Chemical Composition
8.1 Counting by Weighing
• One very important aspect of chemistry is
the synthesis of new substances. One of
the first things a chemist needs to do when
making a new substance is identifying it…
what is its composition? What is its
chemical formula?
• Here we will learn how to determine a
compound’s formula…
Counting by Weighing
• A strategy used to determine amounts of
substances
• The Chemistry of Jelly Beans! (205)
• Example:
– If you have a bag of jelly beans weighing 543
g and you know that, on average, one jelly
bean weighs 1.47 g, about how many jelly
beans are in the bag?
• 543 g of jelly beans x (1 jelly bean / 1.50g) =
• = 362 jelly beans!
• Because atoms are so tiny, the normal units
of mass are much too large to be convenient.
• Scientists have defined a smaller unit of
mass called the atomic mass unit (amu).
• 1 amu = 1.66 x 10-24 g
• The average atomic mass is the atomic
mass of all the isotopes of a particular atom
in relation to their percent abundance.
The Mole
• The mole is a unit used to describe the
number of atoms. It is abbreviated mol.
• One mole of anything consists of 6.02 x
1023 of that substance.
• Avogadro’s Number = 6.02 x 1023
• Named after Amadeo Avogadro
• National Mole Day – October 23 from 6:02
a.m. to 6:02 p.m.
The Mole
• Because… One mole of anything consists
of 6.02 x 1023 of that substance.
• 1 mol of eggs = 6.02 x 1023
• 1 mol of paper = 6.02 x 1023
• 1 mol of jelly beans = 6.02 x 1023
Using the “Mole”
• In order to calculate the amount of grams
or moles of a certain substance, we must
use the substances MOLAR MASS
• Determining Molar Mass
– Use the Periodic Table
– Molar Mass of
• O = 16.00 g/mol
• H = 1.01 g/mol
• C = 12.01 g/mol
• Calculate the number of moles in a 10.0
gram sample of aluminum.
– 10 g Al x (1 mol / 26.06 g)
– Grams cancel
– Left with mol Al
• Calculate the number of atoms in the
above sample of aluminum.
– # Mol Al x (6.02 x 1023 atoms / 1 mol)
– Mol cancel
– Left with atoms
• Calculate the number of moles and the
number of atoms in a 5.68 mg sample of
silicon.
– CONVERT mg to g
– 5.68 mg x (1 g /1000 mg)
– # g x (1 mol / 26.08 g) = mol Si
– # mol Si x (6.02 x 1023 atoms / 1 mol) = atoms
Si
The Mole
• Calculate the number of moles and atoms
for each set of information:
– 25.0 g sample of calcium
– 57.7 g sample of sulfur
• Calculations Video
• The molar mass of any substance is the
mass of 1 mole of the substance.
• The molar mass is obtained by summing
the masses of the component atoms.
• Calculate the molar mass of SO2.
• Calculate the molar mass of C2H3Cl
• Calculate the molar mass of CaCO3
• Calculate the molar mass of Na2SO4
Atomic Masses
• Calculate the molar masses of the
following compounds:
– AlCl3
– H2O
– C6H6
– NH4NO3
– Mg(ClO3)2
– CsO2
Grams
Moles
Moles
Divide by Formula
Mass
Grams
Multiply by Formula
Mass
Molecules
Moles
Moles
Molecules
Divide by Avogadro’s
Number
Multiply by Avogadro’s
Number
Practice!
Percent Composition
• So far we have discussed the composition
of compounds in terms of number of
atoms, it is also useful to know a
compounds composition in terms of the
masses of the elements.
• We can do this from the formula of a
compound by comparing the mass of each
element present in 1 mole of the
compound
Percent Composition
masselement
Mass % 
100
masscompound
Percent Composition
• Example: Determine the % Composition of
Ethanol (C2H5OH)
– 2 C’s
6 H’s
1O
– Step 1: Calculate the molar mass of each element in
the compound and the total molar mass of the whole
compound
• 2 x 12.01 = 24.02 g/mol C
• 6 x 1.01 = 6.06 g/mol H
• 1 x 16.00 = 16.00 g/mol O
46.07 g / mol C2H5OH
– Step 2: Determine the % of each element present by
dividing the molar mass by the total molar mass of the
whole compound
• (16.00 / 46.07) x 100% = 34.73% O
• Calculate the percent composition of C6H6.
• Calculate the percent composition of
C10H14O.
• Calculate the percent composition of
C14H20N2SO4.
• An empirical formula is the lowest whole
number ratio of atoms in a compound.
• A molecular formula gives the composition
of the compound.
• It is possible for more than one compound to
have the same empirical formula but it is
impossible for more than one compound to
have the same molecular formula.
STEP:
Obtain the mass of each element present
(in grams).
Determine the number of moles of each
type of atom present. (Divide grams by
formula mass).
Divide the moles by the smallest number.
Get whole numbers (by multiplying)
• Find the empirical formula of sample of vanadium oxide
which contains 0.3546 grams of vanadium and 0.2784
grams of oxygen.
– Convert g  mol
• 0.3546 g V x (1 mol / 50.94 g V) = 6.96 x 10-3 mol V
• 0.2784 g O x (1 mol / 16.00 g O) = 0.0174 mol O
– Divide by the smallest number of moles
• 6.96 x 10-3 mol V / 6.96 x 10-3 mol = 1 V
• 0.0174 mol O / 6.96 x 10-3 mol = 2.5
• **Too far to round so multiply both by 2 to get to the nearest
whole #’s
– Write out the formula: V2O5
• Find the empirical formula of a sample of lead arsenate
which contains 1.3813 g of lead, 0.00672 g of hydrogen,
0.4995 g of arsenic, and 0.4267 g of oxygen.
– Convert g  mol
– Divide by the smallest number of moles
– Write out the formula:
• Find the empirical formula of a compound which
contains 65.02% platinum, 9.34% nitrogen,
2.02% hydrogen, and 23.63% chlorine.
• Find the empirical formula of a compound which
has 63.68% carbon, 12.38% nitrogen, 9.80%
hydrogen, and 14.14% oxygen.
• **Hint: When %’s are given, assume that the
total mass is 100 grams so the mass = the %
– Example: 65.02% Pt = 65.02 g Pt
Calculation of Empirical Formulas
• Additional Examples
– Example 8.11
– Example 8.12
– Example 8.13
– Example 8.14
– Self Check Exercise 8.8
– Self Check Exercise 8.9
– Self Check Exercise 8.10
• Worksheets from Mrs. Harlan
1. Find empirical formula.
2. Calculate empirical formula mass.
3. Divide molecular formula mass by
empirical formula mass.
4. Multiply empirical by whole number to get
molecular formula.
5. Calculate molecular formula as a doublecheck
• Determine the empirical and molecular formulas of a compound
which has a molecular mass of 98.96 g/mol and contains
71.65% Cl, 24.27% C, and 4.07% H.
• Calculate empirical formula
–
–
–
–
71.65 g Cl x (1 mol / 35.45 g) =2.02 / 2.02 = 1 Cl
24.27 g C x (1 mol / 12.01 g) = 2.02 / 2.02 = 1 C
4.07 g H x (1 mol / 1.01 g) = 4.03 / 2.02 = 2 H
CClH2
• Determine the empirical formula mass
–
35.45 + 12.01 + 2(1.01) = 49.48 g/mol
• Divide the molecular mass by the empirical formula mass
– 98.96 / 49.48 = 2
• Multiply the empirical formula by the above number
– 2(CClH2) = C2Cl2H4
• Determine the empirical and molecular formulas
of a compound which has a molecular mass of
283.88 g and contains 43.64% P and 56.36% O.
– Calculate empirical formula
– Determine the empirical formula mass
– Divide the molecular mass by the empirical
formula mass
– Multiply the empirical formula by the above
number
Calculation of Molecular Formulas
• Example 8.15
• Self Check 8.11
• Worksheet from Mrs. Harlan