1. CHAP 3 Stoichiometry

2. Key terms Atomic mass – average mass of the atoms of an element. (aka average atomic mass) based on the standard mass of Carbon-12 Mass spectrometer – instrument used to determine relative masses of atoms by deflection in a magnetic field Mole (mol) – number equal to the number of carbon atoms in Carbon-12 Avogadro’s number – 6.022x10 23 units of a substance in 1 mole Molar mass – mass of 1 mole of a substance

3. Calculations Using dimensional analysis – units must be diagonal to cancel out. 1 mole = 6.022 x 10 23 particles (atom, molecules,etc) 1 mole = ? grams of element

4. Finding molar mass If element = mass on periodic table Sodium (Na) = 22.99 grams/mol If compound = # each element * mass on periodic table C 6 H 12 O 6 = C - 6 * 12.0 = 72.0 H - 12 * 1.08 = 12.96 O – 6 * 16.0 = 96 Total = 180.96 grams/mol If a hydrate – mass includes the mass of water times the coefficient.

5. Percent Composition Find how much of each element are present in a compound - %’s. Check to be sure percents add up to 100% Or mass percent of a particular ion or part of the compound. % by mass = Mass of element/ion Total mass of compound

6. Empirical Formulas (simplest formula) Start with grams of element – convert from % to grams if needed (based on 100 grams) Divide each gram by the molar mass of that element. Chose the smallest mole value and divide each mole value by the smallest. If whole numbers, these become subscripts for final formula. If decimal value that has easy fraction equivalent, multiply all by the denominator to make whole number. Then use as subscripts for formula.

7. Molecular Formula Molecular formula = (empirical formula) X X = molecular formula mass (molar mass) empirical formula mass then distribute or multiply each subscript on the empirical formula

8. Basic Reaction Types Combustion Reaction – hydrocarbon type compound reacts with oxygen to produce carbon dioxide and water C x H y + O 2  CO 2 + H 2 O Decomposition Reaction – (1 reactant) breaks down into simpler elements/compounds AB  A + B Synthesis reaction (combination) – simple elements/compounds combine to form a larger compound (1 product) A + B  AB Double displacement/replacement – ions switch partners AD + BF  AF + BD Single displacement/replacement – an element replaces a similar element. (element alone on both sides) XY + M  MY + X

9. Chemical Equations Reactants  products Be sure all elements appear on both sides – law of conservation of mass Identify reaction type to help. Balance the equation by adding coefficients Add symbols to describe elements Solid (s), liquid (l), gas (g), aqueous solution (aq)

10. Reaction Stoichiometry Must start with a balanced equation Use coefficients in calculations – mole ratio of substances. Mole ratio comes from the coefficients of the balanced equation Limiting reactant – what substance if used up first – limits the extent of the reaction. Start with 2 givens Convert both givens to the same product Compare both numbers – smaller number is produced by the limiting reactant (LR). Other is called excess reactant (ER)

11. Yield Percent yield = Actual yield * 100% theoretical yield Actual yield = what you get (from calculation or experiment) Theoretical yield = what should get – maximum amount produced from limiting reactant ** Should not be more than 100%