Chapter 3: Calculations with Chemical
Formulas and Equations
MASS AND MOLES OF SUBSTANCE
3.1 MOLECULAR WEIGHT AND FORMULA
WEIGHT
- Molecular weight: (MW) sum of the atomic
weights of all the atoms in a molecule of the
substance.
- H2O = 18.0 amu (2 x 1.0 amu for two H atoms
plus 16.0 amu from one O atom)
Formula weight (FW) – sum of the atomic
weights of all atoms in a formula unit of a
compound. Whether molecular or not.
Sodium chloride, NaCl (formula unit) formula
weight of 58.44 amu (22.99 amu from Na plus
35.45 amu from Cl).
NaCl is ionic (no molecular weight)
3.2 THE MOLE CONCEPT
- Developed to deal with the enormous
numbers of molecules or ions in samples of
substances.
- Mole: the quantity of a given substance that
contains as many molecules or formula units
as the number of atoms in exactly 12 g of
carbon-12.
The number of atoms in a 12-g sample of
carbon-12 is called Avogadro’s Number (NA).
= 6.02 x 10^23
A mole of a substance contains Avogadro’s
number of molecules (6.02 x 10^23)
Mole (much like dozen or gross) refers to a
particular number of molecules.
1 mole of ethanol = 6.02 x 10^23 ethanol
molecules
When talking about ionic substances we are
referring to the number of formula units not
molecules.
One mole of sodium carbonate Na2CO3
Contains:
6.02 x 10^23 Na2CO3 units
Which contains:
2 x 6.02 x 10^23 Na+ ions
1 x 6.02 x 10^23 CO32- ions
Specify
Mole of oxygen atoms means 6.02 x 10^23 O
atoms
Mole of oxygen molecules means 6.02 x 10^23
O2 molecules
(2 x 6.02 x 10^23 O atoms)
Molar mass: mass of one mole of the substance.
Carbon-12 has a molar mass of exactly 12g/mol
For all substances, the molar mass in grams per
mole is numerically equal to the formula weight in
atomic mass units.
Ethanol, molecular formula C2H5OH
MW = 46.1 amu
Molar mass = 46.1 g/mol
MOLE CALCULATIONS
- Conversion from mol to g
- Ethanol is 46.1 g/mol
1 mol C2H5OH = 46.1 g Cr2H5OH
To convert grams to moles
• You need to prepare acetic acid from 10.0g of
ethanol, C2H5OH. Convert 10.0g C2H5OH to
moles C2H5OH by multiplying by the
appropriate conversion factor.
3.3 MASS PERCENTAGES FROM THE FORMULA
- Mass percentage of A as the parts of A per
hundred parts of the total, by mass.
3.4 ELEMENTAL ANALYSIS: PERCENTAGES OF
CARBON, HYDROGEN, AND OXYGEN
- To determine the formula of a new compound is
to obtain its percentage composition.
- Ex. Burn a sample of the compound of known
mass and get CO2 and H2O.
- Next relate the masses of CO2 and H2O to the
masses of carbon and hydrogen.
- Then calculate the mass percentages of C and H.
- Determine the mass percentage of O by
difference.
3.5 DETERMINING FORMULAS
- Empirical formula (simplest formula) –
formula of a substance written with the
smallest integer subscripts.
- For most ionic substances, the empirical
formula is the formula of the compound.
- This is often not the case for molecular
substances.
- Hydrogen peroxide: molecular formula is H2O2
- The empirical formula (just tells you the ratio
of numbers of atoms in the compound) HO
• Why do ionic compounds usually have the
same formula and empirical formula?
• No multiple proportions like molecules.
• Compounds with different molecular formulas
can have the same empirical formula.
• They will also have the same percentage
composition.
• Acetylene C2H2
• Benzene C6H6
• Same empirical formula CO, different
molecular formulas, different chemical
structures.
To obtain the molecular formula of a substance,
you need to know:
1. The percentage composition, from which the
empirical formula can be determined
2. Molecular weight
EMPIRICAL FORMULA FROM THE COMPOSITION
- You can find the empirical formula from the
composition of the compound by converting
masses of the elements to moles.
• Ex. 3.10
MOLECULAR FORMULA FROM EMPIRICAL
FORMULA
- Molecular formula of a compound is a
multiple of its empirical formula.
- Acetylene C2H2 is equivalent to (CH)2
- Benzene C6H6 is equivalent to (CH)6
Molecular weight = n x empirical formula weight
• N = number of empirical formula units in the
molecule.
• Molecular formula is obtained by multiplying
the subscripts of the empirical formula by n
n = molecular weight/empirical formula weight
• After you find the empirical formula, calculate
its empirical formula weight.
• From an experimental determination of its
molecular weight, you can calculate n and
then the molecular formula.
• Ex. 3.11
STOICHIOMETRY: QUANTITATIVE RELATIONS IN
CHEMICAL REACTIONS
Stoichiometry – the calculation of the quantities
of reactants and products involved in a
chemical reaction.
Relationship between mass and moles.
3.6 MOLAR INTERPRETATION OF A CHEMICAL
EQUATION
N2 + 3H2 = 2NH3
How much hydrogen is required to give a
particular quantity of ammonia?
One N2
One mole of N2 reacts with
Three H2
three moles of H2 to give two
Two NH3
moles of NH3.
Write pg. 81
3.7 AMOUNTS OF SUBSTANCES IN A CHEMICAL
REACTION
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