1. Conceptual Statement #2: Units of Matter Communication in Chemistry Chemical Formulas Net charge = 0 (+) charged cation first (-) charged anion last Chemical Equations Law of Conservation Of Matter and Energy reactantsproducts Number and types of atoms In the reactant Equals Number and type of atoms in products Types of Chemical Reactions synthesis decomposition Single replacement Double replacement combustion

2. Writing a Chemical Formula Naming Different Types of Compounds Stock system Name of Cation Including Roman Numeral – Followed by anion If binary, anion ends in -ide Covalent bonds Prefix - name of cation Followed by anion ending in -ide Acid/Base Binary acid Hydro-name of anion-ic acid Other acid Anion-ic acid

3. Interpreting a Chemical Formula Vocabulary Chemical symbol Cation (+) on Left Anion (-) on Right subscript Tells how many parentheses Shows groupings coefficient Tells how many Sets of this Compound or Molecule Ca 3 (PO 4 ) 2 has 3 Ca and 2 (PO 4 ) ------------------------------------- Each Ca 3 (PO 4 ) 2 has 3 Ca, 2 P, and 8 Oxygen

4. Naming Compounds Naming Ionic Compounds with more than one possible charge using the Stock System Name of the cationCharge of the cationName of the anion FeCl 2 Fe +2 Cl -1 Iron (II) chloride

5. How to Balance a Chemical Equation 2. Make sure 1.Write balanced chemical formulas 2.Use subscripts to balance formulas 1.Balance the equation so the number and type of atoms are the same in the Reactants and Products. 2.Use coefficients to balance equations Check your work! Check Your Work ! Step #1 Step #2 Step #3

6. How to Find the Formula of a Compound in the Laboratory In a sample, find the molecular masses of each atom In a 100g sample, Ca = 40%, C =12%, O = 48% 40g(1mole/40gCa) 1mole Ca 12g(1mole/12g C) 1 mole C 48g(1mole/ 16g O) 3 mole O CaCO 3 Atomic mass Change % To g

7. Molecular and Empirical Formulas are related by a mole ratio. For a substance With 92%C and 7.7% H The mole ratio of C:H is 1:1 If the Molecular Mass Is 26g/mole 26/13 = 2 C2H2 If the Molecular Mass Is 78g/mole 78/13 = 6 C6H6 C= 12 g/mole H = 1.0 g/mole 92g/12g/mole = 7.7mole C 7.7g/1.0g/mole = 7.7mole H The mole ratio is 1:1 The empirical formula weight Is 12 +1 = 13g/mole

8. How to Calculate % Composition Begin with the formula: C6H12O 6 6 mole C(12.0 g/mole) 72 g C 12 mole H (1.01g/mole) 12.1 g H 6 mole O(16.0g/mole) 96.0 g O Find the gram formula mass of each element Find how many of Each atom you have. ++= 180.1g Divide each by the sum (the gram formula weight 72.0g/180.1g x100% 12.1g/180.1g x 100% 96.0g/180.1g x 100% 40.0% C6.72% H53.3% O Gram Formula weight

9. Stoichiometry: Conversion factors are used to convert from (units) Given to (units) Unknown. Mole of Given Volume of Gas at STP Mass in grams Particles (atoms, Molecules 22.4 L/mole 6.02 x 10 23 particles / mole Gram Formula mass Mole of Unknown Volume of Gas at STP Mass in grams Particles (atoms, Molecules 22.4 L/mole Gram Formula mass 6.02 x 10 23 particles / mole Mole ratio

10. Vocabulary Atom Avagadro’s number Balanced Chemical Equation Balanced Chemical Formula Chemical formula Combustion Reaction Compounds Covalent compounds Decomposition Reaction Double Replacement Reaction Dimensional analysis Empirical Formula Ionic Charges Ionic Compounds Law of Conervation of Mass and Energy Law of Definite Proportions Liter Molecular compounds Molecular formula Moles Oxidation Number Percent Composition Single Replacement Reaction Standard Temperature and Pressure Stock system Synthesis Reaction Transition Elements