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Review of Chemistry 11. Prior Knowledge Needed for Chemistry 12. Naming molecules and writing formulas Balancing Chemical Equations Counting Significant figures Calculating with significant figures Mole calculations Basic Stoichiometry. Naming Molecules and Writing Formulas.
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Review of Chemistry 11
Prior Knowledge Needed for Chemistry 12 • Naming molecules and writing formulas • Balancing Chemical Equations • Counting Significant figures • Calculating with significant figures • Mole calculations • Basic Stoichiometry
Naming Molecules and Writing Formulas
What do you remember? Categorize the following according to what elements are involved and the types of intramolecular bonds: HCl, NaCl, CO2, Cu(ClO) 2, CaO, HF, Ca(OH) 2, HIO3, Sr(NO3) 2, Pb(NO3) 2, AgNO3, KBr, NaOH, N2O3, FeO, CH3COOH,
Analyze each pair (name = formula) and come up with some rules… Hyroiodic Acid = HI Nitrous Acid = HNO2 Nitric Acid = HNO3
Copper(II) nitride = Cu(NO3)2 Sodium phosphate = Na2PO4 Gold(II) oxide = AuO
Sulphur diflouride = SF2 Dinitrogen trioxide = N2O3
Balancing Chemical Equations “BCE”
General Guidelines • Balance any polyatomic ions first • Balance metals • Balance nonmetals • Balance “O” and “H” last • Recount both sides to make sure!!
Counting Significant figures
Answers… 100 = 1 0.001 = 1 0.010 = 2 10.04 = 4 1.0 x 102 = 2 197 = 3 0.0000200910 = 6 100.20 = 5 10.0 = 3 1090 = 3
What are the general rules for what zero’s are significant?
Answers… Adding and Subtracting… 10.00 + 2.0 = 12.0 36.98 – 0.0013= 36.98 General Rule?
Answers… Multiplying and Dividing… 190 / 3 = 60 1403 / 7.0 = 2.0 x 102 0.09700 x 1.2 = 0.12 General Rule?
Mole Calculations
Answer… How many grams of NaCl are in 4.50 moles of NaCl? # g NaCl = 4.50 mol x 58.5 g = 263 g NaCl 1 mol What steps were used to answer this question?
Answer… How many moles of NH4Br are in 15.87 g of NH4Br? # mol NH4Br= 15.87g x 1 mol =0.162 mol NH4Br 97.9 g What steps were used to answer this question?
Basic Stoichiometry
Answers… Given the following balanced equation, answer the questions following it: 9Na + 4ZnI2 8NaI + NaZn4 a.) If 2.50 moles of Na are reacted, how many moles of ZnI2 will be consumed? # mol ZnI2 = 2.50 mol Na x 4 mol ZnI2 = 1.11 mol ZnI2 9 mol Na b.) If you completely react 526.68 g of ZnI2, what mass of NaZn4 will be produced? # g NaZn4 = 526.68g of ZnI2x 1 mol ZnI2 x 1 mol NaZn4 x 284.6g NaZn4 = 117.4g NaZn4 319.2 g ZnI2 4 mol ZnI2 1 mol NaZn4