Working with units

• DO NOT GET CARELESS…EVER!– Always keep track of your units

• I will take points off of anything you give me if you leave the units off

• I will stare at you when you give me an answer without units until you put them in your answer

• IT REALLY IS THAT IMPORTANT!!!!

Dimensional Analysis

• You will frequently have to convert between units during your life (Yes, your entire life!)

• To do this, we will need to employ DIMENSIONAL ANALYSIS

• Step 1: What do you have? What do you need?• Step 2: What is the conversion factor?• Step 3: Setup a calculation that cancels your

given units and puts your target units on top.

Dimensional Analysis: Example

• A sample of an alloy has a density of 7.9 kg/cm3. What is the density in mg/m3?

Dalton’s Atomic Hypothesis1. All atoms of a particular element are

identical2. Atoms of different elements have different

masses3. A compound is a specific combination of

atoms of more than one element4. In a chemical reaction, atoms are neither

created nor destroyed; they exchange partners to produce new substances

These tenets were first proposed 202 years ago and they are all true!

The Periodic Table

C: Compounds

• Compounds. Huh?• A compound is an electrically neutral

substance that consists of two or more different elements with their atoms present in a definite ratio

• Compounds: Terminology– Binary: Consists of only 2 elements– Organic: Contains Carbon and hydrogen– Inorganic: No Carbon

Ionic and Molecular Compounds

• Ionic Compound: Ions form compound that is electrically neutral– Usually formed by the reaction of a metal and a nonmetal

Na(s) + Cl(g) NaCl(s)

• Molecular Compound: Binary molecular compounds are usually formed by the reaction of 2 nonmetals

2H2(g) + O2(g) 2H2O (l)

E: Moles and Molar Masses

• The Mole: The most important concept you’ve never heard about

A mole of objects contains the samenumber of objects as there are carbon atoms in

12.0 g of Carbon-12

6.022x1023 somethings/mole = Avogadro’s Number

• 1 mole of any element always has 6.0221x1023 atoms in it

• 1 mole of elephants always has 6.0221x1023 elephants in it.

• In equations, you will frequently see the number of moles represented by the variable “n”

• 1 kmol = 1000 mol• 1 mol = 1x10-6 mol• # of objects = n(NA) where NA is Avogadros’ number

Molar Mass• The molar mass is the mass of one mole of material

– Also referred to as molecular weight or formula weight for compounds

• The molar mass of an element is the mass of one mole of its atoms

• The molar mass of a compound is the mass of a mole of its molecules.

• Molar mass (M) is usually given in g/mole

Mass of Sample = nM

Determination of Chemical Formulas

• For hundreds of years, people knew that drinking willow bark tea would belp cure a headache

• When scientists identified a compound form Willow bark that was biologically active, they had to determine its structure before they could synthesize it.

• Determining the chemical formula is the first step in the structure determination process…

Chemical Formulas

• We will deal with 2 types of chemical formulas:

1. Empirical formula: Shows the relative numbers of each element present in the compound (Emp. Form. Of Hydrogen Peroxide?)

2. Molecular formula: The precise number of atoms of each element present in the compound (Mol. Form. Of Hydrogen Peroxide?)

Mass Percentage Composition

• The mass percentage composition of a compound tells us the percentage (by mass) that a given element comprises of the compound.

• We frequently determine the mass % composition using combustion analysis

Mass % Composition Summary:

Mass % composition is found by calculating the fraction of the total mass contributed by each element present in a compound and expressing the fraction as a percentage.

Determining Empirical Formula from Mass % Data

• To convert the mass % composition obtained from a combustion analysis into an empirical formula, we must convert the mass % of each type of atom into the relative number of atoms.– To do this, assume that we have 100g of

sample– The mass % will then be in grams

Determining Molecular Formulas

• Once we know the empirical formula, we need one more piece of information to determine the molecular formula: The Molar Mass

• Say we know the empirical formula of a compound is C3H4O3.– All we know about this compound at this point is the ratio of

the 3 elements. – We don’t know the exact number of each type of atom in the

molecule.– Is the Molecular Formula C6H8O6, C12H16O12 or C18H24O18?

G: Molarity

• Hands down one of the most important concepts you need to master is you are going to stay in the sciences. Period.

• The Molar Concentration, c, of a solute in solution is the number of moles of solute divided by the volume of the solution (in liters).– Also referred to as Molarity

Molarity

The symbol M is used to denote the molarity of the solution

1M NaCl = 1 mole NaCl per liter of H2O

G4: Dilutions• Frequently in the laboratory, you will need to make dilutions

from a stock solution.• This involves taking a volume from the stock and bringing it to

a new volume with solvent.• In order to perform these dilutions, we can use the following

equation:c1V1 = c2V2

Where: c1 = Stock concentrationV1 = Volume removed from stockc2 = Target conc of new sol’nV2 = Volume of new solution

Law of Conservation of Matter

• “Matter can neither be created nor destroyed” – Antoine Lavoisier, 1774

If a complete chemical reaction has occurred, all of the reactant atoms must be present in the product(s)

Law of Conservation of Mattera)

b)

•Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated

•6 molecules of Cl2 react with 1 molecule of P4

•3 molecules of Cl2 react with 2 molecules of Fe

Example of Using Stoichiometric Coefficients

Balancing Chemical Reactions

• Let’s look at Oxide Formation• Metals/Nonmetals may react with

oxygen to form an oxide with the formula MxOy

• Example 1: Iron reacts with oxygen to give Iron (III) Oxide

Fe (s) + O2 (g) → Fe2O3 (s)

How do we solve it?

• Step 1: Look at the product. There are 3 atoms of oxygen in the product, but we start with an even number of oxygen atoms.– Let’s convert the # of oxygens in the product

to an even number

Fe (s) + O2 (g) → Fe2O3 (s)

Result: Fe (s) + O2 (g) → 2Fe2O3 (s)

How do we Solve It?

• Then, balance the reactant side and make sure the number/type of atoms on each side balance.

Fe (s) + O2 (g) → 2Fe2O3 (s)

Balanced Equation: 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)

How do we Solve It?

• Example 2: Sulfur and oxygen react to form sulfur dioxide.

S (s) + O2 (g) → SO2 (g)

• Step 1: Look at the reaction. We lucked out!

Balanced Equation: S (s) + O2 (g) → SO2 (g)

How do we Solve It?

• Example 3: Phosphorus (P4) reacts with oxygen to give tetraphosphorus decaoxide.

P4 (s) + O2 (g) → P4O10 (s)

• Step 1: Look at the reaction. The phosphorus atoms are balanced, so let’s balance the oxygens.

Balanced Equation: P4 (s) + 5O2 (g) → P4O10 (g)

How do we Solve It?

• Example 4: Combustion of Octane (C8H18).

C8H18 (l) + O2 (g) → CO2 (g) + H2O (g)• Step 1: Look at the reaction. Then:

– Balance the Carbons

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

How do we Solve It?

• Step 2: Balance the Hydrogens

C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (g)

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

• Step 3: Balance the Oxygens– Problem! Odd number of oxygen atoms

• 12.5 Oxygens on reactant side– Solution: Double EVERY coefficient (even those with a value of ‘1’)

How do we Solve It?

• Step 3 (cont’d): Balance the Oxygens

2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)

C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)

• Step 4: Make sure everything checks out

Review of Balancing Equations