1.2 FORMULAS
1. Define MOLAR MASS:
 UNITS: grams per mole…written  ________________
 The term molar mass can be used for each of the following:
a) ATOMIC mass
Examples
b) MOLECULAR mass
(covalently bonded compounds)
Examples
c) FORUMULA mass
(ionically bonded compounds)
Examples
*WHAT IS A HYDRATE?
 an ionic compound with _____________ molecules incorporated into their crystal structure
 Hydrates have a specific number of water molecules chemically bonded to each formula unit. The bond between the
water molecules and the formula unit is relatively weak.
Examples:
MgSO4 ∙ 7H2O
 magnesium sulfate heptahydrate
Ba(OH)2 ∙ 8H2O 
CaCl2 ∙ 2H2O

*The raised dot denotes a hydrated compound.
*Anhydrous compounds  compounds with _______water molecules incorporated into them (e.g. CuCl2, MgSO4)
RELATIVE Atomic Mass, Ar
 The relative atomic mass of an element is the mass expressed on the ______________ __________. This is a
_____________ number because, technically, it is a ratio. *Remember  anything with “relative” in front of it is
__________________.
 Relative to what?...one twelfth (1/12) the mass of a __________ atom ; one twelfth of a ________ atom equals the
mass of a __________ atom. Therefore, the mass of a fluorine atom, for example, is about ________times heavier than a
H-1 atom.
 Why is the relative atomic mass of hydrogen not exactly 1.00?...due to the existence of ___________________.
Hydrogen has _____ isotopes [H-1(normal hydrogen , H-2(deuterium), H-3(tritium)]. Remember, atoms of the same
element that contain different numbers of ______________are called isotopes. Most elements are made up of two or more
isotopes. The only elements with one naturally occurring isotope are beryllium, sodium, aluminum, and phosphorus. The
relative atomic mass found on the periodic table is the weighted average mass of all the naturally occurring isotopes for
that element (relative to 1/12 the mass of a C-12 atom.)
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ISOTOPIC ABUNDANCE  the relative amount in which each isotope is present in nature
e.g. Magnesium has three naturally occurring isotopes. A sample of magnesium is always made up of these three isotopes
in the same proportion.
Isotope
24
12 Mg
25
12 Mg
26
12 Mg
Mg-24
Isotopic
Abundance
79%
Mg-25
10%
Mg-26
11%
*Calculate the relative atomic mass of magnesium.
*What is Relative Molecular Mass, Mr? ….same as a molar mass calculation (only difference?) *What is Relative
Formula Mass?
Practice Calculations:
206
207
1. Lead occurs naturally as four isotopes. These isotopes are 204
82 Pb (1.37%, 204.0), 82 Pb (26.26%, 206.0 ), 82 Pb (20.82%,
207.0), and 208
82 Pb (51.55%, 208.0). Calculate the relative atomic mass of lead. (Ans. 207.2 u)
2. Hydrogen is found primarily as two isotopes in nature: H-1 (Ar = 1.0078) and H-2 (Ar = 2.0140). Calculate the
percentage abundance based on hydrogen’s relative atomic mass (i.e. the value given on the periodic table).
3. Chlorine exists as two isotopes – Cl-35 and Cl-37. If the average atomic mass is 35.45, calculate the percentage
abundance for each isotope.
4. Thallium consists of thallium-203 and thallium-205. Using the value form the periodic table, determine the relative
abundance of each isotope.
EMPIRICAL FORMULA (“empirical” (Greek = emperirikos) means “by ____________________”)
*Empirical Formula defined (___________________ formula)  the lowest whole number ratio of the elements in the
compound
Complete the following table:
Name of Compound
Molecular (actual)
Empirical (simplest)
Lowest ratio of
Formula
formula
elements
hydrogen peroxide
H 2O 2
glucose
C6H12O6
*benzene
C6H6
*acetylene (ethyne)
C2H2
**aniline
C6H7N
**water
H2O
*The same empirical formula can represent more than one compound.
**Many compounds have molecular formulas that are the same as their empirical formulas.
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What is the relationship between the empirical formulas of C6H6 and C2H2 and their molar masses?
Formula
Molecular, C6H6
Related Molar Mass (g mol-1)
Empirical, CH
Formula
Molecular, C2H2
Ratio
M .M .Molecular

M .M . Empirical
Related Molar Mass (g mol-1)
Empirical, CH
Ratio
M .M .Molecular

M .M . Empirical
*Note that the ratio of the molar masses is the same as the ratio of the subscripts in the two formulas.
PERCENTAGE COMPOSITION
* Determine the percentage composition(by mass) of :
a) water
b) carbon dioxide
DETERMINING A COMPOUND’S EMPIRICAL FORMULA FROM PERCENTAGE COMPOSITION
EXAMPLE: A compound is 17.6% hydrogen and 82.4% nitrogen by mass. Determine the empirical formula of the
compound.
Steps:
Rhyme (memory aid):
1) Percent to Mass
2) Mass to Mole
3) Divide by Small
* 4) Multiply ’till Whole
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MOLECULAR FORMULA
*Molecular Formula defined (________________formula)  describes the actual number of each element that make up
a molecule or formula unit
*Being able to determine the molecular formula of a compound is an important skill often engaged by many different
kinds of analytical researchers and technicians. (e.g. forensic scientists, sport doping control)
*RECALL: Many compounds can have the same empirical formula.
*Examples: The empirical formula for all of the following compounds is _______________.
Compound
Molecular Formula
formaldehyde
acetic acid
lactic acid
erythrose
ribose
glucose
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6
*Whole-number
multiple
1
2
3
4
5
6
Molar Mass
(g/mol)
30.03
60.05
90.08
120.10
150.13
180.16
*The subscripts in the empirical formula are multiplied by this number(i.e. the number in the third column) in order to
attain the subscripts for the molecular formula.
Determining a Molecular Formula
Example:
*The empirical formula of butane, the fuel used in disposable lighters, is C2H5. In an experiment, the molar mass of
butane was determined to be 58 g/mol. What is the molecular formula of butane?
Steps:
1)
2)
3)
Determine the molar mass of the empirical formula.
Divide the molar mass of the compound by the molar mass of the empirical formula.
Multiply all the subscripts in the empirical formula by the number obtain in Step (2).
APPLICATION: The Carbon-Hydrogen Combustion Analyzer
*Applications  forensic science, food chemistry, pharmaceuticals and academic research
 Allows chemists to determine the percentage composition of compounds that are made up of carbon, hydrogen and
oxygen.
*How does it work?
 Relies on the fact that the complete combustion (reaction with __________) of a hydrocarbon yields only two products
 _____ and _____.
 The products, CO2 and H2O, are absorbed and then weighed. The percentage composition of the unknown can then be
determined.
Example:
A 1.000 g sample of a pure compound, containing only carbon and hydrogen, was combusted in a carbon-hydrogen
combustion analyzer. The combustion produced 0.6919 g of water and 3.338 g of carbon dioxide.
a) Calculate the masses of the carbon and the hydrogen in the sample.
b) Determine the empirical formula of the compound.
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