Working with units

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/cm3. What is the density in mg/m3? 

4. Dalton’s Atomic Hypothesis • All atoms of a particular element are identical • Atoms of different elements have different masses • A compound is a specific combination of atoms of more than one element • 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 2H2(g) + O2(g)  2H2O (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 6.022x1023 somethings/mole = Avogadro’s Number

9. 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

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

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 H2O 

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: c1V1 = c2V2 Where: c1 = Stock concentration V1 = Volume removed from stock c2 = Target conc of new sol’n V2 = 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 Cl2 react with 1 molecule of P4 • 3 molecules of Cl2 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 MxOy • Example 1: Iron reacts with oxygen to give Iron (III) Oxide Fe (s) + O2 (g) → Fe2O3 (s)

24. How do we solve it? Fe (s) + O2 (g) → Fe2O3 (s) • 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 Result: Fe (s) + O2 (g) → 2Fe2O3 (s)

25. How do we Solve It? Fe (s) + O2 (g) → 2Fe2O3 (s) • Then, balance the reactant side and make sure the number/type of atoms on each side balance. Balanced Equation: 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)

26. 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)

27. 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)

28. 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)

29. How do we Solve It? C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g) • Step 2: Balance the Hydrogens C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (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? C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g) • Step 3 (cont’d): Balance the Oxygens 2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g) • Step 4: Make sure everything checks out

31. Review of Balancing Equations