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”
Why?
General Guidelines
• Balance any polyatomic ions first
• Balance metals
• Balance nonmetals
• Balance “O” and “H” last
• Recount both sides to make sure!!
Al + O2 Al2O3
C2H6 + O2 CO2 + H2O
Counting Significant
figures
Answers…100 = 1
0.001 = 10.010 = 210.04 = 4
1.0 x 102 = 2197 = 3
0.0000200910 = 6100.20 = 5
10.0 = 31090 = 3
What are the general rules for what zero’s are
significant?
Calculating with Significant
Figures
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?
General Rules…
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
What steps were taken to answer these questions?
