### Announcements - 9/17/02 Another Example

```Announcements - 9/17/02
•No online Quiz this week ☺
•Exam #1 – Wed 7pm 235 Marsh
-bring calculator (no PDAs!) and pen/pencil(s)
-see Info Page for details
•Demo Today!
1
Another Example
A solution of sodium chloride was added to a solution of
silver nitrate, forming a precipitate of silver chloride
1. sodium choride + silver nitrate → silver chloride + sodium
nitrate
2. NaCl + AgNO3 → AgCl + NaNO3
Formulas
3. NaCl + AgNO3 → AgCl + NaNO3
Balanced
4. NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) →
AgCl (s) + Na+ (aq) + NO3- (aq)
Ag+ (aq) + Cl- (aq) → AgCl (s)
Net Ionic Equation
2
1
Demo!
How can we tell an ionic compound
from a molecular compound?
Measure conductivity (in sol’n)!
3
Let’s Get Quantitative
Recall:
1 mole = 6.02 x 1023 particles
Atomic Masses (Weights): mass/atom (amu)
Molecular/Formula Weights: mass/compound (amu)
Molar Masses: mass of a mole of atoms or
compound (g)
4
2
Why Fractional Molar Masses?
Need to consider the natural
abundances of isotopes
Example: Chlorine
75.5%
35Cl
+ 24.5%
37Cl
(0.755)(34.97) + (0.245)(36.97) = 35.45 g/mol
-This is a weighted average; 1 mol of Cl will
have a mass of 35.45 grams
5
General Strategy
Be Careful!
moles of: atoms? molecules?
6
3
Mole-Based Calculations
How many grams of Phosphorous are
there in 0.010 mol P2O5?
Strategy: mol P2O5 → mol P → g P
0.010 mol P2O5 x 2 mol P x 30.974 g P = 0.61948 g P
1 mol P2O5
1 mol P
Round to:
0.62 g Phosphorous
7
How Many Atoms?
How many Phosphorous atoms are there
in 0.010 mol P2O5?
Strategy: mol P2O5 → mol P → #P atoms
0.010 mol P2O5 x 2 mol P x 6.022 x 1023 P atoms
1 mol P2O5
1 mol P
=
= 1.20440 x 1022 P atoms
= 1.2 x 1022 P atoms
8
4
Strategy: Empirical Formula
from Percent Composition
9
Empirical Formula from
Percent Composition
What is the empirical formula for a binary
compound which is found to be:
56.4% Oxygen (by mass)
43.6% Phosphorous (by mass)?
Strategy: % → grams → mol
(% is a relative measure, so DEFINE a sample size (100 g))
In a 100-g sample:
56.4 g O x 1 mol O
43.6 g P
15.999 g O
x
1 mol P
30.974 g P
=
3.525 mol O
=
1.4076 mol P
10
5
Empirical Formula continued
This gives:
P1.4076O3.525
Dividing: PO2.50
→ P2O5