Chemical Composition
8.1 Counting by Weighing
 Which method is more effective?
 Counting jelly beans vs. weighing jelly bean
total mass
 Average mass = ------------------
number of samples
8.2 Atomic Mass: Counting Atoms by
Weighing
 Atomic mass unit (amu): unit of mass for an atom
 1 amu = 1.66 x 10-24 g
Example
 Calculate the mass in amu of a sample of
 aluminum contains 75 atoms
 Carbon contains 62 atoms
 Calculate the number of
 sodium atoms present in a sample that has 1172.49 amu
 Copper atoms present in a sample that has a mass of 1779.4
amu
8.3 The Mole
 SI unit for an amount of substance
 Countable units
 Can be obtained from a balanced equation
E.g C(s) + O2(g)  CO2(g)
1 atom of C reacts with 1 molecule O2 to yield 1 molecule of
CO2
OR
1 mol of C reacts with 1 mole O2 to yield 1 mole CO2
8.3 The Mole
 Described the number of objects present in a dozen
 Based on C-12
 1 mol of anything = 6.02 x 1023 units of that substance
(atoms, molecules, particles, dollars etc..)
 Avogadro’s number
 1 mol of an atom = atomic weight of an atom in grams
 Use as a conversion factor
 E.g Atomic weight of Na = 22.99 amu
 1 mol Na = 22.99 g Na
Steps in determination the
mole
 Determine the final unit
 Determine the atomic mass of an atom
 Write out an equivalence
 Express equivalence as a conversion factor
Example
 Calculate the number moles of atoms and the number of
atoms in
 2.50 g sample of calcium
 57.7 g sample of sulfur
8.4 Molar Mass
 Like density, it’s a ratio of two numbers.
 Mass per one mole of substance

Unit => g/mol
 E.g if 0.400 mol of a compound has a mass of 17.6 g, what is
the molar mass of the compound?
 Replaced with molecular weight because molar mass
described the concept more accurately
 Formula weight is used for ionic compound
Example
 What is the molar mass of CH4?
 Calculate molar mass of sulfur dioxide, a gas produced
when sulfur containing fuels are burned.
 Calculate the molar mass for each of the substance below
(record to two decimal places)
 Water
 Ammonia
 Barium hydroxide
Calculating the Mass from Moles
 Recall
 Molar mass is the mass per one mole of a substance
 1 mol of substance = # g
 Use as a conversion factor
 E.g
if the molar mass of NaCl = 58.44 g/mol
Example
 Calcium carbonate (also called as calcite), is the principal
mineral found in limestone, marble, chalk, pearls and the
shells of marine animals such as clams
 Calculate the molar mass of calcium carbonate
 A certain sample of calcium carbonate contains 4.86 mol.
What is the mass in grams of this sample?
Example
 Calculate the mass of 4.85 mol of acetic acid, HC2H3O2.
Vinegar is a dilute solution of acetic acid
 Calculate the molar mass of sodium sulfate. A sample of
sodium sulfate with a mass of 300.0 g represents what
number mole of sodium sulfate?
 How many water molecules are in a 10.0g sample of water?
8.5 Percent Composition
 Percent composition – the mass present of each element
present in a compound
 % composition =
 To calculate the percent composition (percentage composition)
of a compound
 Calculate the molar mass of the compound
 Calculate the total mass of each element present in the formula of the
compound
 Calculate the percent composition (percentage composition):
% by weight (mass) of element = (total mass of element present ÷
molecular mass) x 100
Example
 Carvone is a substance that occurs in two forms, both
of which have the same molecular formula (C10H14O)
and molar mass. One type of carvone gives caraway
seeds their characteristic smell; the other is responsible
for the smell of spearmint oil. Calculate the mass of
percent of each element in carvone.
 Determine the mass percent of each element in sulfuric
acid.
8.6 Formula of Compounds
 Formula of a compound represents the relative numbers of
various types of atoms present.
 E.g CO2, H2O etc..
 Empirical formula: simplest formula
 Molecular formula: the actual formula of a compound
8.7 Calculation of Empirical Formula
 In each case below, the molecular formula for a
compound is given. Determine the empirical formula
for each compound
 C6H6. This is the molecular for benzene; a liquid commonly
used in industry as a starting material for many important
products
 H2O2. This is hydrogen peroxide, a substance commonly
diluted with water and used as a disinfectant
 CCl4. This is carbon tetrachloride, an organic solvent
8.7 Calculation of Empirical Formula
 Step 1: Obtain the mass of each element present (in grams)
 Step 2: Determine the number of moles of each type of atom
present
 Step 3: Divide the number of mole of each element by the
smallest number of moles to convert the smallest number to 1.
If all of the numbers so obtained are integers, these are the
subscripts in the empirical formula. If one or more of these
number are not integers, go on to step 4
 Step 4: Multiply the numbers you derived in step 3 by the
smallest integer that will convert all of them to whole numbers.
This set of whole numbers represents the subscripts in the
empirical formula
Example
 In a lab experiment it was observed that 0.6884 g of lead
combines with 0.2356 g of chlorine to form a binary compound.
Calculate the empirical formula of this compound
 When a 2.00 g sample of iron metal is heated in air, it reacts with
oxygen to achieve a final mass of 2.573 g. Determine the
empirical formula for this iron oxide
Calculating Empirical Formula from
Percent Composition
 Cisplatin, the common name for a platinum compound that
is used to treat cancerous tumors, has the composition
(mass percent) 65.02% platinum, 9.34 % nitrogen, 2.02%
hydrogen, and 23.63% chlorine. Calculate the empirical
formula for csiplatin
8.8 Calculation of Molecular Formulas
 Molecular formula – gives the actual numbers of
atoms in a molecule
 Molecular formula = n x empirical formula
 may be the same as the empirical formula
 a multiple of the empirical formula (or # of empirical unit) is
defined:
 Multiple ( # of empirical unit) or
n=
Example
 A compound used as an additive for gasoline to help
prevent engine knock shows the following percent
composition
71.65 % Cl
24.27 % C
4.07 % H
The molar mass is known to be 98.96 g/mol. Determine
the empirical formula and the molecular formula for
this compound
 A compound containing carbon, hydrogen, and oxygen
is found to be 40.00% C and 6.700% by mass. The
molar mass of this compound is between 115 g/mol
and 125 g/mol. Determine the molecular formula for
this compound