No Slide Title

Download Report

Transcript No Slide Title

CHAPTER
THE LANGUAGE OF CHEMISTRY
Objectives

Elements, compounds, and mixtures

The metric system and conversions

Significant figures

Scientific notation

Unit-conversion method

Mass, volume, density, temperature, heat
1.1 The Composition of Matter

Chemistry is a science that studies the composition
and properties of matter




Matter is anything that takes up space and has mass
Mass is a measure of the amount of matter in a sample
Chemistry holds a unique place among the sciences
because all things are composed of chemicals
A knowledge of chemistry will be valuable whatever
branch of science you study
Material Properties


Characteristics or properties of materials distinguish one
type of substance from another
Properties can be classified as intensive or extensive



Intensive properties are independent of the amount of
substance, such as color and density
Extensive properties dependent on the amount of substance,
such as mass and volume
Properties can be classified as physical or chemical


Physical properties can be observed and measured without
changing the chemical identity of the substance, such as melting
point, color, density, odor...
Chemical properties involve a chemical change and result in
different substances, such as flammability
Physical property
Density
Chemical property
Chemical Reaction
Iron + Chlorine  Iron Chloride
Lithium
Oil
Water
Physical Change and Chemical Change

Physical Change: the substance changes in its

Chemical Change, also called Chemical Reaction,
physical appearance without altering its identity,
such as change of physical state from liquid to gas,
or from liquid to solid.
the substance is transformed into chemically
different substances:

Decomposition of H2O into its elements H2 and O2
Physical change
Chemical change
Iron + Oxygen  Rust
Sugar + heat  Caramel
Wood + fire  Ashes
Ways of classifying matter

Two of the principal ways of classifying matter are
according to:
1. its physical state (gas, liquid, or solid) called
state of Matter
2. its composition as pure substance, or mixture.
Pure Substances



Pure Substances are matter with fixed composition
and distinct properties.
Water and salt are pure substances
We can classify pure substances as either elements
or compounds…
Elements

Elements are substances that cannot be
decomposed by chemical means into simpler
substances:


Sodium, Chlorine, Carbon…
Each element is assigned a unique chemical symbol




Most are one or two letters
First letter is always capitalized
All remaining letters are lowercase
Names and chemical symbols of the elements are listed
on the inside front cover of the book
Some Chemical Symbols
Compounds



Compounds are substances formed from two or
more different elements combined in a fixed
proportion by mass
The physical and chemical properties of a
compound are, in general, different than the
physical and chemical properties of the elements of
which it is comprised
Elements and compounds are examples of pure
substances whose composition is the same,
regardless of the source
Sample Elements and Compounds
Hg
Br
I
C12H22O11
Cu
NaHCO3
Cu
Red P
Cd
Mixtures


Mixtures consist of varying amounts of two or more
elements or compounds, also called components
Homogeneous mixtures, also called solutions, are
uniform throughout


Air (gaseous), Gasoline (liquid), and Brass (solid) are all
solutions
Heterogeneous mixtures consist of two or more phases.
They do not have uniform composition, properties and
appearance.

Salad dressing, Sand, Wood are heterogeneous mixtures
Separating Mixtures

Solids can be separated
from liquids using




Liquids can be separated
using



Filtration
Centrifugation
Decantation
Decantation (if immiscible)
Distillation (if miscible)
Use physical properties to
separate substances:



Solubility
Sublimation
Magnetism…
Summarizing the Categories of Matter
Concept Checklist






Elements and compounds
Pure substances and mixtures
Homogeneous and heterogeneous mixtures
Physical changes / properties
Chemical changes / properties
Chemical symbols
Exercises

Physical or chemical properties?





Water boils at 100°C
Oxygen reacts with iron to form rust
Magnesium dissolves in hydrochloric acid
When heated in air, magnesium changes from a shiny
metal to a white powder that no longer conducts electricity
Pure substance, compound, or mixture?




Air
Table salt (sodium chloride)
Aluminum foil
Blood
Exercises

What is the chemical symbol?






Sodium
Chlorine
Iodine
Oxygen
Hydrogen
Which element is it?




Fe
Ag
N
C
Ne
F
Si
K
1.2 Measurement and the metric system

Quantitative observations help identify substances:





Measurements involve a comparison to a unit of
measurement



Numerical, how much…
Called measurements
Crucial in scientific problem solving
Necessary to test and refine hypotheses
6 feet tall, using the foot unit
A number without units is NOT a measurement
Measurements involve uncertainty depending on the
device used for the measurement
Measurements always include units

SI units are the standard in science


Based on the metric system. Standardized to ensure
compatibility b/w different countries
Conversion b/w units by moving decimal point
Derived SI units…

Area = length x width


Volume = LxWxD


(units for volume?) = mxmxm
 SI unit for volume is m3
Velocity = distance/time


(units for area?) = m x m  SI Unit for area is m2
(units for velocity?) = m / s
 SI Unit for area is m/s
Units undergo the same kinds of mathematical
operations that numbers do
Construct SI units of any convenient size using
decimal multipliers


Mathematical manipulation of the basic units
involves simply moving the decimal point: multiply or
divide by multiples of ten called decimal multipliers
Decimal multipliers are represented by Greek or
Latin prefixes:

kilo = 1000 times or 103  kilometer (km) = 1000 meters

deci = 1/10 times or 10-1  decigram (dg) = 0.1 grams (g)
Decimal multipliers that serve as SI prefixes
Exercises

Express:

2 ms in second (s)

5 cm in meters (m)

100 mL in liters (L)

1300 g in milligrams (mg)

0.813 kg in grams (g)
1.3 Measurement, Uncertainty, and Sig. Fig.




Mistakes are due to some preventable mix-up made
during measurement
Errors are due to inherent limitations in the
instruments and procedures used to make the
measurement
You can’t eliminate errors, but you can minimize
them when you know their source…
To have a scientific meaning, measurements need
to be reported with their degree of error (uncertainty
or variability):

4 ∓ 0.1g
4 ∓ 0.001g
4 ∓ 0.0001g
Estimating readings from a scale

Take 2 thermometers:

Left scale is 1° apart:




Right scale is 0.1 ° apart:



24 °C < T < 25 °C
Estimate to 1/10 °C of the smallest division
T is about 24.3 °C
We are certain of the 10th digit
We can estimate to 1/100 °C
Record all the digits, up and including the
first estimated digit:


Left thermometer: 24.3 °C
Right thermometer: 24.32 °C
Uncertainty in Measurements
Significant figures convention



More significant figures = Greater precision
How do you count significant figures?
When do zeros count:




Zeros to the right of the decimal count
Zeros to the left of the first nonzero digit never count
0.002040
0 does not count
0 counts
Zeros at the end of a number without a decimal point are
assumed not to be significant
45,000
Tip: use scientific notation…
Exercises

How many significant figures?

46.94

307

16.0

0.008

0.02030
1.4 Scientific Notation

Scientific notation, or exponential notation, simplifies
the manipulation of very large and very small
numbers:
(0.000000000000000000000000000000663 x 30,000,000,000) ÷ 0.00000009116
Can you keep track of all the zeros?
Same problem in scientific notation:
(6.63 x 10-31 x 3.0 x 10-10) ÷ 9.116 x 10- 8
Scientific Notation Tutorial
Writing in scientific notation



A number (N) is represented by a coefficient and an
exponent factor (10x)
The coefficient is based on the number (N) with the
decimal point moved right behind the first non-zero
digit (standard form)
The exponent factor (10x) is based on how the
decimal point was moved:



If moved to the left then (x) is positive
If moved to the right then (x) is negative
The value of (x) is equal to the number of places the
decimal point was moved
Examples
 233.
becomes 2.33 x 102
 0.0034
becomes 3.4 x 10-3
(1/X = X-1, and 1/s = s-1)
 3.
becomes 3 x 100
(100 = 1, it is always omitted)
Exercises


Express in scientific notation

0.0007068

8,234.1
Convert to non-exponential form

2.3 x 10-4

1.78 x 103
1.6 Calculations and Significant Figures

When several measurements are obtained in an
experiment they are usually combined in some way
to calculate a desired quantity.




Area of a rectangular carpet = length x width
If one of these measurements is very precise and the other
is not, we can’t expect too much precision in the calculated
area.
We follow certain rules according to the kinds of
arithmetic being performed
Ignore exact numbers (those based on counting not
measuring) when applying these rules
Multiplication and division

The number of significant figures in the answer
should not be greater than the number of significant
figures in the least precise measurement:
Addition and subtraction

The answer should have the same number of
decimal places as the quantity with the fewest
number of decimal places:
Rounding numbers

When rounding, look at the digit following the one
that is to be last:

If it is less than 5, drop it and all the following digits

If it is 5 or more, increase by 1 the number to be rounded

Retain all digits until your calculation is complete. Round
off the final result only.
Exercises

Calculate the volume of a cube that is 8.5cm on a
side:


Addition:


V=LxWxD= 8.5 x 8.5 x 8.5 = 614.125 cm3 = 610 cm3
1.9375 + 34.23 + 4.184 = 40.3515 = 40.35
Subtraction:

94.935 m – 7.6 m = 87.335 m = 87.3 m
1.7 The Unit-Conversion Method

The unit-conversion method, or dimensional
analysis, is the method used by scientists to help
them perform the correct arithmetic to solve a
problem.


Treat a numerical problem as one involving a conversion
of units from one kind to another.
To do this, we use one or more conversion factors to
change the units of the given quantity to the units of the
desired quantity:
(given quantity) x (conversion factor) = (desired quantity)
The unit-conversion method cntd.


A conversion factor is a fraction formed from a valid
relationship or equality between units and is used to switch
from one system of measurement and units to another.
To construct a valid conversion factor, the relationship
between the units must be true:



Incorrect: 3ft = 41 in
Correct: 1 in = 2.54 cm
We will use the unit conversion method extensively
Converting between English and Metric Units
What is wrong with the following setup?
1.4 day
x 1 day x
24 hr
60 min
1 hr
x 60 sec
1 min
The unit setup is what’s wrong
1.4 day
x 1 day x
24 hr
60 min
1 hr
Final Units = day2 sec/ hr2
Not the final units needed
x 60 sec
1 min
Steps to Problem Solving








Read problem
Identify data
Write down a unit plan from the initial unit to the
desired unit
Select conversion factors
Change initial unit to desired unit
Cancel units and check
Do math on calculator
Give an answer using significant figures
Exercises

Convert 74.1 mL into liters:


Find your initial units (mL) and your final units (L)
Use the correct conversion factor 1L = 1000 mL
1L
74.1mL x
 0.0741 L
1000mL

How many pounds are in 752.4 g?

How many seconds are in 7 days?
Solutions

There are 453.59g in a pound:
1 lb
752.4 g x
 1.659lb
453.59g

There are 60s in a minute, 60 minutes in an hour, 24
hours in a day:
24 hr 60 mn 60 s
7 days x
x
x
 604800s
1 day
1 hr
1 mn
1.8 Mass and Volume


Substances can be identified using their physical
and chemical properties
Two important physical properties:


Mass is a measure of the quantity of matter present in an
object, and can be measured on a balance or scale.
Volume is a measure of how much space it occupies. It
can sometimes be determined by measurement if the
object’s dimensions followed by a calculation of its volume
based on a suitable formula:


Volume (box) = L x W x H.
It would be incorrect to assume that small objects have low
mass, or that large objects have high mass
Units used in laboratory

Mass:



In the SI, the base unit for the measurement of mass is the
kilogram (kg),
The gram (g) is a more conveniently sized unit for most
laboratory measurements.
One gram is one thousandth of a kilogram:



1 kg = 1000 g, so 1 g must equal 0.001 kg
One milliliter of water has a mass of about 1 gram.
One liter of water has a mass of about 1 kilogram.
Devices used to measure mass
Triple beam balance
Modern analytical balance
Old tow pan beam balance
Learn to use the triple beam balance
Method to measure the mass of a liquid


The raw data collected are given as follows:
Empty
Container
Container &
Liquid
25 g
75 g
The mass of the liquid is determined by subtracting
the mass of the empty container from the mass of
the container with liquid:
Mass of the Liquid = 75 g - 25 g = 50 g
Units used in laboratory…

Volume:





Volume is a derived unit with
dimensions of (length)3
The derived SI unit for volume is the
cubic meter, m3
The traditional metric unit of volume
used in lab is the liter (L).
One liter is exactly 1000 cubic
centimeters or 1000 milliliters:
1L = 1000cm3 = 1000mL
Devices for measuring liquid volumes
Measuring volume by displacement
A solid displaces a matching volume of water when the solid
is placed in water
Volume (mL) of water displaced = 33 mL - 25 mL = 8 mL

Volume of object (cm3) = 8 mL x 1 cm3 =
1 mL
33 mL
25 mL
8 cm3
Exercises

Express a mass of 2.87kg in grams:



Find initial units (?) and final units (?)
Use the correct conversion factor: 1kg = ? g
Express the quantity 3.97L in milliliters


Find initial units (?) and final units (?)
Use the correct conversion factor: 1L = ? mL
Solutions

There are 1000g in a kg
1000g
2.87kg x
 2870g
1kg

There are 1000 mL in a L
1000 mL
3.97 L x
 3970 mL
1L
1.9 Density

Density compares the mass of an object to its
volume
mass
g
g
D


3
volume mL cm
Note: 1 mL = 1 cm3

You can see, by examining this formula, that objects
of high density will have a lot of mass occupying a
relatively small volume
More about density
Uses:
 Evaluating the purity of solids and liquids
 Estimating the amount of solid dissolved in a
solution (urine and kidney function)

Density increases as amount of solid dissolved increases
Properties:
 Volume depends on temperature, so density
depends on temperature as well


As temperature increases, volume increases (slightly), and
density decreases
Object floats on water if its density is less than that
of water, and vise a versa
Density Connections
Mass
Volume
kg
L
g
mL (cm3)
mg
Specific Gravity



Specific gravity is a unitless
connection between mass and
volume
The specific gravity of a substance
is defined as the ratio of the density
of the substance to the density of
water.
The specific gravity tells us how
much denser than water a
substance is.
1.10 Temperature




Particles are always moving. When you heat water, the
water molecules move faster.
When molecules move faster, the substance gets hotter.
When a substance gets hotter, its temperature goes up.
Temperature measures the hotness or coldness of an
object
Temperature is determined by using a thermometer that
contains a liquid that expands with heat and contracts
with cooling.
Fahrenheit Formula
Temperature Conversions
A person with hypothermia has a body temperature of
29.1°C. What is the body temperature in °F?
°F
=
1.8 (29.1°C) + 32
exact tenth's
exact
=
52.4 + 32
=
84.4°F
tenth’s
Exercises:
 What °F temperature is equivalent to 100 °C
 Convert 98.6 °F into both Celsius and Kelvin teperatures
1.11 Heat and Calorimetry




Heat is energy that flows from something warm
to something cooler
A hotter substance gives Kinetic Energy to a
cooler one
When heat is transferred (lost or gained), there
is a change in the energy within the substance
Heat is measured in calories or joules
1 kcal = 1000 cal
1 calorie = 4.18J
1 kJ = 1000 J
Learning Check


A. When you touch ice, heat is transferredfrom
1) your hand to the ice
2) the ice to your hand
B. When you drink a hot cup of coffee, heat is
transferred from
1) your mouth to the coffee
2) the coffee to your mouth
Learning Check

When you heat 200 g of water for 1 minute, the
water temperature rises from 10°C to 18°C.
400 g
200 g

If you heat 400 g of water at 10°C in the same pan
with the same amount of heat for 1 minute, what
would you expect the final temperature to be?
1) 10 °C
2) 14°C
3) 18°C
Specific Heat




Why do some foods stay hot longer than others?
Why is the beach sand hot, but the water is cool on
the same hot day?
Different substances have different capacities for
storing energy
It may take:
- 20 minutes to heat water to 75°C.
- 5 minutes to heat the same mass of aluminum
- 2 minutes to heat the same mass of copper
Specific Heat Values

Specific heat is the amount of heat needed to raise
the temperature of 1 g of a substance by 1°C
water
aluminum
glass
silver
olive oil
cal/g°C
1.00
0.22
0.163
0.057
0.471
J/g°C
4.18
0.90
0.68
0.24
1.97
Learning Check H3
A. A substance with a large specific heat
1) heats up quickly 2) heats up slowly
B. When ocean water cools, the surrounding air
1) cools
2) warms 3) stays the same
C. Sand in the desert is hot in the day, and cool at
night. Sand must have a
1) high specific heat
2) low specific heat
Measuring Heat
Requires

Grams of substance

Temperature change T = Tfinal – Tinitial

Specific heat of the substance

heat = mass x temp. change x specific heat
grams x
T
x
Cp
Heat Calculations

A hot-water bottle contains 750 g of water at 65°C. If
the water cools to body temperature (37°C), how
many calories of heat could be transferred to sore
muscles?
heat = g
x T
x Sp. Ht. (H2O)
750 g
x 28°C x 1.00 cal
g°C
= 21 000 cal
Heat Calculations

How many kcal are needed to raise the temperature
of 120 g of water from 15°C to 75°C?
1) 1.8 kcal
2) 7.2 kcal
3) 9.0 kcal
Calculating Specific Heat

What is the specific heat of a substance if 334J of
heat added to 52g of that substance cause its
temperature to rise from 16°C to 48°C?
Heat
Cp 
mass x T

Heat = 334J, mass=52g, ΔT = (48-16)°C = 32°C
Cp = 334 / (52x32) = 0.201 J / g.°C