## 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"

**Traditional Fields of Chemistry**

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:

English | Equation |
---|---|

Pressure P is proportional to amount n | P ∝ n |

Pressure P is proportional to temperature T | P ∝ T |

Pressure P is proportional to recipical of volume1/V | P ∝ 1/V |

*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,
CO_{2}. Furthermore, the sugar and CO_{2} 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,
H_{2}O.

### 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 | |||||
---|---|---|---|---|---|---|

Yotta | Y | 10^{24} | deci | d | 10^{-1} | |

Zetta | Z | 10^{21} | centi | c | 10^{-2} | |

Exa | E | 10^{18} | milli | m | 10^{-3} | |

Peta | P | 10^{15} | micro | μ | 10^{-6} | |

Tera | T | 10^{12} | nano | n | 10^{-9} | |

Giga | G | 10^{9} | pico | p | 10^{-12} | |

Mega | M | 10^{6} | femto | f | 10^{-15} | |

Kilo | k | 10^{3} | atto | a | 10^{-18} | |

Hecto | h | 10^{2} | zepto | z | 10^{-21} | |

Deka | da | 10^{1} | yocto | y | 10^{-24} |

Example: 1000 m = 1 km; 10 cm = 10^{4}μm, etc.

### 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

and

** 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.

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
$$T=\frac{1}{2}m{v}^{2}$$
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 m^{2} / s^{2}

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 $$T=\frac{1}{2}m{v}^{2}$$ In order to make the correct units, speed should be in the m/s unit. $$\begin{array}{l}T=\frac{1}{2}\left(1400kg\right){\left[\left(\frac{115km}{1hr}\right)\left(\frac{1000m}{1km}\right)\left(\frac{1hr}{60\mathrm{min}}\right)\left(\frac{1\mathrm{min}}{60s}\right)\right]}^{2}=709313.2J=7.1\times {10}^{5}J=7.1\times {10}^{2}kJ\\ T=\frac{1}{2}\left(12000kg\right){\left[\left(\frac{38km}{1hr}\right)\left(\frac{1000m}{1km}\right)\left(\frac{1hr}{60\mathrm{min}}\right)\left(\frac{1\mathrm{min}}{60s}\right)\right]}^{2}=668518.5J=6.7\times {10}^{5}J=6.7\times {10}^{2}kJ\end{array}$$ 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,**then round up.

*and more numbers trailing*- 2.424522 in 4 sig fig → 2.425

**If the first digit next to the last digit is 5,**then round down.

*and no number trailing*- 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/cm^{3} |

Concentration | mole/L | oz/gal |

Volume is a special case | ||

Volume | m^{3} | in^{3} |

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.

Let's start with something very simple.

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/cm

^{3}. How many g are there in 30.0 cm

^{3}?

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/cm^{3}, we can convert g to cm^{3} and cm^{3} to g; any way we want it.

In our case, we are looking for gram quantity given the volume of 30.0 cm^{3}. So, we set up our equation as,

^{3}

Since we need to cancel the cm^{3}, we place cm^{3} 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 = 10^{9} 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. $$?nm=570cm\left(\frac{1m}{100cm}\right)\left(\frac{{10}^{9}nm}{1m}\right)=570\times {10}^{7}nm=5.70\times {10}^{9}nm$$

How about this?

** 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 10^{5}cm, and the 10^{5} 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 10

^{8}m/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, $$?km=1yr\left(\frac{365days}{1yr}\right)\left(\frac{24hr}{1day}\right)\left(\frac{60\mathrm{min}}{1hr}\right)\left(\frac{60s}{1\mathrm{min}}\right)\left(\frac{2.998\times {10}^{8}m}{1s}\right)\left(\frac{1km}{{10}^{3}m}\right)=9.42\times {10}^{12}km$$

## 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 10^{13} in the numerator, and the denominator
is 10^{-3}. Thus combining the powers of 10, the value is 4*2*6*6*3 x
10^{10}. So, 8*36*3 = 8*108 = 864 ≈ 900, and combining with the
powers of 10, you get 900 x 10^{10} = 9 x 10^{12}.