Chemical Formulas
Chemistry 1
Oxidation Numbers
 “oxidation states”
 The number of electrons that must be added to or
removed from an atom in a combined state to
convert the atom into the elemental form.
Assigning Oxidation States
Rules for Assigning Oxidation States
The Oxidation State of…
 An atom in an element is zero
 Na(s), O2(g), Hg(l)
 A monatomic ion is the same as its charge
 Na+, Cl-
 Fluorine is -1 in its compounds
 HF, PF3
 Oxygen is usually -2 in its compounds
 H2O, CO2
 Exception: Peroxides (containing O22-) in which oxygen is -1
 Hydrogen is +1 in its covalent compounds
 H2O, HCl, NH3
Oxidation Numbers - Model
Assign oxidation numbers to the atoms of each element in the
following compounds.
 PI3
 HF
 CS2
 As2S3
 NO2-
Oxidation Numbers Practice
Assign oxidation numbers to the atoms of each element in the
following compounds.
 CI4
 HCl
 SO3
 IO3
 H2CO3
 SO42-
Percent Composition
 The percentage by mass of each element in a compound.
massofelementincompound
x100% = %ElementinCompound
massofcompound
1.
Using the periodic table determine the molar
mass of the compound.
2.
Take each mass for each element (molar mass
times number of atoms) and divide it by the
total mass of the compound.
3.
Multiply this value by 100 to obtain a
percentage.
4.
Percentages should add up to equal 100%
Steps For Calculations
Percent Composition
 Determine the percent composition in the following
compounds:
 PbBr4
 Ba(OH)2
Percent Composition
 Determine the percent composition in the following
compounds:
 NaOH
 (NH4)3PO4
 CCl2F2
 Pb(NO3)2
Bite the Bubble Lab
 See handout
Review Practice
 Empirical Formula: The formula with the smallest
whole number ratio of the elements in the
compound (simplest formula)

1.
Note: May not always be the same as the molecular
formula.
To calculate, convert masses of elements to
moles.
 If a question gives the percentages of each compound,
assume you have 100 g of the sample and then each
percentage becomes the mass of the element.
2.
Next, divide each mole by the smallest one.
3.
This gives the ratio of atoms, which you will use
to write the formula.
Empirical Formula
Empirical Formulas
The percent composition of a compound
was found to be 63.5% silver, 8.2% nitrogen,
and 28.3% oxygen. Determine this
compound’s empirical formula.
Name the compound.
Empirical Formulas
 A 170.00g sample of a compound contains 29.84 g
sodium, 67.49 g chromium and 72.67 g oxygen.
What is the empirical formula?
 Name the compound.
Empirical Formula
 200.00 grams of an organic compound is known to
contain 83.884 grams of carbon, 10.486 grams of
hydrogen, 18.640 grams of oxygen and the rest is
nitrogen. What is the empirical formula of the
compound?
Molecular Formulas
 Molecular Formulas: A formula that specifies the actual number
of atoms of each element in one molecule of the substance.
-Molecular formula is a multiple of the empirical formula.
Steps for Solving:
 Determine the Empirical Formula first!
 Use the formula :
molecularweight
N=
empiricalformulaweight
N= The number we will multiply the subscripts in the empirical
formula by to obtain the new molecular formula.
Empirical and Molecular Formulas
 A compound of boron and hydrogen has a percent
composition of 78.14% boron and 21.86% hydrogen.
If the molar mass is 27.6, what is the empirical and
molecular formula?
Empirical and Molecular Formulas
 A compound with the formula C2H5O is found to
have a molar mass of 90g. What is the molecular
formula of the compound?
Individual Practice
Unit Packet Part 4 and 5
 Page 251
 # 2, 6-8, 10, 11, 16-18, 21, 23-25, 32, 33, 36-41, 44, 4750
 Test will cover…
 New
 Ionic, Molecular, and Acid Nomenclature
 Oxidation Numbers and Percent Composition
 Empirical and Molecular Formulas
 Chemical Bonding
 Periodic Table Trends
Chapter Review