1. 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!!!!

2. 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.

3. Dimensional Analysis: Example A sample of an alloy has a density of 7.9 kg/cm 3. What is the density in mg/m 3 ? 

4. Dalton’s Atomic Hypothesis 1.All atoms of a particular element are identical 2.Atoms of different elements have different masses 3.A compound is a specific combination of atoms of more than one element 4.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!

5. The Periodic Table

6. 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

7. 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 2H 2 (g) + O 2 (g)  2H 2 O (l)

8. E: Moles and Molar Masses The Mole: The most important concept you’ve never heard about A mole of objects contains the same number of objects as there are carbon atoms in 12.0 g of Carbon-12 6.022x10 23 somethings/mole = Avogadro’s Number

9. 1 mole of any element always has 6.0221x10 23 atoms in it 1 mole of elephants always has 6.0221x10 23 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(N A ) where N A is Avogadros’ number

10. 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 

11. 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…

12. 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?)

13. 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 

14. 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.

15. 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 

16. 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 C 3 H 4 O 3. –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 C 6 H 8 O 6, C 12 H 16 O 12 or C 18 H 24 O 18 ? 

17. 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

18. Molarity  The symbol M is used to denote the molarity of the solution 1M NaCl = 1 mole NaCl per liter of H 2 O

19. 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: c 1 V 1 = c 2 V 2 Where: c 1 = Stock concentration V 1 = Volume removed from stock c 2 = Target conc of new sol’n V 2 = Volume of new solution 

20. 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)

21. Law of Conservation of Matter a) b) Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated 6 molecules of Cl 2 react with 1 molecule of P 4 3 molecules of Cl 2 react with 2 molecules of Fe

22. Example of Using Stoichiometric Coefficients

23. Balancing Chemical Reactions Let’s look at Oxide Formation Metals/Nonmetals may react with oxygen to form an oxide with the formula M x O y Example 1: Iron reacts with oxygen to give Iron (III) Oxide Fe (s) + O 2 (g) → Fe 2 O 3 (s)

24. 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) + O 2 (g) → Fe 2 O 3 (s) Result: Fe (s) + O 2 (g) → 2Fe 2 O 3 (s)

25. How do we Solve It? Then, balance the reactant side and make sure the number/type of atoms on each side balance. Fe (s) + O 2 (g) → 2Fe 2 O 3 (s) Balanced Equation: 4Fe (s) + 3O 2 (g) → 2Fe 2 O 3 (s)

26. How do we Solve It? Example 2: Sulfur and oxygen react to form sulfur dioxide. S (s) + O 2 (g) → SO 2 (g) Step 1: Look at the reaction. We lucked out! Balanced Equation: S (s) + O 2 (g) → SO 2 (g)

27. How do we Solve It? Example 3: Phosphorus (P 4 ) reacts with oxygen to give tetraphosphorus decaoxide. P 4 (s) + O 2 (g) → P 4 O 10 (s) Step 1: Look at the reaction. The phosphorus atoms are balanced, so let’s balance the oxygens. Balanced Equation: P 4 (s) + 5O 2 (g) → P 4 O 10 (g)

28. How do we Solve It? Example 4: Combustion of Octane (C 8 H 18 ). C 8 H 18 (l) + O 2 (g) → CO 2 (g) + H 2 O (g) Step 1: Look at the reaction. Then: –Balance the Carbons C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + H 2 O (g)

29. How do we Solve It? Step 2: Balance the Hydrogens C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + 9H 2 O (g) C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + H 2 O (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’)

30. How do we Solve It? Step 3 (cont’d): Balance the Oxygens 2C 8 H 18 (l) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O (g) C 8 H 18 (l) + 12.5O 2 (g) → 8CO 2 (g) + 9H 2 O (g) Step 4: Make sure everything checks out

31. Review of Balancing Equations