# Chapter 1 Chemistry: The Study of Change

## Chemistry

Study of "everything" — central science

Examples:
drugs
textiles
fabrication of computer chips
metabolism
many many more

My definition: "Study of behavior of e- in atoms and molecules"

These are dissolving (or smearing) quickly because of collaboration and new research

 Analytical Quantitation of chemicals — many instrumentations Biochemistry Study of biological processes Inorganic Study of molecules containing all but carbon Organic Study of molecules containing carbon Physical Understand chemistry through physical means

## Scientific Method

Scientific Method — obtained based upon experiments or observations

Processes in Scinece
Experimental observation tentative explanation hypothesis Law

Law explains empirical facts.

For example. You will learn about gases in Chapter 9. In that chapter, three facts emerge from experiments:

EnglishEquation
Pressure P is proportional to amount nP n
Pressure P is proportional to temperature TP T
Pressure P is proportional to recipical of volume1/VP 1/V
putting all these together

PV = nRT, Ideal Gas Law

Explains how gases behave.

Two levels of understanding: Law vs. Theory
Law = empirically derived
Theory = explains Law

Theory explains underlying mechanism at deeper and more fundamental level.

Therefore, colloquial "in theory" is not how we use theory in science. MUCH DEEPER!

For example: Carbon has 4 bonds Facts from experiments

Quantum mechanics explains why carbon has 4 bonds.

## Classification of Matter

Matter = occupies space and has mass

We can classify Matter into several levels, as shown in the following figure.

Substance = definite composition and distinct properties

Mixtures consist of Hetergenous and Homogeneous types. Heterogenous mixture is made up of two different and distinct phase separation between substances. For example, oil and vinegar layers of your favorite Italian dressing, have two distinct separation b/w oil layer and aqueous layer. On the other hand, homogenous mixture is like your softdrink before you open the bottle; can not distingush b/w water, sugar, and carbon dioxide, CO2. Furthermore, the sugar and CO2 are distributed throughout the water. Once you open the cap, however, it starts to fizzle, there you created phase separation, in this case you can consider it as heterogenous, but homogenous between water and sugar.

Pure substances can be classified by Elements and Chemical compounds. Chemical compunds are made up of one type of molecules. For example, water is a chemical compound having definite proportion, two Hydrogens and one Oxygen, represented by a chemical formula, H2O.

### Chemical Elements

Element (or atom) is the basic building block of chemistry.

Fundamental Substances — chemically unalterable or broken down to simpler substance

Over 90 different naturally occuring elements listed in the Periodic Table of Elements

Group = elements in the same column
Period = elements in the same row

Each group has similar chemical properties

For example, Carbon belongs to group 14 so as silicon. Both can form up to 4 bonds.

Majority of elements are metals, i.e. conducts electricity and has shiny colors

 Group 1 Alkali metals Group 2 Alkali Earth metals Groups 3 - 12 Transition metals Groups 13 - 18 Main group elements Elements 57 - 71 Lanthanides Elements 89 - 103 Actinides Group 15 Pnictogens Group 16 Chalcogens Group 17 Halogens Group 18 Noble gas

## Three States of Matter

Gas — No definite shape. Sparsely populated.
Liquid — No definite shape. Viscosity.
Solid — Definite shape. High density.

Example: Steam = gas, water = liquid, and ice = solid

## Properties of Substances

Two types:

Physical property — change occurs w/o changing chemical properties
Melting, boiling, density, solubility, among others
Chemical property — changes arrangement of atoms in molecule
From carbon, such as the pencil lead and fluorine gas C-F bond containing Teflon — very tough material used for coating on a cooking surface of a pan — resistant toward acid and base

Another way: Extensive vs. Intensive properties

Intensive — independent of amount (or size)
Density, melting point, boiling point — good for ID
Extensive — depends on amount (or size)
Mass, volume, energy

!!! One can make extensive property into intensive by by another extensive property.

Mass (extensive) Density (intensive) = Mass/Volume

## Measurement

In this class, all quantities measured follows the standard unit, called SI unit. The following table lists some of the important ones.

 Mass kilograms kg Lenth meter m Temperature Kelvin K Amount of substances mole mol Time second s Electrical Current ampere A Luminous Intensity candela cd

Scientific notation — used to report measured values
It is quite nuisance to write a very large or very small number by using all those zeros. For example, 0.000 000 888. This can be written compactly by using scientific notation. 0.000 000 888 is expressed as 8.88 x 10-7. In this notation, it is said to have 3 significant digits (these three numbers you have confidence in)

Prefixes for SI units

Larger prefix Smaller prefix
YottaY1024 decid10-1
ZettaZ1021 centic10-2
ExaE1018 millim10-3
PetaP1015 microμ10-6
TeraT1012 nanon10-9
GigaG109 picop10-12
MegaM106 femtof10-15
Kilok103 attoa10-18
Hectoh102 zeptoz10-21

Example: 1000 m = 1 km; 10 cm = 104μm, etc.

Below is a movie called "Powers of Ten" gives you perspectives on distance that span of 40 orders of magnitude! (The order is the number that is raised on 10. Thus, if you have the spans over 100 times or 102, then we say that it spans two orders of magnitude.)

### Unit Conversion

Converting the measured values into another unit. In this course, we will delve into this subject much deeper than just a simple conversion, such as

Conversion b/w mass: g kg
Conversion b/w length: mile cm
Temperature conversion: °F °C

$T( ∘ C )= 5 8 ( T( ∘ F )−32 )$

You can covert the temperature in the Celsius scale to what is known as the absolute temperature scale.

$T( K )=T( ∘ C )+273.15$ where K is called the Kelvin unit.

Example: What is the temperature at which both the Fahrenheight and Celcius scales have the same numerical value?

Since T(°F) and T(°C) are equal,

T(°F) = T(°C) = x

and using above equation, by substituting x, you get,

$x=\frac{5}{9}\left[x-32\right]$

Now, solve for x. You get x = -40. Therefore, -40°C = -40°F.

+ If you want to know how it is derived, click here.

We will come back to the unit conversion little later with a powerful technique, called Dimensional Analysis.

### Energy

Definition of energy: Capacity to do work or supply heat.

$E=w+q$ where w is the work, and q is the heat.

Another way, energy can be divided into kinetic energy, T, and potential energy, V, $E=T+V$ where $T= 1 2 m v 2$ with m = mass of object that is moving and its velocity is v.

The potential energy, V, is the energy depending on position. Intuitively, if I drop a ripe tomato from 30 cm in height vs. dropping from the roof of a building, you know you can get a bigger splatter by dropping from the rooftop. Before dropping, your tomato has potential energy (associated with gravity) that depended on the height.

Energy unit: we use joules (J) 1 J = 1 kg m2 / s2

Chemists like to use another unit of energy: calorie (cal) 1 cal = 4.184 J

Example: Which one has more kinetic energy, a 1400 kg car moving at 115 km/hr or a 12000 kg truck moving at 38 km/hr?

Calculate kinetic energy for both cases using $T= 1 2 m v 2$ In order to make the correct units, speed should be in the m/s unit. $T= 1 2 ( 1400kg ) [ ( 115km 1hr )( 1000m 1km )( 1hr 60min⁡ )( 1min⁡ 60s ) ] 2 =709313.2J=7.1× 10 5 J=7.1× 10 2 kJ T= 1 2 ( 12000kg ) [ ( 38km 1hr )( 1000m 1km )( 1hr 60min⁡ )( 1min⁡ 60s ) ] 2 =668518.5J=6.7× 10 5 J=6.7× 10 2 kJ$ So, the car has more kinetic energy, but they are close!

## Handling Numbers

There is always error associated with your measurement.

Systematic error — inherent error, for example, in instrument
Random error — imprecise measurement, for example, your shaky hands!

Precision — degree of reproducibility
How a series of measurements deviate from each other
when deviation is small, it is said to be precise.
Accuracy — how close a measurement is to the true value

### Significant Figuers

Degree of condidence — meaningful # of digits in reported values

Is it meaningful to say π=3.14159265359979? It really depends on how you carried out your measurement.

This is dependent on the number of digits that you are using in the calculations.

Before we move on, if the reported values appear as:
8.8 x 10-2 g, it is said to have two significant figures.
8.80 x 10-2 g, it is said to have three significant figures.
8.800 x 10-2 g is said to have 4 sig fig, and so on.

If a reported value is 800 m, is it 800 m ± 1 m? ± 10 m?
How would you report the following situation?

### Significant figures in calculations

Multiplication and division — A x B = C. Use significant figure of whichever the smaller one (A or B) to the value of C.
Addition and subtraction — A - B = C. Use the smaller of the two decimal places of A or B in reporting C.

### Rounding

If 0 - 4: If the first digit next to the last digit in the significant figure is 0 - 4, then remove all numbers smaller than the last significant digit.
2.4268 in 2 sig fig → 2.4

If 6 - 9: If the first digit next to the last digit in the significant figure is 6 - 9, then round up (add 1 to the last digit).
2.4268 in 3 sig fig → 2.43

If the first digit next to the last digit is 5, and more numbers trailing then round up.
2.424522 in 4 sig fig → 2.425

If the first digit next to the last digit is 5, and no number trailing then round down.
2.445 in 3 sig fig → 2.44

## Dimensional Analysis

Derived units — units that are composed of more than one fundamental unit. For example,

 Velocity m/s miles/hr Density g/mL kg/cm3 Concentration mole/L oz/gal Volume is a special case Volume m3 in3

We will do all kinds of conversions throughout this course. So, please follow closely!

We use dimensional analysis. The name of the game is to set up an equation such that you cancel out the given unit with conversion factors. It is more difficult to read it than to just do it!

The following figure shows the anatomy of dimesional analysis.

In this equation, the unit you are converting to appears on left with a question mark, and is equated to a certain given unit. In order for you to convert the given unit to desired one is to use a conversion factor, which usually comes from derived unit. If you place a right unit in the denominator in the parenthesis, you can cancel the unit with the given unit. By placing a unit in the denominator dictates what unit to put in the numerator.

Density and velocity are the typical examples of conversion factor. The former a has unit such as g/mL. Let's say that the density is 0.5 g/mL. It means for the purpose of dimensional analysis, 0.5g is equivalent to 1 mL. Or the speed limit on Major Deegan Expressway is 40 mi/hr in the Bronx. For our purpose, it has no deeper meaning, other than 40 mi is equivalent ot 1 hr. So, use derived units as your conversion factor.

Converting length

Example:What is 5.7 m in cm?

Since 1 m = 100 cm according to the prefix table I gave you above, so each meter is 100 cm. In dimenstional anaylsis, you write down the question in the mathematical form as,

?cm = 5.7 m

Now, you have to cancel the m with the denominator of the conversion factor, in this case 1 m = 100 cm. Then, $?cm=5.7m\left(\frac{100cm}{1m}\right)$

$?cm=5.7\overline{)m}\left(\frac{100cm}{1\overline{)m}}\right)=570cm$

So far, so good. Right?

Let's try using density.

Example:Say that the density of some metal is 22.0 g/cm3. How many g are there in 30.0 cm3?

OK, the definition of density is d = m/v, but we don't really care if we know that or not. Since the unit is given as g/cm3, we can convert g to cm3 and cm3 to g; any way we want it.

In our case, we are looking for gram quantity given the volume of 30.0 cm3. So, we set up our equation as,

?g = 30.0 cm3

Since we need to cancel the cm3, we place cm3 of the density, and the leftover goes to the numerator (22.0g). Then, $?g=30.0c{m}^{3}\left(\frac{22.0g}{1c{m}^{3}}\right)=660g$

Let's do little more complicated one.

Example:What is 570 cm in nm?

Again by using the prefix table, we know 1 m = 100 cm, and we know that 1 m = 109 nm.

As before we set up the equation as,

?nm = 570 cm

In order to cancel cm in the given unit, use use 1m=100cm conversion, $?nm=570cm\left(\frac{1m}{100cm}\right)$

We now have m unit left on the numerator, which doesn't match with our goal of conversion to nm. So, we add another parenthesis so that $?nm=570cm\left(\frac{1m}{100cm}\right)\left(\frac{{10}^{9}nm}{1m}\right)$

Now we are left with nm, which matches with the unit we're looking for.

All you need to do now is to compute the number. $?nm=570cm\left(\frac{1m}{100cm}\right)\left(\frac{{10}^{9}nm}{1m}\right)=570×{10}^{7}nm=5.70×{10}^{9}nm$

Example:How many miles are there in 320000 cm, if 1 mi = 1.609 km?

Since mile to km conversion is given, everything else can come from the prefix table.

?mi = 3200000 cm

to start our calculation. We have cm in the given, so 100 cm = 1 m, then m to km, then km to mi? Sounds like a plan, $?mi=320000cm\left(\frac{1m}{100cm}\right)\left(\frac{1km}{1000m}\right)\left(\frac{1mi}{1.609km}\right)=1.988813mi=2.0mi$

Considering the fact that 320000 cm = 3.2 x 105cm, and the 105 would cancel with 100 and 1000 in the denominator, hence you only need to calculate 3.2/1.609, which give you the answer.

Let's try a difficult one.

Example:How many km are there in 1 light year. A light year is the distance that light travel in a year. The speed of light is 2.998 x 108m/s.

The difficulty is not in the conversion, but to think about what is equated to what initially. It is clear that we are looking for distance in km. Then, what is the given unit? Since we know the speed of light which convert distance and time. We are looking for distance, then the given must be in the form of time. Yes, we should equate km to 1 year for which light travels.

?km = 1 yr

Then, I can set up the following, $?km=1yr\left(\frac{365days}{1yr}\right)\left(\frac{24hr}{1day}\right)\left(\frac{60\mathrm{min}}{1hr}\right)\left(\frac{60s}{1\mathrm{min}}\right)\left(\frac{2.998×{10}^{8}m}{1s}\right)\left(\frac{1km}{{10}^{3}m}\right)=9.42×{10}^{12}km$

## Real-World Problem Solving

Back of an envelope or order of magnitude estimation are used in real-life situations. These are the ball park figures of accurate numerical calculations. Such is useful when trying to figure out the desired value in the context of the discussion.

Using the example immediately above on how many km there are in 1 light year, we can see that 4*2*6*6*3 x 1013 in the numerator, and the denominator is 10-3. Thus combining the powers of 10, the value is 4*2*6*6*3 x 1010. So, 8*36*3 = 8*108 = 864 ≈ 900, and combining with the powers of 10, you get 900 x 1010 = 9 x 1012.