Definitions
Solution: A homogeneous mixture
Solute: The one present in smaller amount
Solvent: The one present in greater amount
Concentration: Amount of Solute
Amount of Solvent
Some units for concentration
g, g, mg, mg
ml L L ml
The most common (and useful) units are
mol = Molarity = Molar Concentration
L
THE FOLLOWING ARE ONLY FOR AQUEOUS SOLUTIONS & DO NOT APPLY TO GASES
M= mol
L
mol= M(L)
L= mol
M
NOTE: Think triangle of density/mass/volume
and now an informational video..
Good luck!
Nov 18, 2009
at
8:33 PM
Today was the last class before mid-term exams. In class, we discussed the empirical formula and went over what would be on the exam.
You need to know:
SI system
Classification
Lab Safety & Labs
Here are some videos to help!
You need to know:
Nomenclature
- Binary Ionic
- Multivalent
- Polyatomic
- Acids/Bases
- Hydrates
- Molecular Compounds
- Classical Naming System
- Mole Conversion Table
- Mole to mass and volume (gases atSTP)
- Density
- Number of Molecules
- Atoms
SI system
Classification
Lab Safety & Labs
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Empirical Formulas
Nov 17, 2009
at
10:16 PM
Molecular
ex. (in-class) A sample of an unknown compound is analyzed and found to contain 8.4g of C and 2.1g of H, and 5.6g of O.
C8.4H2.1O5.6 <--------- this is wrong because it is not in whole numbers!
ex. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?
Here's a helpful hint for putting your ratios into whole numbers:
If the ratio ends in...
~0.5 multiply by 2
~0.33 or ~0.66 multiply by 3
~0.25 or ~0.75 multiply by 4
~0.2, ~ 0.4, ~0.6, ~0.8 multiply by 5
- P4O10
- C10H22
- C6H18O3
- C5H12O
N2O4
Empirical
P2O5
C5H11
C2H6O
C5H12O
NO2
-Empirical formulas gives the whole number ratios of elements in a compound
-Molecular formulas give the actual numbers
-therefore, the empirical formula is the simplest form| Element | Mass (g) | Atomic Mass | Moles |
| C | 8.4 | 12.0 g/mol | 8.4g ÷ 12.0 g/mol = 0.7 mol |
| H | 2.1 | 1.0 g/mol | 2.1g ÷ 1.0 g/mol = 2.1 mol |
| O | 5.6 | 16.0 g/mol | 5.6g ÷ 16.0 g/mol = 0.35 mol |
ex. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?
| Element | Mass (g) | Atomic Mass | Moles |
| Ca | 13.5 | 40.1 g/mol | 13.5g ÷ 40.1 g/mol = 0.337 mol |
| O | 10.8 | 16.0 g/mol | 10.8g ÷ 16.0 g/mol = 0.269 mol |
| H | 0.675 | 1.0 g/mol | 0.675g ÷ 1.0 g/mol = 0.675 mol |
Here's a helpful hint for putting your ratios into whole numbers:
If the ratio ends in...
~0.5 multiply by 2
~0.33 or ~0.66 multiply by 3
~0.25 or ~0.75 multiply by 4
~0.2, ~ 0.4, ~0.6, ~0.8 multiply by 5
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