Water, Water Everywhere
Water as Liquid - Rainwater
Water as Solid - Iceberg
Snow and Snow Flakes
Water Vapor (Steam)
Water Cycle
Water Cycle
Water Molecule
Hydrogen Bonding in Water
What Makes Soils & Rocks Have Different
Colors
Where does Chemistry fit in?
• Chemistry is about the study of matters
and the changes they undergo.
• Chemistry probes the fundamental units
of matter in order to understand how and
why they are what they are.
• Chemists always ask questions and try to
find the answers.
The Central Science
• Chemistry is regarded as the central science.
• Chemistry is essential in understanding the
various aspects of living and non-living things;
• It is essential in understanding natural and
unnatural processes of nature.
What is Matter?
• The materials of the universe
 anything that has mass and occupies space
What Type of Change?
• Physical or Chemical processes;
• Physical Change:
A process that alters the state of a substance, but
not its fundamental composition.
• Chemical Change:
A process that alters the fundamental composition
of the substance and, therefore, its identity.
The Scientific Approach
1.
2.
3.
4.
5.
6.
Making Observations/collecting Data
Formulating Hypotheses
Testing the Hypotheses
Revising the Hypothesis
Summarizing Hypotheses into a Theory
Summarizing observations or natural
behavior into a Scientific Law
Steps in the scientific Method?
1. Identify the Problems and ask Questions
2. Develop a Hypothesis based on observations
3. Test The Hypothesis
(Design & Perform Experiments)
4. Collect more Data
5. Analyze Results
6. Make a Conclusion
7. Suggest further studies on the subject.
Definitions of Terms in Scientific Methods
1. Hypothesis:a plausible or logical statement that attempts to explain
the observation or data.
2. Theory :a set of (tested) hypotheses that explain a certain
behavior of nature.
3. Scientific Law :a concise statement about a natural phenomenon or
behavior.
Measurements
The Number System
• Decimal Numbers:
384,400
0.08206
• Scientific Notations:
3.844 x 105
8.206 x 10-2
Units of Measurements
• Units give meaning to numbers.
Without Unit
384,400 ?
0.08206 ?
144 ?
With Units
384,400 km (very far)
384,400 cm (not very far)
0.08206 L.atm/(K.mol)
144 eggs
English Units
Mass: ounce (oz.), pound (lb.), ton;
Length: inches, feet, yards, miles;
Volume: pints, quarts, gallons, in3, ft3, etc.;
Area: acre, hectare, in2, ft2, yd2, mi2.
Metric Units
1.
2.
3.
4.
Mass: milligram (mg), gram (g), kilogram (kg),
Length: cm, m, km, mm, mm, nm,
Area: cm2, m2, km2
Volume: mL(cm3), dL, L,, m3.
SI Units
1.
2.
3.
4.
5.
6.
7.
8.
Mass = kilogram (kg)
Length = meter (m)
Area = square meter (m2)
Volume = cubic meter (m3)
Temperature = Kelvin (K)
Energy = Joule (J)
Charge = Coulomb (C)
Time = second (s)
Prefixes in the Metric System
• Prefix
Symbol
10n
Decimal Forms
Giga
G
109
1,000,000,000
Mega
M
106
1,000,000
kilo
k
103
1,000
deci
d
10-1
0.1
centi
c
10-2
0.01
milli
m
10-3
0.001
micro
m
10-6
0.000,001
nano
n
10-9
0.000,000,001
—————————————————————
Accuracy and Precision
in Measurements
• Accuracy
The agreement of an experimental value with the
“true” or accepted value;
• Precision
The reproducibility of measurements of the same
type;
Accuracy and Precision
Errors in Measurements
• Random errors
1. values have equal chances of being high or low;
2. may be minimize by taking the average of several
measurements of the same kind;
• Systematic errors
1. Errors due to faulty instruments;
2. Reading is either higher or lower than the correct
value by a fixed amount;
3. Weighing by differences can eliminate systematic
errors of the faulty instruments.
Significant Figures
• All non-zero digits
Example: 453.6 has 4 significant figures.
• Captive zeros
Example: 1.079 has 4 significant figures.
• Trailing zeros if the decimal point is shown
Example: 1080 has 3 significant figures, but 1080. or
1.080 x 103 has 4 significant figures.
• Leading zeros are not significant figures
Example: 0.02050 has 4 significant figures
How many significant figures?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
0.00239
0.01950
1.00 x 10-3
100.40
168,000
0.082060
144 eggs in a carton
Express one thousand as a value with
two significant figures.
Rounding off Calculated values
• In Multiplications and Divisions
Round off the final answer so that it has the same
number of significant figures as the one with the least
significant figures.
Examples:
(a) 9.546 x 3.12 = 29.8 (round off from 29.78352)
(b) 9.546/2.5 = 3.8 (round off from 3.8184)
(c) (9.546 x 3.12)/2.5 = 12 (round off from 11.913408)
Rounding off Calculated values
• In Additions and Subtractions
Round off the final answer so that it has the same
number of digits after the decimal point as the data
value with the least number of such digits.
Examples:
(a) 53.6 + 7.265 = 60.9 (round off from 60.865)
(b) 53.6 – 7.265 = 46.3 (round off from 46.335)
(c) 41 + 7.265 – 5.5 = 43 (round off from 42.765)
Mean, Median & Standard Deviation
• Mean = average
Example:
• Consider the following temperature values:
20.4oC, 20.6oC, 20.3oC, 20.5oC, 20.4oC, and 20.2oC;
(Is there any outlying value that we can throw away?)
• No outlying value, the mean temperature is:
(20.4 + 20.6 + 20.3 + 20.5 + 20.4 + 20.2) ÷ 6 = 20.4oC
Mean, Median & Standard Deviation
Median:
1. the middle value (for odd number samples) or
2. average of two middle values (for even number)
3. when values are arranged in ascending or descending
order.
For the following temperatures:
20.4oC, 20.6oC, 20.3oC, 20.5oC, 20.4oC, and 20.2oC,
the median = 20.4oC
Mean, Median & Standard Deviation
• Standard Deviation, S =
( X i - X )2
( n  1)
(for n < 10, Xi = sample value; X = mean value)
[Note: calculated value for std. deviation should have
one significant figure only.]
For above temperatures, S = 0.1; Mean = 20.4 ± 0.1 oC
Calculating Mean Value
• Consider the following masses of pennies (in grams):
2.48, 2.50, 2.52, 2.49, 2.50, 3.02, 2.49, and 2.51;
• Is there outlying value?
Yes; 3.02 does not belong in the group – can be discarded
• Outlying values should not be included when calculating the
mean, median, or standard deviation.
• Average (mean) mass of pennies is,
(2.48 + 2.50 + 2.52 + 2.49 + 2.50 + 2.49 + 2.51) ÷ 7 = 2.50 g;
Calculating Standard Deviation


2
(
X
X
)
(
X
X
)
_________________________
i
i
-0.02
0.0004
-0.00
0.0000
0.02
0.0004
-0.01
0.0001
0.00
0.0000
-0.01
0.0001
0.01
0.0001___
Sum:
0.0011
------------------------------------------
S
(X
i
- X)
2
( n  1)
0.0011

(7 - 1)
 0.0135 0.01
Mean and Standard Deviation
• The correct mean value that is consistent with
the precision is expressed as follows:
2.50 ± 0.01
Using Q-test to retain or reject questionable values
• Calculate Qcalc. as follows:
• Qcalc. = | Questionable value- nearestneighbor |
(Highest value - Lowest value)
• Compare Qcalc with Qtab from Table-2 at the chosen
confidence level for the matching sample size;
• If Qcalc < Qtab, the questionable value is retained;
• If Qcalc > Qtab, the questionable value is can rejected.
Rejection Quotient
• Rejection quotient, Qtab, at 90% confidence level
• ———————————————————
• Sample size
Qtab ___
•
4
0.76
•
5
0.64
•
6
0.56
•
7
0.51
•
8
0.47
•
9
0.44
•
10
0.41
——————————
Performing Q-test on Sample Data
• Consider the following set of data values:
0.5230, 0.5325, 0.5560, 0.5250, 0.5180, and 0.5270;
• Two questionable values are: 0.5180 & 0.5560 (the lowest and
highest values in the group)
• Perform Q-test at 90% confidence level on 0.5180:
| 0.5180- 0.5230| 0.0050

 0.13 0.56
(0.5560- 0.5180) 0.0380
• Qcalc. = 0.13 < 0.56
• (limit at 90% confidence level for sample size of 6)
• We keep 0.5180.
Performing Q-test on questionable value
• Calculate rejection quotient for 0.5560:
| 0.5560- 0.5325| 0.0235

 0.618 0.56
(0.5560- 0.5180) 0.0380
• Qcalc. = 0.618 > 0.56
• (limit at 90% confidence level for a sample of 6 is
0.56)
• We reject 0.5560.
Calculate the mean using acceptable values
(0.5230 0.5325 0.5250 0.5180 0.5305)
Mean, X 
5
2.6290

 0.52580
5
• Re-write the mean value to be consistent with the precision:
• Mean = 0.526 ± 0.005
Calculating Standard Deviation
__
Xi
(Xi - X )
0.5230
0.5325
0.5250
0.5180
0.7270
-0.0028
0.0067
-0.0008
-0.0078
0.0012
__
(Xi - X )
2
• 
•
•
•
•
•
__
7.8 x 10-6
4.5 x 10-5
6.4 x 10-7
6.1 x 10-5
1.4 x 10-6
S = 1.16 x 10-4
S( X i - X ) 2
1.16x 10-4
S

 0.00539 0.005
(n - 1)
4
Mean value must be consistent with the precision
Standard deviation:
1.
2.
3.
4.
should have one significant digit only;
It shows where the uncertainty appears in mean value;
That is, which digit on the mean contains error;
The mean value should be rounded off at the digit where
it becomes uncertain.
Thus, the mean consistent with the precision will be:
0.526 ± 0.005
(the mean value is precise up to the third decimal place)
Problem Solving by Dimensional Analysis
• Value sought = value given x conversion factor(s)
Example:
How many kilometers is 25 miles? (1 mi. = 1.609 km)
Value sought: ? km; value given = 25 miles;
conversion factor: 1 mi. = 1.609 km
? km = 25 mi. x (1.609 km/1 mi.) = 40. km
Unit Conversions
(1) Express 26 miles per gallon (mpg) to
kilometers per liter (kmpL).
(1 mile = 1.609 km and 1 gallon = 3.7854 L)
(Answer: 11 kmpL)
(2) If the speed of light is 3.00 x 108 m/s, what is
the speed in miles per hour (mph)?
(1 km = 1000 m and 1 hour = 3600 s)
(Answer: 6.71 x 108)
Temperature
•
Temperature scales:
1. Celsius (oC)
2. Fahrenheit (oF)
3. Kelvin (K)
Reference temperatures: freezing and boiling point
of water:
Tf = 0 oC = 32 oF = 273.15 K
Tb = 100 oC = 212 oF = 373.15 K
Temperature Conversion
• Fahrenheit to Celsius:
(T oF – 32 oF) x (5oC/9oF) = T oC
Example: converting 98.6oF to oC;
(98.6 oF – 32 oF) x (5oC/9oF) = 37.0 oC
Temperature Conversion
• Celsius to Fahrenheit:
ToC x (9oF/5oC) + 32 oF = T oF
Example: converting 25.0oC to oF;
25.0 oC x (9oF/5oC) + 32 oF = 77.0 oF
Temperature Conversion
• Celsius to Kelvin: T oC + 273.15 = T K
• Kelvin to Celsius: T K – 273.15 = T oC
Examples:
25.0 oC to Kelvin = 25.0 + 273.15 = 298.2 K
310. K to oC = 310. – 273.15 = 27 oC
Temperature Conversion
1) What is the temperature of 65.0 oF expressed in
degrees Celsius and in Kelvin?
(Answer: 18.3 oC; 291.5 K)
2) A newly invented thermometer has a T-scale that
ranges from -50 T to 300 T. On this thermometer, the
freezing point of water is -20 T and its boiling point is
230 T. Find a formula that would enable you to
convert a T-scale temperature to degrees Celsius.
What is the temperature of 92.5 T in Celsius?
(Answer: 45.0 oC)
Density
• Density = Mass/Volume
(Mass = Volume x density; Volume = mass/density)
Units: g/mL or g/cm3 (for liquids or solids)
g/L (for gases)
SI unit: kg/m3
Examples: density of water = 1.00 g/mL (1.00 g/cm3);
in SI unit = 1.00 x 103 kg/m3
Determining Volumes
• Rectangular objects: V = length x width x thickness;
• Cylindrical objects: V = pr2l (or pr2h);
• Spherical objects: V = (4/3)pr3
• Liquid displacement method:
the volume of object submerged in a liquid is equal
to the volume of liquid displaced by the object.
Density Determination
Example-1:
A cylindrical metal rod that is 1.00 m long and a
diameter of 1.50 cm weighs 477.0 grams. What is the
density of metal?
Volume = p(1.50 cm)2 x 100. cm = 177 cm3
Density = 477.0 g/177 cm3 = 2.70 g/cm3
Density Determination
Example-2:
A 100-mL graduated cylinder is filled with 35.0 mL
of water. When a 45.0-g sample of zinc pellets is
poured into the graduate, the water level rises to 41.3
mL. Calculate the density of zinc.
Volume of zinc pellets = 41.3 mL – 35.0 mL = 6.3 mL
Density of zinc = 45.0 g/6.3 mL = 7.1 g/mL (7.1 g/cm3)
Density Calculation #1
• The mass of an empty flask is 64.25 g. When filled
with water, the combined mass of flask and water is
91.75 g. However, when the flask is filled with an
alcohol sample the combined mass is found to be
85.90 g. If we assume that the density of water is 1.00
g/mL, what is the density of the alcohol sample?
(Answer: 0.787 g/mL)
Density Calculation #2
A 50-mL graduated cylinder weighs 41.30 g when
empty. When filled with 30.0 mL of water, the
combined mass is 71.25 g. A piece of metal is
dropped into the water in the graduate, which
causes the water level rises to 36.9 mL. The
combined mass of graduate, water and metal is
132.65 g. Calculate the densities of water and
metal.
(Answer: 0.998 g/mL and 8.9 g/mL, respectively)
Classification of Matter
Matter
Homogeneous
Mixture
Heterogeneous
Mixture
Pure
Substances
Elements and
Compounds
Classification of Matter
• Mixture: matter with variable composition
• Homogeneous mixture:
One that has a uniform appearance and composition
throughout the mixture;
• Heterogeneous mixture:
One that has neither uniform appearance or composition –
the appearance and composition in one part of the mixture
may differ from the other part;
• Substance: matter with a fixed composition
Substances
• Element:
Composed of only one type of atoms – it cannot be
further reduced to simpler forms.
• Compound:
Composed of at least two different types of atoms
combined chemically in a fixed ratio – it may be
broken down into simpler forms (or reduced to the
elements)
Physical Changes
Examples:
1. melting,
2. freezing,
3. evaporation,
4. condensation,
5. sublimation,
6. dissolution.
Chemical Changes
Examples:
1. combustion (burning),
2.
3.
4.
5.
6.
decomposition,
chemical combination (synthesis),
fermentation,
corrosion,
oxidation and reduction,
(any chemical reactions)