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Transcript
```Chemical Composition Notes
Conversions
Moles are a chemist’s dozen:
 3 dozen eggs x 12 = 36 eggs
 Moles x (conversion factor) = new unit

Conversion factors are
o Atomic mass or molar mass
 Units of grams per mole (g / mol)
 Used for converting between grams and moles
 Units of atoms per mole (atoms / mol)
 Used for converting between moles and atoms
Here are the equations
Moles x AM = grams
moles x Av# = atoms
Grams / AM = moles
atoms / Av# = moles
Don’t forget Av# = 6.022 x 1023
Percent Composition
Designated as a fraction 
Part
Whole
Example: CH3O
AM = 31.034
The part is each element times its quantity
The whole is the AM for the compound
C
CH 3O
3H
CH 3O
O
CH 3O



Make sure you use the mass of each
12.011
(100) = 38.703% Carbon
31.034
3.0237
(100) = 9.7432% Hydrogen
31.034
15.999
(100) = 51.553% Oxygen
31.034
Determining Empirical Formulas
Steps:
1.
3.
2.
Convert grams to moles (gives ugly mole ratio)
Divide by all numbers by the smallest (sets smallest to 1)
If decimals are present, multiply all numbers to create integers
4.
Apply integers to the formula
Given either mass of percentages of constituent elements it works the same
Example: An unknown sample was found to have a percent composition as follows:
47.0% K, 14.5% C, and 38.5% O. Find the empirical formula.
Given %, assume a 100 g sample so:
47.0 g K
14.5 g C
38.5 g O
This is a mass ratio, but all analysis is done by mole ratio so convert each into moles
47.0
14.5
38.5
= 1.20 mol K
= 1.21 mol C
= 2.41 mol O
39.098
12.011
15.999
These are the right units, but we need to reduce our numbers
Divide all by the smallest reducing the least down to one while holding the ratio
1.20
1.21
2.41
= 1K
= 1C
= 2O
1.20
1.20
1.20
So the final compound is KCO2
Determining Molecular Formulas
Steps:
1. Fine the empirical mass (molar mass of the EF)
2. Molar mass / Empirical mass = multiplier
3. Empirical formula (multiplier) = molecular formula
Example: Given the empirical formula above, determine the molecular formula if the
molar mass of the compound is 166.214 g/mol
First find the empirical mass (atomic mass of the empirical formula)
39.098 + 12.011 + 2 (15.999) = 83.107 g/mol
Now divide the atomic mass by the empirical mass
166.214 / 83.107 = 2
This is your multiplier (n) value
KCO2 x 2 = K2C2O4
Vocabulary, definitions,
abbreviations, and units.
Abbrev
Units
 Atomic mass: mass of one mole of an atom
AM
g/mol
 Molar mass: mass of one mole of a molecule
MM
g/mol
 Mole: 6.022 x 1023 atoms make up one mole
mol
 Avogadro’s Number: 6.022 x 1023
Av#
 Mass: what asking for mass, calculate grams
g
 Atom: basic building block of matter
atom
 % composition: the % by mass of each element in a compound
 Empirical formula: reduced form of a chemical formula
EF
 Empirical mass: molar mass of an empirical formula
EM
 Molecular formula: actual chemical formula
MF
atoms / mol
% (element)
g/mol
```
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