1. DO NOW: 1. Define Stoichiometry 2. If 2 moles of nitrogen gas reacts to form ammonia, how many moles of hydrogen is required? Grams? How many moles of ammonia is produced? Grams?

2. Do Now: Compare and contrast 1. An element and a compound 2. A substance and a mixture 3. Homogeneous and heterogeneous mixtures 4. List 5 methods of separating mixtures or decomposing compounds

3. Elements and compounds Compare Contrast Both stable forms of matter Represented by symbols (elements) or formulas (compounds) Both have fixed composition Elements are the smallest stable form of matter that can exist by itself. Cannot be broken down into stable components Cannot be broken down by ordinary physical or chemical means Only one type of atom Displayed on periodic table Compounds are 2 or more types of atoms chemically bonded Properties significantly (usually) different from component elements Can be broken down into stable component atoms

4. Substances and Mixtures Compare Contrast Both forms of matter Both composed of atoms and molecules Substances have fixed composition Can be represented by symbols or formulas Mixtures are physically put together Not bonded Each substance retains physical and chemical properties (most) No fixed composition (varies)

5. Homogeneous and Heterogeneous Compare Contrast Both forms of matter Both physical combinations of component substances Both NOT chemically combined Components of both retain (most) physical and chemical properties Homogeneous Uniform composition (and properties) for given sample (different samples have differing composition and properties) Appears to be in single phase Heterogeneous Not uniform throughout Separate, distinct regions visible Varying properties and composition WITHIN same sample

6. Separation Methods Mixtures Compounds Filtration solid-liquid phase heterogenous only Distillation liquid-liquid, solid liquid homogenous. Based on differences in boiling point (volatitily) Chromatography Paper, liquid, gas Separation due to differences in polarity / mass / size / attraction/bonding between mobile and immobile media Mechanical Sieve/Mesh Magnets Physical Electrolysis (electric current) 2H 2 O  2H 2 + O 2 Heat 2Ag 2 O  4Ag + O 2 Reaction of compound with more active metal/non-metal Ca + CuCl 2  Cu + CaCl 2 Light 2H 2 O 2  2H 2 O + O 2

7. 3.1: ATOMIC MASS A. C-12, the Relative Standard 1. C-12 is assigned a mass of exactly 12 atomic mass units (amu) 2. Masses of all elements are determined in comparison to the carbon -12 atom (12C) the most common isotope of carbon 3. Comparisons are made using a mass spectrometer B. Atomic Mass (Average atomic mass, atomic weight) 1. Atomic masses are the average of the naturally occurring isotopes of an element 2. Atomic mass does not represent the mass of any actual atom 3. Atomic mass can be used to "weigh out" large numbers of atoms. Average atomic mass = Σ (% of each isotope)(atomic mass of each isotope) 100

8. Mass Spectoscopy Used to identify the varying masses of isotopes in a naturally occurring sample of any given element (also compounds--later) How it works Vaporized atoms are bombarded with electrons to form positive ions Positive ions are accelerated using an electric field Ions are deflected in the magnetic field according to their mass and charge (always positive) Lighter/more highly charged particles are deflected MORE Comparisons of % deflected at each point are made Detected and converted to a readable output (mass spectra)

9. Plots charge/mass ratio (x-axis) vs. relative abundance (y- axis) Peaks represents isotope with that mass and relative percent (consistent with relative abundance) Used to solve for average atomic mass Average Atomic Mass = Σ (% of each isotope)(atomic mass of each isotope) 100

10. Example What element is this likely to be? Analyze # isotopes? 4 Relative abudance? 88 most abudant Average atomic mass closest to? <88 Element id? Strontium

11. Example What element is this likely to be? Analyze – # isotopes? 10 – Relative abudance? 118 and 121 most abudant – Average atomic mass closest to? Between 118-121 – Element id? Tin

12. Note: Not to scale on either axis Identify diatomic element. Analyze # isotope? 3 Relative abundance 16 most abudant Most likely to be? oxygen Why so many peaks? -O 2 can form from any combination of the isotope 16-16 (32), 16-17 (33), 16-18 (34), 17-17 (34), 17-18 (35), 18-18 (36)

13. Is the above a complete/true mass spectrum of water? Why or why not? What other peaks/relative abundance would you expect? 2, 3 (Extremely tiny) 24, 23, 22, 21, 20, 19 (quite small)

14. 3.2: The MOLE A. Avogadro's number 1. 6.022 x 10 23 units = 1 mole 2. Named in honor of Avogadro (he did NOT discover it) B. Measuring moles 1. An element's atomic mass expressed in grams contains 1 mole of atoms of that element a. 12.01 grams of carbon is 1 mole of carbon (any random sample) b. 12 grams of carbon-12 is 1 mole of carbon-12 (specific isotope)

15. 3.3: MOLAR MASS A. Molar Mass (Gram molecular weight) 1. The mass in grams of one mole of a compound a. Mass of _______________ particles 2. The sum of the masses of the component atoms in a compound a. Molar mass of ethane (C2H6): Mass of 2 moles of C = 2(12.01 g) Mass of 6 moles of H = 6(1.008 g) 30.07 g 3. Examples: a. Determine the molar mass of calcium carbonate. b. A certain sample contains 4.50 moles of calcium carbonate. Determine the mass of the carbonate ion present.

16. 3.4: PERCENT COMPOSITION A. Calculating any percentage 1. "The part, divided by the whole, multiplied by 100" B. Percentage Composition 1. Mass of each element expressed as a percent of the total mass of the compound. 2. Calculate the percent of each element in the total mass of the compound C. Examples: 1. Determine percent composition of Carvone (C 10 H 14 O). 2. Determine percent composition of penicillin (C 14 H 20 N 2 SO 4 ). (#atoms of the element)(atomic mass of element) x 100 (molar mass of the compound)

17. 3.5: Determining the Formula of A Compound A. Empirical Formula: simplest whole number ratio of atoms of each element in a compound. 1. Entirely different and unrelated compounds, with entirely different molecular formulas may have the same empirical formula. 2. Example: Benzene, C 6 H 6 ; Ethene, C 2 H 2 ; 1,3,5,7- cyclooctatetraene, C 8 H 8 are very different compound but each have a 1:1 ratio of C to H atoms in their molecular formula, so all have an empirical formula of CH.

18. B. Determining the empirical formula: 1. Determine the percentage of each element in your compound 2. Treat % as grams, and convert grams of each element to moles of each element 3. Find the smallest whole number ratio of atoms. Find the smallest number of moles calculated and divide ALL the results of the calculations, by that number. NOTE: avoid rounding up or down too much at this stage, and be lenient with significant figures. 4. If the ratio is not all whole number, it should include a recognizable decimal (.25,.333,.500). Multiply each by an integer (4, 3, 2 as appropriate) so that all elements are in whole number ratio Example: Determine empirical formula of compound composed of 71.65% Cl, 24.27% C and 4.07% H.

19. C. Molecular formula: tells us exactly how many atoms of each element are present in the compound, rather than just the simplest whole number ratio. 1. Simple multiple of the empirical formula 2. Indicate ACTUAL formula 3. May or may not indicate information about the actual structure of category of compound (more later)

20. B. Determining the molecular formula 1. Find the empirical formula mass (of previously established empirical formula) 2. Divide the known (given) molecular mass by the empirical formula mass, deriving a whole number, n 3. Multiply the empirical formula by n to derive the molecular formula Example: The molar mass of the sample (previous slide) is known to be 98.96 g/mol

21. 3.6: CHEMICAL EQUATIONS A. Chemical reactions 1. Reactants are listed on the left hand side 2. Products are listed on the right hand side 3. Atoms are neither created nor destroyed (conservation of mass) a. All atoms present in the reactants must be accounted for among the products, in the same number b. No new atoms may appear in the products that were not present in the reactants

22. B. The Meaning of a Chemical Reaction 1. Physical States a. Solid - (s) b. Liquid - (l) c. Gas - (g) d. Dissolved in water (aqueous solution) - (aq) 2. Relative numbers of reactants and products a. Coefficients give atomic/molecular/mole ratios WHY? multiplying ALL coefficients by the same value (avogadro’s number) does not affect the relationship or ratio.

23. 3.7 Balancing Chemical Equations A. Determine what reaction is occurring 1. It is sometimes helpful to write this in word form: Hydrogen + oxygen  water B. Write the unbalanced equation 1. Focus on writing correct atomic and compound formulas H2 + O2  H2O C. Balance the equation by inspection 1. It is often helpful to work systematically from left to right 2H2 + O2  2H2O D. Include phase information 2H2 (g) + O2 (g)  2H2O (l)

24. 3.8 Stoichiometric Calculations: Amounts of Reactants and Products A. Balance the chemical equation B. Convert value given (generally mass, volume, particles, density or other data) of reactant or product to moles C. Compare moles of the known to moles of the desired substance 1. A ratio derived from the coefficients in the balanced equation D. Convert from moles back to required value (mass, volume, particles, density or other)

25. 3.9 Calculations Involving a Limiting Reactant A. Concept of limiting reactant (limiting reagent): " I want to make chocolate chip cookies. I look around my kitchen (I have a BIG kitchen!) and find 40 lbs. of butter, two lbs. of salt, 1 gallon of vanilla extract, 80 lbs. of chocolate chips, 200 lbs. of flour, 150 lbs. of sugar, 150 lbs. of brown sugar, ten lbs. of baking soda and TWO eggs. It should be clear that it is the number of eggs that will determine the number of cookies that I can make." 1. The limiting reactant controls the amount of product that can form

26. B. Solving limiting reactant problems 1. Convert grams of reactants to moles 2. Use stoichiometric ratios to determine the limiting reactant 3. Solve as before, beginning the stoichiometric calculation with the grams of the limiting reactant C. Calculating Percent Yield 1. Actual (Experimental) yield - what you got by actually performing the reaction 2. Theoretical yield - what stoichiometric calculation says the reaction SHOULD have produced Actual Yield x 100% = percent yield Theoretical Yield