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Chemistry -Moles

Chemical Quantities
Atomic Mass Unit
• Defined as unit of mass used to express atomic
and molecular weights, equal to one-twelfth of
the mass of an atom of carbon-12.
• It is equal to approximately 1.66 x 10-27 kg or
1.66 x 10-24 g.
Relative Atomic Mass
• Based on the 12C scale.
• On the 12C scale, atoms of the isotope carbon-12
are assigned a relative atomic mass (relative
isotopic mass) of exactly 12.
• The relative atomic mass of an element is the
average mass of one atom of the element
relative to one-twelfth the mass of one atom of
• Using the 12C scale, the relative atomic mass of
hydrogen is 1.008 and Mg = 24.312.
Relative Atomic Mass
• The relative atomic mass of carbon is 12.011.
This is because naturally occurring carbon
contains a few atoms of carbon-13 and carbon14 mixed in with those of carbon-12.
• Each element exists as a mixture of isotopes.
Number on the bottom
of each square in the
periodic table is the
average weight of the
element (in amu).
Atomic Mass
• Atomic masses are
determined on a relative
• The standard scale
references the carbon-12
isotope = 12.000 amu
• All other atomic masses are
determined relative to
Relative Molecular/Formula Mass
• The relative molecular mass of an element or
compound is the sum of the relative atomic
masses of all the atoms in its molecular formula.
Ex: Relative molecular mass of H2O (H = 1.008; O
= 15.999) = 18.015 (No units)
• Metal compounds such as sodium chloride and
copper sulphate consist of giant structures
containing ions, not molecules. For ionic
compounds, chemists use the term ‘relative
formula mass’.
Relative Molecular/Formula Mass
• The relative formula mass of a compound is the
sum of the relative atomic masses of all the atoms
in its formula.
Relative formula mass of CuSO4 (Cu = 63.54; S =
32.064; O = 15.999) = 159.60 (no unit)
• Using atomic mass to determine no of atoms:
Ex: How many atoms of lead are present in 1.00 g of
lead sample? (1 amu = 1.66 x 10-24 g)
Ans: 2.9 x 1021 atoms
Using mass spectra to calculate relative atomic
Chlorine consists of a mixture of two isotopes
with relative isotopic masses of 35 (35Cl) and 37
(37Cl) in the proportion of 3:1.
Of every four chlorine atoms, three are chlorine35 and one is chlorine-37.
Thus the average mass of a chlorine atom on the
12C scale, which is the relative atomic mass of
chlorine is: (3/4 × 35) + (1/4 × 37) = 35.5
There are two naturally occurring isotopes of
copper (Cu). 63C has an isotopic mass of
62.9296 amu. 65Cu has an isotopic mass of
64.9278 amu. Calculate the percent
abundance of each isotope of copper.
MCu = 63.546 amu
Ans: 63Cu: 69.15 and 65Cu: 30.85%
The Mole
• It is more convenient to use another unit for large
quantities of particles such as atoms or molecules.
• Chemists measure the amount of a substance in terms
of moles.
• One mole of an element or a compound has a mass
equal to its relative atomic mass, its relative molecular
mass or its relative formula mass in g.
• Molar mass (M) is the mass of one mole of an element
or a compound. The unit for molar masses is g mol-1.
• Ex: One mole of copper is 63.54 g, one mole of water
is 18.015 g; and molar mass of CuSO4 is 159.60 g.
The Avogadro Constant
• Amount of substance (mol) = mass of substance
(g)/molar mass (g mol-1)
• Avogadro constant (6.023 × 1023 mol-1) is the
number of atoms, molecules or formula units in
one mole of any substance.
• 1 mol of iron (55.8 g) contains 6.023 × 1023 Fe
• 1 mol of water (18.0 g) contains 6.023 × 1023 H2O
• 1 mol of CuSO4 (159.6 g) contains 6.023 × 1023
CuSO4 formula units.
The Avogadro Constant
• The relative atomic mass in grams (molar mass) of
any element contains 6.023 × 1023 atoms.
• No of atoms, molecules or formula units = amount
of substance in moles × Avogadro constant
• Avogadro’s constant provides the connecting
relationship between molar masses and atomic
Chemical Formula
The subscripts in a chemical formula indicate the
number of atoms of each element present in a
The subscripts in a chemical formula can also
indicate the number of moles of atoms of each
element present in one mole of a compound.
Ex: In one molecule (mol) of glucose (C6H12O6),
there are 6 atoms (mol) of carbon, 12 atoms
(mol) of hydrogen, and 6 atoms (mol) of oxygen.
The Avogadro Constant
• How many molecules of bromine are present in
0.045 mole of bromine gas?
Ans: 2.7 x 1022
• How many moles of carbon atoms are present in
1.85 moles of glucose?
Ans: 11.1 moles
Molar Mass
• The atomic mass of a carbon-12 atom is 12.00
• The atomic mass of one mole of carbon-12
atoms is 12.00 g
• One mole of any element is the amount of atoms
(molecules or ions) that is equal to its atomic
mass (in grams)
• This mass contains 6.023 х 1023 particles of that
Molar Mass
When the number of grams of a substance equals
the formula mass of that substance, Avogadro’s
number of molecules of that substance are present.
• 3 mg of sodium chloride was present per cup of
coffee. How many moles of sodium chloride are
in each cup of coffee?
Ans: 5.13 x 10-5
• Ethylene glycol (antifreeze) has the formula
C2H6O2. How many molecules are present in a
3.86 × 10-20 g sample?
Ans: 375
Ans: 12.6
• How many moles of sodium ions are present in
0.100 mol of sodium carbonate?
Ans: 0.200 mol
• A chemist has a jar containing 388.2 g of iron
filings. How many moles of iron does the jar
Ans: 6.951 mol
• A student needs 0.366 mol of zinc for a reaction.
What mass of Zn in g should the student weigh?
Ans: 23.9 g.
How many moles of lithium are there in 1.204
1024 lithium atoms?
Ans: 1.999 moles.
How many boron atoms are there in 2.00 g of
Ans: 1.11 x 1023 atoms
• How many moles of carbon dioxide are in 66.0 g
of dry ice, which is solid CO2?
Ans: 1.50 moles
Percent Composition
• It is the percent by mass of each element in a
• Can be determined
• By its chemical formula
• Molar masses of the elements that compose
the compound
• The percent of each element contributes to the
mass of the compound:
mass of each element
mass percent of each 
element in a compound molar mass of the compound
Percent Composition
• What is the percent composition of each
element in NH4OH?
Ans: Molar mass of NH4OH = 35.05g
% N: 39.97; % H: 14.38; %O: 45.65
• What is the % composition of ammonium
Ans: % N: 29.2; % C: 12.5; % H: 8.3; % O:
Empirical Formulae
• The simplest ratio of elements in a
• An empirical formula shows the
simplest (smallest) whole number ratio
for the atoms of each element in a
formula unit of a compound.
• If the percent composition is known, an
empirical formula can be calculated.
Empirical Formulae
To Determine the empirical formula:
1) Calculate the moles of each element
Use molar mass and atomic mass
2) Calculate the ratios of the elts to each other
3) Find the lowest whole number ratio
 Divide each number of moles by the smallest
number of moles present
Note: The subscripts in a formula are expressed as
whole numbers, not as decimals. If the number(s)
are NOT whole numbers, multiply each number
by the same small integer (2, 3, 4, 5, or 6) until a
whole number is obtained.
Empirical Formula
Ex: When 10.00 g of ethene was analysed, it was found to
contain 8.57 g of carbon and 1.43 g of hydrogen. What is
its empirical formula?
Moles of carbon = 8.57/12.00 = 0.714
Moles of hydrogen = 1.43/1.00 = 1.43
Ratio of combined atoms = 0.714/0.714 and 1.43/0.714 =
Empirical formula of ethene = CH2
This formula shows only the simplest ratio of carbon
atoms to hydrogen atoms. The actual formula could be:
CH2, C2H4, C3H6, C4H8, etc. as all these formulae give
CH2 as the simplest ratio of atoms.
Molecular Formula
• Experiments show that the relative molecular
mass of ethene is 28, which corresponds to an
actual formula of C2H4 and not CH2. The
relative molecular mass of a substance can be
obtained using a mass spectrometer.
• A molecular formula shows the actual number
of atoms of each element in one molecule of a
• The empirical formulae of some compounds
like ethene can be calculated using combustion
data in place of their composition by mass.
• The combustion data is obtained by burning the
compound in oxygen to produce carbon dioxide
and water.
• From the masses of carbon dioxide and water
produced, it is possible to calculate the masses
of carbon and hydrogen combined in the
• These masses of carbon and hydrogen can then
be used to calculate the empirical formula of the
Ex: A sample of a cpd containing only C and H was burnt
completely in oxygen. All the C was converted to 3.38 g
of CO2 and all the H was converted to 0.692 g of H2O.
Determine the empirical formula of the compound.
Ans: One mole of CO2 contains one mole of carbon.
44 g of CO2 contains 12 g of carbon.
Mass of C in 3.38 g CO2 = (12 × 3.38)/44 = 0.92 g
One mole of H2O contains 2 moles of H.
18 g of H2O contains 2 g of H.
Mass of H in 0.692 g of H2O = (2 × 0.692)/18 = 0.077 g
Empirical formula = CH
No of moles
0.92/12 = 0.077 0.077/1 = 0.077
Divide by smallest
Empirical and Molecular Formulae
• Calculate the empirical formula and the mass
of the empirical formula
• Divide the given molecular mass by the
calculated empirical mass
• Answer is a whole number multiplier
• n represents a whole number multiplier from 1
to as large as necessary.
molar mass ( g / mol)
empirical formula mass ( g / mol )
Empirical and Molecular Formulae
Multiply each subscript in the empirical formula
by the whole number multiplier to get the
molecular formula
Empirical and Molecular Formula
Lactic acid has a molar mass of 90.08 g and has the
following percent composition:
40.0% C, 6.71% H, 53.3% O
Determine the empirical and molecular formula of lactic
acid? (Ans: CH2O, C3H6O3)
• Assume a 100.0 g sample size, convert percent numbers
to grams
• Convert mass of each element to moles
• Divide each mole quantity by the smallest number of
• Obtain the value of n (whole number multiplier)
• Multiply the empirical formula by the multiplier
Stoichiometric Reactions
Given the reaction:
2 KClO3 → 2 KCl + 3 O2
How many (i) moles and (ii) molecules of
O2 will be produced if 0.400 mol of KClO3
is heated?
Ans: (i) 0.600 mol; (ii) 3.61 x 1023
Stoichiometric Reactions
• Theoretical yield represents the maximum
amount of a product that may be produced from a
given set of reactants.
• If ammonia, NH3, is burned in air, the following
reaction takes place:
4 NH3 + 3 O2 → 2 N2 + 6H2O
How many g of water will be produced with 51.0
g of NH3 in the presence of excess oxygen?
Ans: 81 g
Ans: 16
Ans: 13.8