Chapter 1: Fundamental Concepts

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Transcript Chapter 1: Fundamental Concepts

Chapter 1: Matter, Measurement, and Problem
Solving
1. Chemistry
the “central science”
the study of matter and its changes
2. The Scientific Method
OBSERVATION
(empirical facts (data),
then “laws”)
OBSERVE
Gather data,
“laws”
Hypothesis
EXPLANATION
(hypothesis and theory)
TEST
Experiment
3. Matter
Matter occupies space and has mass
(mass is the amount of matter, weight is the force
of gravitational attraction on the mass)
THEORY
Properties of Matter
1. States of matter; solid, liquid, gas
2. Physical Properties vs. Chemical Properties
A physical change does not change the chemical makeup.
A chemical change (reaction) changes the substance.
~Would changing the state be a physical or chemical change?
3. Intensive vs. Extensive Properties
Extensive depends on sample size, intensive does not.
volume
mass
color
mp, bp
conductivity
~Which is better for determining an unknown substance?
Classification of Matter
Elements are
Compounds are
substances that
cannot be
decomposed by
chemical means into
simpler
substances.
substances formed
from two or more
elements combined
in a fixed proportion
by mass.
solution; a
homogeneous
mixture of
substances
Atom: smallest component of an element
Molecule: particle that is combination of two or more atoms
smallest component of a compound
Energy
• All reactions require or generate energy.
• Law of Conservation of Energy
– Energy cannot be created or destroyed but can be
changed from one form to another.
Energy
Kinetic
Energy due to movement,
dependent on mass and velocity
K.E. = ½ mv2
Potential
Stored energy, can be
converted to kinetic energy
Potential energy in chemicals is called chemical energy.
• Heat and Temperature
Heat, or thermal energy, is a form of energy (internal motions of atoms
and molecules).
Temperature is a measure of the intensity of heat (average kinetic energy).
Units of Measurement
1.
Systèm International Units (Table 1.1)
Base Units
length
m meter (39.37 inches)
mass
kg kilogram (2.205 lb)
time
s
second
temp
K
kelvin
Derived Units e.g. Volume = length x length x length (e.g, m3)
1 mL = cm3 and 1 L = 1000 mL = 1000 cm3
2.
The Metric System (Table 1.2)
KNOW ALL of the Decimal Multipliers and SI Prefixes from tera to
atto!!!
e.g. k = kilo = 103
1 km = 103 m
n = nano = 10-9
1 nm = 10-9 m
3.
Metric - English Conversions (Table 1.3)
KNOW at least one conversion for each:
length 1 inch = 2.54 cm or 1 m = 39.37 inches
mass
1 kg = 2.205 lb or 1 lb = 454 g
volume 1 L = 1.057 qt or 1 gal = 3.786 L
Units of Measurement
4. Temperature Scales (Figure 1.12) -- Know how to convert!
–
–
–
Fahrenheit
Celsius
Kelvin
°F TF = (9/5) TC + 32
°C TC = (5/9)(TF - 32)
K
TK = TC + 273.15
Reliability of Measurements
“true value” - measured value =
ERROR
Uncontrolled
conditions
Poorly defined
measurements
Limited
reproducibility
e.g temp in this room…
Uncertainties are indicated
through using significant figures.
Using the first thermometer, the
temperature is 24.3 ºC (3 significant digits).
Using the more precise (second)
thermometer, the temperature is 24.32 ºC (4
significant digits)
Example Problem
• In analyzing a sample of polluted water a chemist measured
out 25.00 mL of water with a pipet. At another point 25 mL
was measured in a graduated cylinder. What is the
difference between the two measurements?
Example Problem
In analyzing a sample of polluted water a chemist measured
out 25.00 mL of water with a pipet. At another point 25 mL
was measured in a graduated cylinder. What is the
difference between the two measurements?
Answer:
25 mL from a graduated cylinder means between 24 and 26
mL. From the pipet 25.00 mL indicates a range between
24.99 to 25.01. Therefore the pipet measures volume with
greater precision.
Calculations and Significant Figures
1.
Accuracy and Precision
Accuracy - how close to the “true” value? (systematic errors)
Precision - how reproducible is the measurement? (random
errors)
2.
Significant Figures
# of “significant figures” shows degree of uncertainty in
measurement
e.g. a certain distance, in inches, could be 11.1 or 11.08, or
11.083 depending on how carefully it was measured (3, 4, or
5 sig figs)
3.
Exact Numbers
Values that are exactly counted or defined can be assumed to
have an infinite number of sig figs, e.g.
- 25 people
- 1 foot = 12 inches
- 1 inch = 2.54 cm
Rules for Significant Figures
•
•
Non zero integers always count as sig figs
Zeros
Leading Zeros - those preceding all non zero integers
and do not count. e.g. 00.0035 (2 s.f)
Captive Zeros - those between non zero integers,
count as sig figs , 1.008 = 4 s.f’s
Trailing Zeros - are at the end of a number and are
significant when a decimal point is present, e.g. 0.120
--trailing zeros before decimal point: ambiguous
(e.g. 300 could be 1, 2, or 3 sig figs). Use scientific notation!
•
Exact Numbers – calculations not obtained
through use of measuring device are all
significant, 4/3  r 3
Sig Fig Example
~Overall, this number has how many sig fig?
Calculations with Sig Figs
• Calculations with Sig Figs
– multiplication and division
• Look for factor with fewest # of sig figs
– addition and subtraction
• Look for value with fewest # of decimal places
• Example:
12.5 + 1.247
3.6 x 0.004215
= "905.95755"
= 906 = 9.1 x 102
13.7
0.015
• Sig Fig rules are applied in the same order as the
mathematical operations!
• During calculations, keep the number in your calculator;
only round off at the end!
Example Problems
4.56 x 1.4
=
6.38
Corrected
Limiting Term
only has 2 Significant Figure
12.11 + 18.0
+ 1.013
Limiting Term
(One decimal place)
=
6.4
2 Significant Figures
31.123
Corrected
31.1
One Decimal Place
Sample Problem
Complete the following calculation. Write the answer in proper scientific
notation, rounded to the correct number of significant figures, and with
proper units.
(5.4 kg + 127 kg) (0.03700 m)
3.474 x 10-3 kg/m2
=
Sample Problem
Complete the following calculation. Write the answer in proper scientific
notation, rounded to the correct number of significant figures, and with
proper units.
(5.4 kg + 127 kg) (0.03700 m)
3.474 x 10-3 kg/m2
=
1.41 x 103 m3
Unit Conversions
“UNIT ANALYSIS”
given quantity x conversion factor(s) = desired quantity
(starting units)
(target units)
e.g. “48 inches is 4 feet” how is this shown in a calculation?
(48 inches) x (1 ft/12 inches) = 4.0 ft
Now, what is 48 inches in meters?
(48 in) x (2.54 cm/in) x (1 m/100 cm) = 1.2 m
Another example: convert 25 miles/gallon to km/L:
25 mi
5,280 ft
12 in
2.54 cm
10-2 m
gal
mi
ft
in
cm
Finally, correct sig fig!
1 km
1 gal
103 m
3.786 L
11 km/L
=
10.6 km
L
Density and Specific Gravity
• density
d = mass/volume (usually g/cm3 or g/mL)
– e.g. density of water is 1.00 g/cm3 or 1.00 g/mL
– density of iron is 7.86 g/mL
• specific gravity
dsubstance/dwater (a dimensionless quantity)
– e.g. specific gravity of iron is 7.86
– (i.e. iron is 7.86 times more dense than water)
density of
substance
=
specific gravity
of substance
x
density of H2O
• equivalence factor -- just like a conversion factor, used in
unit analysis, e.g 7.86 g/mL
given quantity x equivalence factor(s) = desired quantity
(starting units)
(target units)
Sample Problems
1. An ocean dwelling dinosaur has an estimated body volume
of 1.38 x 106 cm3 and a mass of 1.24 x 106 g. What is its
density?
Answer = 0.899 g/cm3 (note sig fig!!)
2. The density of table salt is 2.16 g/mL at 20 oC. What is its
specific gravity?
specific gravity = (2.16 g/mL)/(1.00 g/mL) = 2.16
Sample Problems
• At the Athens Olympics, Justin Gatlin won the gold medal in
the men’s 100 (treat as 3 sig fig) meter race in a time of 9.85
seconds. The temperature at the time was 23 °C. Calculate
his average speed over that distance in units of miles/hour.
• Butane (the fuel in your typical gas grill), has a specific
gravity of 0.579. The new tank you just bought from WalMart contains 5.00 gallons of butane. Calculate the mass of
the butane in pounds.
Sample Problems
• At the Athens Olympics, Justin Gatlin won the gold medal in
the men’s 100 (treat as 3 sig fig) meter race in a time of 9.85
seconds. The temperature at the time was 23 °C. Calculate
his average speed over that distance in units of miles/hour.
• Answer: 22.7 miles/hour
• Butane (the fuel in your typical gas grill), has a specific
gravity of 0.579. The new tank you just bought from WalMart contains 5.00 gallons of butane. Calculate the mass of
the butane in pounds.
• Answer: 24.1 lb
Sample Problem
• Extremely tiny clusters of atoms or molecules are often
called nano-particles due to the small scale of their
dimensions. The specific gravity of gold is 19.3 and the
mass of one gold atom is 197 amu (where 1 amu = 1.66 x
10-24 g). Calculate the number of gold atoms in a sphereshaped nanoparticle of gold that is 2.000 nanometers in
diameter. [Hint: recall that the volume of a sphere is
(4/3)r3 where r is the sphere’s radius and  = 3.14159]
Sample Problem
• Extremely tiny clusters of atoms or molecules are often
called nano-particles due to the small scale of their
dimensions. The specific gravity of gold is 19.3 and the
mass of one gold atom is 197 amu (where 1 amu = 1.66 x
10-24 g). Calculate the number of gold atoms in a sphereshaped nanoparticle of gold that is 2.000 nanometers in
diameter. [Hint: recall that the volume of a sphere is
(4/3)r3 where r is the sphere’s radius and  = 3.14159]
Answer: 247 atoms per nanoparticle