1. http://www.skanschools.org/webpages/rallen/files/notes%20-%20unit%205%20-%20moles%20&%20stoich_2012_key.pdf

2.  Monoatomic  one atom of an element that’s stable enough to stand on its own (very rare)- not bonded to anything  Diatomic (diatoms)  elements whose atoms always travel in pairs (N 2, O 2, F 2 etc.) bonded to another atom of the same element

3.  What would be the mass of one molecule of oxygen (O 2 )?  O 2  subscript tells you the total # of atoms in molecule/compound ▪ O 2 = 2 x 16 amu = 32 amu

4.  Formula Mass  the mass of an atom, molecule or compound in atomic mass units (amu)  Gram Formula Mass  the mass of one mole of an atom, molecule or compound in grams  Mole  6.02 x 10 23 units of a substance

5.  Step 1: calculate the GFM for the compound  Ex. CaCl 2  Ca = 1 x 40.08 = 40.08  Cl= 2 x 35. 453 = 70.906 Formula  % composition by mass = mass of part x 100 Mass of whole “Parts” * Find the percent composition to the nearest 0.1%

6.  A hydrate is a crystalline compound in which ions are attached to one or more water molecules  Ex. Na 2 CO 3 10H 2 O  Notice how water molecules are built into the formula  Substances without water built into the formula are called anhydrates

9.  How many moles are in 4.75 g of sodium hydroxide (NaOH)?  Na  1 x = Total  O  1 x =  H  1 x = 22.98 15.99 1.007 GFM = 39.977 g/mol

10.  Plug in the given value and the GFM into the “mole calculations” formula and solve for moles  # of moles = 4.75 g = 39.977 g/mol.119 mol

12.  What is the mass of 4.5 moles of KOH? Mass KOH = 4.5 moles x 56.087 g/mol mass KOH = 252.39 g

13.  A chemical equation is a set of symbols that state the products and reactants in a chemical reaction  Reactants  the starting substances in a chemical reaction (left side of arrow)  Products  a substance produced by a chemical reaction (right side of arrow)

14.  Ex.  2Na + 2H 2 O  2NaOH + H 2  Chemical equations must be balanced  Law of conservation of mass: mass can neither be created nor destroyed in a chemical reaction

15.  The number of moles of each element on the reactants side must be the same as the number of moles of each element on the products side  Coefficients and subscripts tell us how many moles of each element we have

16.  Ex. Balanced Equation  C + O 2  CO 2  1 mol of Carbon  2 mol of Oxygen  This means the equation is balanced Each side of the arrow

17.  Ex. Unbalanced Equation  H 2 + O 2  H 2 O  Coefficient= integer in front of an element or compound which indicates the # of moles present  Subscript = the integer to the lower right of an element which indicates # of atoms present  Species= the individual reactants and products in a chemical reaction

18.  Ex. Unbalanced Equation  H 2 + O 2  H 2 O  What do we use to balance equations?  Coefficients  * we never change the subscripts in a formula  Balanced equation: 2H 2 + O 2  2H 2 O

21.  Ex.

22.  When balancing chemical equations, polyatomic ions may be balanced as a single element rather than as separate elements as long as they stay intact during the reaction

23.  Al 2 (SO 4 ) 3 + Ca(OH) 2  Al(OH) 3 + CaSO 4  Polyatomic ions= sulfate and hydroxide  Polyatomic ions remain intact during the reaction, can be considered one unit  Balance the equation: Al 2 (SO 4 ) 3 + 3Ca(OH) 2  2Al(OH) 3 + 3CaSO 4

24.  Type 1: Single Replacement  Reaction where one species replaces another (one species alone on one side and combined on the other)  Ex. 3Ag + AuCl 3  3AgCl + Au  2Cr + 3H 2 SO 4  Cr 2 (SO 4 ) 3 + 3H 2

25.  Type 2: Double Replacement  Reaction where compounds react, switch partners and produce 2 new compounds  Ex.  Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3  Na 3 PO 4 + 3 Ag NO 3  Ag 3 PO 4 + 3NaNO 3  K 2 CO 3 + 2AgNO 3  Ag 2 CO 3 + 2KNO 3

26.  Type 3: Synthesis  Reaction where we take more than one reactant and create one product  Ex.  4Al + 3O 2  2Al 2 O 3  2H 2 + O 2  2H 2 O

27.  Type 4: Decomposition  Reaction where we take one reactant and create 2 products  Ex.  BaCO3  BaO + CO2  2H2O2  2H2O + O2  2Bi(OH) 3  Bi 2 O 3 + 3H 2 O

28.  A chemical equation = recipe for reaction  Coefficients= tell the amount of reactants and products needed  Reactants in an equation react in specific ratios to produce specific amounts of products

29.  Method for solving mole-mole problems  Set up a proportion using your known and unknown values  Cross multiply and solve for your unknown  Always check to make the equation is balanced

30.  Empirical formula= the reduced formula; a formula whose subscripts cannot be reduced any further  Molecular formula= the actual formula for a compound; subscripts represent actual quantity of atoms present

33.  Step 1: always assume you have 100 g sample (the total % for the compound must = 100, so we can just change the units from % to g)  Step 2: convert grams to moles  Step 3: divide all mole numbers by the smallest mole number

34.  ex. a compound is 46.2% mass carbon and 53.8% mass nitrogen. What is its empirical formula?  Step 1: assume 100 g sample  46.2 % C = 46.2 g C  53.8 % N= 53.8 g N

35.  Step 2: Convert grams to moles (have g need moles)  46. 2 g C= 3.85 mol C 53.8 g N = 3.84  12 g/mol C 14 g/mol  * must have whole numbers for subscripts

36.  Step 3: divide each mole number by the smallest mole number (we will round in this step to the nearest integer)  3.85 mol C = 1 3.84 mol N = 1  3.84 mol N 3.84 mol N  Empirical formula: CN

37.  We know how to:  1. find an empirical formula from % mass  2. find an empirical formula from molecular formula  Next how do we find out the molecular formula from an empirical formula?

38.  Ex. a compound is 80.0% C and 20.0% H by mass. If its molecular mass is 75.0 g, what is its empirical formula? What is its molecular formula?  First, determine the empirical formula using the 3 step process

39.  Step 1: assume a 100 g sample  80.0 % C= 80.0 g C  20.0 % H = 20.0 g H  Step 2: convert g to moles  80.0 g C = 6.66 mol C 20.0 g H= 20 mol H  12 g/mol C 1 g/mol

40.  Step 3: divide each by the smallest number of moles and round to the nearest whole  C= 6.66 = 1 H= 20.0 = 3.00  6.66 6.66  Empirical formula: CH 3

41.  Empirical mass (the mass of 1 mol of CH 3 )= 15 g  Molecular mass = 75 g  Molecular mass is 5 times larger than empirical mass  Molecular formula must be 5 times larger than empirical formula

42.  Multiply all subscripts in our empirical formula by 5  Molecular formula (CH 3 )= C 5 H 15