# Chemistry Study Guide

Significant Digits

Rules for significant digits.

1. Non-zero numbers are always significant.

2. Leading zeros are never significant.

•   0.2 has a one significant digit
•   0.0032 has two significant digits

3. Trapped zeros (zeros that are between two integers) are always significant

•  202 has three significant digits
• 2012 has four significant digits

4. Final zeros (to right of the final integer) have a special rule. If there is a decimal point the zeros are significant. If there is not a decimal point, the zeros are not significant

• 200 has one significant digit
•  200. has three significant digits
•  200.0  has four significant digits

5. For Addition/Subtraction line up the decimal points and add or subtract down until you reach the first non-significant digit of any of the numbers.

23.48+1.319=?

Set it up as: This means that the final answer would be 24.80 after you round. Make sure to keep the trailing zero after the decimal since it is significant.

6. For Multiplication/Division simply do the operation and then round your answer to the lowest number of significant digits in the original numbers.

• 78.2 x 12 =?
• 78.2 has three significant digits while 12 was two. So our answer will have two.
• 78.2 x 12 = 938.4 = 940

Example:

How many significant digits does 0849.050 have?

Solution:

0849.050 has six significant digits because the leading zero is never significant while the trapped zero always is and the tailing zero is significant because there is a decimal point.

Example:

Round each number to three significant digits:

1. 50.9415
2. 0.012182
3. 1.00794
4. 131.29300
5. 0200.00

Solution:

1. 50.9
2. .00122
3. 1.01
4. 131
5. 200

Concentration

Concentration is the measure of a specific compound per volume or mass of either a solution or a solvent.

The two most common forms of concentration are Molarity (M) and Molality (m).

Molarity is moles of solute per liter of solution Molality is moles of solute per kilogram of solvent Conversion Dimensional Analysis

If you want to convert between units, we have to have all units be the same in order to add, subtract, multiply, or divide. This is called dimensional analysis.

The objective when doing dimensional analysis is to find a way to relate quantities through ratios. These ratios can be multiplied by other ratios from either unit conversion tables or given information in the problem to cancel unwanted units. A basic understanding of the algebraic principle of cross fraction cancellation is needed. The end goal is to get a value in the desired
units.

Example 1:

What is the density of HCl in kg/m<sup>3 of a 1.1 M solution of HCl?

(The molar mass can be found on the periodic table as 36.46 g HCl/ 1 mol) *Note that density = mass/volume Example 2:

Given the following balanced chemical reaction and 7 grams of NaBr find the number of grams of Br<sub>2 produced.

(The molar mass of NaBr can be found by adding up the mass of Na and Br from the periodic table. It comes out to be 102.89 g NaBr/mol. It is important to remember to switch from moles of NsBr to the melon of Br<sub>2</sub> by using the mole ratio from the chemical equation.) Example 3:

Given the following balanced chemical reaction, 1.5 L of methane in excess oxygen, and that the density of methane is 0.717 kg/m<sup>3 determine the moles of H<sub>2</sub>O produced by the combustion reaction.

(The molar mass of methane can be found by adding up the mass of C and four h from the periodic table. It comes out to be 16.04 g methane/mol. It is important to remember to switch from moles of methane to the moles of water by using the mole ratio from the chemical equation.) Balancing Chemical Equations

Step 1: Count the number of each element on each side.

Step 2: Balance elements appearing the least on each side.

Step 3: Repeat Steps 1 and 2 until all remaining elements that appear more often are also balanced.

Step 4: Check that every element appears the same number of times on either side of reaction.

Example:

• Balancing the following  chemical reaction: Solution:

Step 1: There are 19 C, 40 H, and 2 O on the left and 1 C, 2 H, and 3 O on the right shown by the table below.

Step 2: Both C and H only appear in one molecule on either side of the equation so we will balance them first which gives us the equation: Step 3: The C’s and H’s are balanced on each side, now we can look at the remaining oxygen. There are some multiple of 2 on the left and 58 on the right. Since 58 is a multiple of 2, we know how many oxygen gas molecules we have as shown below: This example can be shown with a chart, as first counting the initial number of elements and proceeding to balance them on each side. Step 4: Recount and make sure it is balanced. A recount shows 19 carbon on each side, 40 hydrogen on each side, and 58 oxygen an each side which means we have finished the problem.

The Ideal Gas Law

PV = nRT

In words, the pressure multiplied by the volume of a contained system is equal to the number of moles of gas at constant temperature times the gas constant. Pay close attention to units on the numbers. All units should be consistent. Temperatures is measured in degrees Kelvin (K).

Example:

A bag of chips is packaged and sealed under standard conditions (1 atm, 298 K). The bag has 0.45 L of gas and can hold up to 0.5 L of gas in addition to the chips before bursting. If you were to leave the bag of chips in your car on a hot day, what temperature (K) could your car get to before the bag explodes? What temperature is this in degree Fahrenheit?

Solution:

Identify values that do not change. P, n, and r are all constant. Charles’s Law Identify known values fro problem: Original V is .45 L, final V is .5 L, and original T is 298 K.

Rearrange equation for unknown. Solve: P<sub>2 = .45 L * 1 atm/.5 L = .9 atm

9 atm > .333 atm so yes, it will explode.

Colligative Properties

A colligative property is something that occurs regardless of the identity of particles in the solute but only the quantity of these particles. Colligative properties include:

• Lowering Vapor-Pressure
• Depressing Freezing-Point
• Elevating Boiling Point

The applications to the colligative properties are quite numerous. Antifreeze in an automobile engine lowers the freezing point to prevent lines freezing and keeps the engine from boiling over as well.

Example:

In the winter salt is put down to lower the freezing point of water to melt ice on sidewalks and roadways. Would sugar work to melt the ice as well? Which would be better to use on roads, calcium chloride or sodium chloride and why?

Solution:

Yes, sugar will dissolve in water and the presence of any dissolved particles will lower the freezing point and melt the ice. The main difference is how many ions are freed in solution per mol to solvent. Glucose, a natural sugar in plants, stays as a single particle in solution. The calcium chloride salt dissolves into a calcium ion and two chloride ions which means for every 1 mol of calcium chloride there are three moles of ions that lower the freezing point of ice. This is more effective than two ions from dissociation of sodium chloride.

Percent Composition

Percent composition can be used to determine the empirical formula for a substance created in a chemical reaction. An empirical formula is the smallest ratio of individual atoms in the compound. The molar mass of the substance can be experimentally determined and using the empirical formula we can determine the molecular formula.

Example:

You are given an unknown compound and determine that it contains 26.58% K, 35.35% Cr, and 38.07% O by mass and has a molecular mass of 292.4 g/mol. What are the empirical and molecular formulas?

Solution:

Step 1: Convert to grams (it is easiest to assume a 100 gram sample).

26.58%*100=26.58 g of K

35.35%*100=35.35 g of Cr

38.07%*100=38.07 g of O

Step 2: Convert to moles of each component with the known molar mass (g/mol) of each element.

26.58 g of K/ 39.10 g/mol = 0.6798 mol K

35.35 g of Cr / 52.00 g/mol = 0.6798 mol Cr

38.07 g of O / 16.00 g/mol = 3.5 mol O

Step 3: Divide by the smallest number in Step 2. In this case we have a tie.

0.6798 mol K / .6798 = 1 mol K

0.6798 mol Cr / .6798 = 1 mol Cr

2.379 mol O / .6798 = 3.5 mol O

Step 4 (empirical formula): Multiply the by the smallest whole number that will bring everything to a whole number. In this case we multiply by 2.

1 mol K *2 = 2 mol K

1 mol Cr*2 = 2 mol Cr

3.5 mol O*2 = 7 mol O

The empirical formula from this analysis is K<sub>2</sub>Cr<sub>2</sub>O<sub>7

Step 5: Find the molar mass of the empirical formula.

(39.10*2)+(52.00*2)+(16.00*7) = 294.2 molar mass which is what was determined experimentally. So the molecular formula is the same as the empirical formula.

Error Analysis

When studying a chemical reaction in a laboratory setting, the amounts of products do not always mirror the chemical equation predicts. This is called experimental error, and error analysis can be performed.

Example:

The boiling point of water was experimentally found to be 98.7 degrees Celsius. The actual boiling point of pure water is 100 degrees Celsius. What is the percent error in the measurement?

Solution:

Error = 98.7 – 100 = -1.3 degrees Celsius Periodic Table of Elements

The Periodic Table is arranged in order in increasing atomic number (and molecular weight) from left to right, up to down, as the bottom right hand corner with the highest atomic number.

There are 3 major groups: Metals, Nonmetals, and Metalloids (the elements that fall on the large black line)

The columns in the periodic table are called groups -Periodic Trends

Atomic Size: decreases from left to right and increases from top to bottom

Ionization Energy: increases from left to right and decreases from top to bottom

Ionic size: increases from top to bottom, from left to right it increases from group 1 through the metals and then decreases from group 3 through the noble gasses.

Electronegativity: decreases from top to bottom, from left to right it increases from left to right (excluding the noble gases, because they have full electron orbitals).

Nuclear charge: increases from top to bottom and increases from left to right.

Shielding: increase from top to bottom and is constant from left to right.  Nomenclature 1

The transition metals are the large group of elements in the periodic table whose charge can change depending on the situation. All the transition metal carry a positive charge. Below is a chart relating the ion symbol to both the stock name and the classical name of the ion. The stock name is commonly used is describing chemical names of compounds. Nomenclature 2

There are also several polyatomic ions that carry a net charge. These are ions with more than one atom in them. They are listed below:

Charge=1- Charge=2- Charge=1+ Nomenclature 3

Naming Compounds:

Naming ionic compound is relatively easy both using symbols and words.

When writing an ionic compound using symbols the total net charge of the molecule must equal zero. This can be achieved by combining ions that are equal in magnitude of charge but opposite in polarity. As you can see in the previous example the 2+ charge and the 2- charge add up to a net charge for the molecule of zero. The combining of ions is not always this easy. Sometimes the charges are not equal in magnitude. For this compound the magnitude of the charge for the two ions was not the same, so to balance the charge 2 ammonium ions are combined with one sulfur ion to get a net charge on the molecule of zero. A final example deals with balancing charge using subscripts on both ions. As you can see, there are subscriptions on both of the ions. This allows the net charge to be zero. Naming ionic compounds using words is simpler than naming by symbols. The positively charged ions (cation) always is written first followed by the negatively charged ion (anion). The cations begin with a capitol letter and the anions begin with  lower case letter. The polyatomic ions are written in the same manner they are written in the charts. The anions that contain a single atom always  has the suffix “-ide”. The following examples are the names of the compounds used above in the same order.

• Copper(II) sulfate
• Ammonium sulfide
• Iron(III) chromate

Nomenclature 4

Molecular Compounds:

When two non-metals form a molecular compound there are special rules for writing their names. Non-metals can combine in multiple ways with each other and the charge alone is not enough to identify the compound so we include prefixes to distinguish how many of each atom is present. The names for these non-metal compounds always end in -ide. It should be noted that if the atom name begins with a vowel, the vowel at the end of the prefix is OMITED from the molecular name. Also if there is only one atom for the first element in the molecular formula the mono- prefix is OMITED from the name. Examples below include instances of this rule. Acids and Bases

-Acids

Bronsted-Lowry definition: a substance that produces hydrogen ions in a solution

Lewis definition: a substance that acts as a proton donor

The measure of the strength of an acid is the ease of the substance to produce a hydrogen ion or donates a proton

-Bases

Bronsted-Lowry definition: a substance that produces hydroxide ions in aqueous solutions

Lewis definition: a substance that accepts protons

The strength is measured by the ease the base is able to produce a hydroxide ion or accepts a proton.

*The greater the pH for a solution is, the more basic the substance. A lower pH correlates with a stronger acid.

Acid/Base Reactions (pH)

The pH of a solution is defined by:

pH = -log([H+])

where [H+] is the concentration of hydrogen ions or hydronium.

The concentration of hydrogen ions in a solution are normally found using equilibrium constants (called Ka for acids or Kb for bases) when dealing with weak acids. A weak acid is one that does not completely dissociate into hydrogen ions and the associated conjugate base. For a strong acid, all of the hydrogen ions separate in the solution so the mole ratio of hydrogen ions to starting moles of acid can be used to directly determine the pH of the solution.

List of common strong acids (that should be memorized):

• Perchloric acid
• Hydrochloric acid
• Hydronium ion
• Hydroiodic acid
• Sulfuric acid
• Hydrobromic acid
• Nitric acid

Example 1:

Find the pH of a 0.001 M, 0.01 M HCl solution. 1 mole HCl = 1 mole H+ Example 2:

Calculate the pH of a 0.1 M aqueous formic acid, given a Ka = 1.8*10^-4

Knowing an acid dissociates to a base + an acid the following equation can be used: An ICE (Initial, Change, Ending) table can be used to help keep track of the total number of moles. The equilibrium constant, Ka, is given by; Complete the ICE table, we can then complete the Ka equation and solve for x.   x = [H+] = 0.004 M. Solving for pH use the -log[H+] = pH Example 3:

Calculate the pH of a 0.075 M aqueous solution of Again, we have a base dissociates to a weak base plus a strong acid. To find the correct number of moles and ICE table is used.  Replace the ending [E] molar values into the Ka equation. x = [H+] = 1.54*10^-9 M. Solving for pH use the -log[H+] = pH will get pH = 8.81.

Equilibrium Reactions

Many reactions that occur naturally also reverse themselves naturally under different circumstances. These reactions never finish completely. Rather than finding a limiting reagent to determine the final amounts of a product, a new method must be utilized for finding equilibrium amounts of a product. Chemical equilibrium is an expression used to describe the
situation when the rate the reaction is occurring in the forward direction is the same as the rate of the reaction occurring in the backwards direction.

-Equilibrium Constants

Using the most general form for a reaction equation we find that the following relationship holds true for equilibrium chemistry. This general formula may seem confusing so an example problem illustrating its use is given.

Example 1:

A mixture at equilibrium has the following measured concentration of gasses. [CO_2] = 0.828 M, [H_2] = 0.828 M, [CO] = 0.672 M, [H_2O] = 0.672 M. Find the equilibrium constant if the balanced equilibrium equation is: Solution 1: Example 2:

Given that [HCl] = 0.780 M, [O_2] = 0.42 M, [Cl_2] = 1.2 M, and Keq = 0.60; find [H_2O] for the following reaction: Solution 2: Rearranging to solve for [H_2O] gives the equation: Solubility Rules

Always Soluble

• Compounds involving Group IA metals, H+, and NH4+ ions.
• Chlorides, bromides, and iodides
• Exceptions: In presence of Cu+, Ag+, Hg 2+, and Pb 2+ they are insoluble
• Nitrates
• Acetates
• Exceptions: In presence of Ag+ or Hg 2+ acetates are insoluble.
• Chlorates
• Sulfates
• Exceptions: In presence of Ca 2+, Sr 2+, Ba 2+, Pb 2+ sulfates are insoluble.

-Always Insoluble

• Sulfides (S^2)
• Exception: If only found with group IA or IIA compounds or NH4+ sulfides are soluble
• Hydroxides (OH-)
• Exception: If only found with Group IA compounds or NH4+, Sr 2+, Ba 2+ hydroxides are soluble.
• Carbonate, Phosphates, and Sulfites
• Exception: If only found with Group IA compounds or NH4+ they are soluble

Covalent Bonding

Covalent bonding is the bonding of atoms through the sharing of electrons

Valance electrons are the electrons in an atom’s outermost orbital. The numbers of these electrons vary depending on the periodic column the atomic is in. These valence electrons determine the charge of an atom as well.

Every atom tries to attain enough electrons in their valence shell to be like a noble gas (8 valence electrons in most cases) so that they will be stable.

Atoms share their valence shell electrons to achieve a full valence shell, this is covalent bonding.

Covalent bonds situate themselves as far away from other bonds as possible around a central atom.

Strongest bonding between atoms. *This is a Lewis Dot Structure depiction of how electrons combine to make covalent bonds. Ionic Bonding

• Bonding between atoms or ions through the combination of charge
• Between metals and non-metals
• Relatively weak bonding compared to covalent bonding
• Ionic bonds tend to dissociate (ions separate) in solvents such as water or ether.

Hydrogen Bonding

• Hydrogen bonding is a form of weak bonding between molecules. It is an attractive force over the linear alignment of Nitrogen, Oxygen, and Fluorine with hydrogen and then another N, O, or F. This phenomenon occurs often in water.
• Much weaker than covalent or ionic bonds but stronger than dipole or Vander Waals bonds interactions.

Dipole Bonding

• Dipole bonding is between two molecules, or opposite charges (a positive and negative).
• Dipoles interact to make an attractive force between them in the same mechanism as ionic bonding does, but this attraction is much less powerful than ionic Bonding and hydrogen bonding.

Vander Waals Forces

• These are forces that occur when molecules are in close proximity with each other.
• This is the weakest force of all the bonds available.
• Small fluctuations in charge over regions of a molecule create temporary dipoles in a molecule. This can cause a near molecule to have an opposing fluctuation and the two molecules will have a small attraction for a short time like dipole bonding but much less powerful.

Partial Pressures

Partial pressures measure the pressure of a single component of gas in a mixture of gasses if that single component was in the same volume alone. Such as if there was a combination of O2 and N2 in a container, the partial pressure of O2 would be equal to the pressure of O2 at the same conditions if it was in the container alone. Example:

Find the total pressure of CO_2 if 0.5 moles of Carbon has a partial pressure of 4 atm combined with 0.5 moles of  O_2 at 2 atm in a closed system.

Solution:

number of moles Na+(n1) x P1 + number of moles Cl-
(n2) x P2 = Ptotal.

0.5* 4+ 0.5 *2= 3 atm

Rate Laws

• The rate of a reaction is the change in concentration of a reactant per unit of time.
• For reactions with a single reactant, the rate of the reaction follows the formula: Where ‘k’ is the constant for the rate reaction and n determines the                  order of the reaction (first, second, or zeroth order)

-For every order of rate reactions there exists an integrated form of the rate reaction that can be used to determine unknowns about concentration, time, or rate constant.

-Integrated First Order Rate Law -Integrated Second Order Rate Law -Integrated Rate Law for a Zero-Order Reaction *The [A] for these laws represents the concentration of the reactant. The ‘k’ represents the rate constant and the‘t’ is for time. The [A]o presents the initial concentration of the reactant.

For reactions with multiple reactions the rate is as follows: Where n, m, and p are respective rate determiners. The values for these can be determined experimentally using measured initial rates for various concentrations of initial reactants.

Redox Reactions

Redox Reaction: A reaction where the oxidation number of an element in a reacting species changes. This indicates that element has undergone either oxidation or reduction.

Example 1: -The redox reaction presented above is not balanced. When balancing a redox reaction both atomic species and charge must be balanced.
-There are two methods for balancing redox reactions. One is called the “oxidation-numberchange method”. The other is called the “half reaction method”.

Oxidation-number-change method

Example 2: First, we must assign oxidation numbers to each atom in the equation. Display the numbers above the atoms as shown below. Remember that the oxidation number simply correspond to the charge of each atom. The whole compound must be a zero charge unless it is an ion (charged). We need to find which atomic species are reduced and which are oxidized. We must also know the degree to which each is oxidized and reduced. Now the final, balanced equation can be written using the coefficients found on the previous step. *It is necessary that this final equation is balanced for both charge and atomic species.

-Half-Reaction Method

A half-reaction is an equation that shows just the oxidation or just the reduction in a redox reaction.

Example 1: First the original unbalanced equation must be written in ionic form. Now write separate half reactions: Balance atoms in the first half reaction: *The water and hydrogen ions had to be added to balance the oxygen atoms that were already in the equation. This is the way to balance in acidic solution. If the solution was basic a combination of water and hydroxide ions would need to be used.

-Balance atoms in the second half reaction: Add enough electrons to each side of each balanced half reaction so that the charge of each half reaction is balanced. *Now each half reaction is balance with respect to both atomic species and charge.

Now we must multiply each half reaction by an appropriate coefficient to make the number of electrons equal in both. Now add the two half reactions together. Now cancel terms appropriately that appear on both sides of the equation. Finally add the spectator ions from the original ionic form of the equation and balance. *This is our final balanced equation!

Calorimetry

The first key concept in calorimetry is that heat and temperature are two different things. Temperature is a measure of energy contained in a body while heat is the transfer of energy. Specific heat is the amount of energy it takes to raise the temperature of 1 gram of a substance 1 degree Celsius. We can use these two concepts to find out the energy output of a chemical reaction.

Example 1:

Your first task is to determine the specific heat of gold by using calorimetry. You take a beaker of hot water and place a 5 gram sample of gold into the water. You measure the temperature until it reads constant. You record that the initial temperature of the gold is 100° C. You are then given a calorimeter (probably a Styrofoam cup) that has 15 grams of water at 25° C. Note: The specific heat of water is 4.18 J/g°C and the heat capacity of the
calorimeter is 0.5 J/°C. You place the gold sample into the calorimeter and wait for the temperature to stabilize and then record that the new temp. is 25.76° C. Use the equation ∆Q=m*C*∆T to determine the specific heat of gold.

Solution 1: Example 2:

Your second task is to use your gold sample and determine the mass of an unknown sample of water using calorimetry. You are using the same calorimeter as before with some unknown amount of water at 25°C. You add the 5 gram sample of gold at 100°C to the water sample. When the temperature stabilizes the final reading shows 25.46°C. Use ∆Q=m*C*∆T to
determine the mass of water present.

Solution 2: Molecular Geometry

When atoms are bonded to a central atom, they form specific shapes based on the number of atoms bonded to the central atom.

Linear: a linear configuration is where the two atoms bonded to the central atom are 180 degrees from each other in a line. Trigonal Planar: Three atoms connected to a central atom. Each bond angle is 120 degrees from the nearest other bond. All bonds and atoms are in the same plane of space. Tetrahedral: Four atoms bonded to a central atom. All bond angles are 109.5 degrees. This bonding configuration is very common for carbon and essential for organic chemistry. Trigonal Bi-pyramidal: Five atoms are bonded to a central atom. Three are bonded in the same plane around the “equator” of the molecule; these bond angles are 120 degrees. The remaining two atoms are bonded at the top and bottom (poles) of the molecule at a bond angle of 90 from the plane the other three atoms make. Octahedral: There are six atoms bonded to a central all evenly space from each other with bond angles of 90 degrees. This is a relatively rare bonding arrangement. Orbital

Now that we have seen the more “macroscopic” view of a molecule we can look at how the bonds between atoms actually take place.

Covalent bonds between atoms share electrons from the valence (outer most) electron shell. Each valence shell has what is called an orbital. Each orbit has a specific shape where the electrons are most likely to be found. These are also called electron clouds.

The number of electrons a neutral atoms has corresponds with how many protons that element has (atomic number). These electrons fill various electron orbitals in the following order. To show the electron configuration of an element, this chart (shown below) works excellent as well as the periodic table. This atomic orbital labels (1s, 2p, 3d, 4f) with the number of electrons assigned to each orbital (or set of orbitals sharing the same label) placed as a superscript. Example 1: Hydrogen
Solution: Hydrogen has one electron in the s-orbital of the first shell, so its configuration is written .

Example 2: Sulfur
Solution: when following the arrows above it’s electron configuration can be written as  .

Each s, p, d, and f orbital has a specific shape.

S-orbitals and P-orbitals D-orbitals 