-2 m)
(1 mm = 10-3 m)
1000 m = 1 km
(1 m = 10-3 km)
1 m = 109 nm = 1012 pm
2.540 cm = 1 inch
1 yd = 3 ft = 12 in.
SI/ ENG
gate
1760 yd = 1 mi
10 cm
3. VOLUME:
1 L = 1000 mL = 1000 cm3
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 Conversion
Problems
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
Eng / SI conversion
135 lb = ? kg
2. State the relationship(s) between the old unit
and the new unit:
1 lb = 453.6 g
103 g = 1 kg
Problem pathway: lb  g  kg
Eng / SI mass
gateway
135 lb = ? kg
Problem pathway: lb  g  kg
3. Multiply the old value, unit by the appropriate
conversion factor(s) to arrive at the new value, unit:
1 lb = 453.6 g
103 g = 1 kg
1 lb = 1 = 453.6 g
453.6 g
1 lb
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 =
1 lb
103 g
61.236 kg
Calculator answer
Not done yet!
All defined conversions within a system are exact;
The “1” in all conversion relationships is exact
3 SF
4 SF
Exact
135 lb X 453.6 g X 1 kg =
1 lb
103 g
Exact,
 SF’s
61.236 kg
Exact
# SF’s allowed, 3
61.236 kg = 61.2 kg (Answer!)
2. LENGTH:
1 m = 100 cm = 103 mm
(1 cm = 10-2 m)
(1 mm = 10-3 m)
1000 m = 1 km
(1 m = 10-3 km)
1 m = 109 nm = 1012 pm
2.540 cm = 1 inch
1 yd = 3 ft = 36 in.
SI/ ENG
gate
1760 yd = 1 mi
Dimensional Analysis and Length Conversion
Problems
Orange light has a wavelength of 625 nm. What would
this length be in centimeters?
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
1m X
109 nm
102 cm =
1 m
625 X 102 - 9 cm
=
625 X 10-7 cm
=
6.25 X 10-5 cm
Group Work 1.4
A football field is 100. yd long. What length is this in
meters? (1 m = 39.37 in)
In water, H2O, the bond length between each H and O is
94 pm. What is this length in millimeters?
10 cm
3. VOLUME:
1 L = 1000 mL = 1000 cm3
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 Conversion
Problems
What is the volume in cm3 occupied by one half gallon
( .50 gal) of Sunkist Orange Juice? What is this value in
in3?
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 in
in3? ( .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
16.39 cm3
= 115.9 in3
= 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 amount
of volume it occupies.
It is a physical property of matter, and can be obtained
for all elements, most compounds and common
solutions in a reference handbook (or on line).
Denser matter “feels heavier” and accounts for the
properties of “floating” and “sinking”
Common Density Values
Dry air: 1.2 g/L at 25
Water:
.917
1.00
.997
Sea Water: 1.025
Antifreeze: 1.1135
oC,
1 atm pressure
g/mL at 0 oC
g/ mL at 4.0 oC
g/mL at 25 oC
g/mL at 15 oC
g/mL at 20oC
Magnesium: 1.74 g/cm3
Aluminum: 2.70 g/cm3
Silver:
10.5 g/cm3
Gold:
19.3 g/cm3
gases
liquids
solids
SOLVING DENSITY PROBLEMS
1. Density itself can be calculated from experimental
values:
Density = mass of object, solution, substance
volume occupied (mL, cm3, L)
Volume can be determined in several ways:
a) direct measurement, liquid
b) liquid displacement, solid
c) measurement of dimensions, calculation
Volume by Displacement:
28.5 mL
Fluid displacement: 28.5 mL
-20.0 mL
8.5 mL
20.0 mL
8.5 mL = 8.5 cm3
A liquid in which the solid does not dissolve is
suitable 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
w
diameter
height
ht
l
SOLVING DENSITY PROBLEMS
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
“conversion factors”
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 in
oz 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
volume
237 mL
mL
= .865 g
mL
Second question: What is this density factor expressed
as oz avoir/ in3?
1. State Question:
.865 g = ? oz avoir
mL
in3
.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
Pathway: g  lb  oz avoir; mL  cm3  in3
.865 g = ? oz avoir
mL
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
Density as a Conversion Factor:
Peanut oil has a density of .92 g/mL. If the recipe calls
for 1 cup of oil (1 cup = 237 mL), what mass of oil, in
lb, 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
1 mL oil
X 1 lb
453.6 g
=
.48068 lb oil
=
.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 mass
of 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 contain
34.29 g of copper. What is the % of Cu in the alloy?
%Cu = 34.29 g Cu
X 100%
79.456 g alloy
= 43.16 g Cu in 100 g alloy
= 43.16% Cu
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 g
of 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 solution
is 1.285 g /mL and the solution is 38.08% H2SO4 by mass?
Analysis of Problem
1.“Describing the Scene”:
Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
2. “Stating the Question”:
What is the mass (in grams) of the H2SO4 in 500. mL of
the battery acid solution
3.“Giving the Relationships”:
if the density of the solution is 1.285 g /mL and the
solution is 38.08% H2SO4 by mass?
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 solution
is 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
Relationships:
1.285 g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
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
Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
How many mL of the acid solution would contain 15.00 g
H2SO4, if the density of the solution is 1.285 g /mL and the
solution is 38.08% H2SO4 by mass?
Question:
Relationships:
15.00 g H2SO4 = ? mL soltn
1.285 g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
Pathway: g H2SO4  g soltn  mL soltn
%
D