CHEM 1411 - Chapter 3
STOICHIOMETRY
Atomic mass , Atomic weight or mass number (A)
Atomic mass of an element is the total number of protons and neutrons present in its atom. The
unit of atomic mass is amu (atomic mass unit).
1 amu = 1.661 x 10-24 g
Average atomic mass of elements (mass number)
It is defined as the average relative mass of one atom of the element, compared to the mass of a
C12 isotope (C12 isotope with atomic mass of 12 amu is taken as the standard one). The unit of
average atomic mass is amu.
Mass spectrometer
This is the instrument used to determine the atomic and molecular masses accurately. The
sample is converted to positive ions by collisions with a stream of high-energy electrons upon
entering the spectrometer. The charged sample is then accelerated using an applied voltage.
The ions are then passed into an evacuated tube and through a magnetic field. The magnetic
field causes the ions to be deflected by different amounts depending on their mass. Based on the
deflections produced, the mass is calculated.
Q: Calculate the average atomic mass of Ne, which is composed of three different isotopes of
atomic masses 20.00 amu, 21.00 amu and 22.00 amu. The natural abundance of each isotope is
90.92%, 0.257% and 8.82% respectively.
Molecular mass
The sum of the atomic masses of the elements present in one molecule of the compound is the
molecular mass.
Formula mass:
It is the sum of the relative masses of all the atoms in the formula unit of that compound.
Formula mass is calculated for the ionic compounds because they do not exist as molecules. They
exist as their ionic pairs. Molecular mass is calculated for covalent compounds.
Formula mass of a compound is identical to its molecular mass, which is the mass in amu, of a
molecule of that compound.
Ex:
NaCl
22.99 amu + 35.45 amu = 58.44 amu
CaCl2
40.08 amu + (2 x 35.45 amu) = 111.0 amu
C6H12O6
(6 x 12.01 amu) + (12 x 1.008 amu) + (6 x 16.00amu) = 180.2 amu
The Mole:
The amount of substance that contains 6.022 x 1023 elementary entities is said to be one mole.
This number is called the Avogadro’s number
One mole contains 6.022 x 1023 elementary entities
1 mole atoms
= 6.022 x 1023 atoms
1 mole molecules
= 6.022 x 1023 molecules
1 mole ions
= 6.022 x 1023 ions
1 mole oranges
= 6.022 x 1023 oranges
1 mole of element = Atomic mass in grams
1 mole of compound = Molecular mass in grams
1 mole of ions = Ionic mass in grams
Number of moles = Mass in grams / molar mass
The Molar mass
Mass of one mole of a substance expressed in grams is the molar mass.
Volume of 1 mole of a gas at STP = 22.4 L
Substance Formula mass or
molar mass
particles per mole
moles
Molecular mass
H2
2.016 amu
2.016 g
6.022 x 1023 molecules of H2
= 2 (6.022 x 1023) atoms of H
1 mole molecules
2 mole atoms
H2O
18.02 amu
18.02 g
6.022 x 1023 molecules of H2O
= 2 (6.022 x 1023) atoms of H
= 1 (6.022 x 1023) atoms of O
1 mole molecules
2 mole atoms
1 mole atoms
NH3
17.03amu
17.03 g
6.022 x 1023 molecules of NH3
= 1 (6.022 x 1023) atoms of N
= 3(6.022 x 1023) atoms of H
1 mole molecules
1 mole atoms
3 mole atoms
Ex.: How many atoms are there in a penny that weighs 3 g of Cu atoms
= (3g Cu) (1mol Cu /63.5g Cu)(6.022 x 1023 Cu atoms/1 mol Cu)
= 3 x 1022 Cu atoms
Percentage Composition of Elements
Percentage composition of Element
=
n x Molar mass of element x 100
Molar mass of Compound
‘n’ is the number of moles of the element present in one mole of the compound.
Calculate the percentage composition of elements in Sulfuric acid.
Empirical Formula
The formula that gives the simplest whole number ratio of the constituting atoms in a molecule
of the compound is called the empirical formula.
Molecular Formula:
The formula that shows the actual number of each kind of atoms present in a molecule
Finding the empirical formula
Q: A compound contains 44.66% Potassium (K), 18.13% Phosphorous (P), 36.63% Oxygen
(O) and 0.5785% Hydrogen (H). Find the empirical formula.
Element
Percentage by mass
# of moles
K
44.66
(44.66/39.10) = 1.142
# of moles
Least value
(1.142/. 5739) = 1.990
Ratio
P
18.13
(18.13/30.97) = 0.5854
(.5854/. 5739) = 1.020
1
O
36.63
(36.63/16.00) = 2.289
(2.289/ .5739) = 3.989
4
H
0.5785
(.5785/1.008) = 0.5739
(.5739/.5739) = 1.000
1
2
Thus, the formula of the compound is K2HPO4.
Molecular Formula = n x Empirical Formula
Molecular Formula mass
n =
Empirical Formula mass
“Law of Conservation of Mass” by Antoine Lavoisier
The law state that mass is conserved in a chemical reaction. It follows that atoms are neither
created nor destroyed during any chemical reaction.
Balancing of Chemical Equation. Equalizing the number of atoms of each element on either
side of the equation is called balancing of chemical equations. Reactants are written on the left
hand side and products are written on the right hand side of the equation. Balancing of
equations is in accordance with the Law of Conservation of Mass.
For balancing, we need to adjust the stoichiometric coefficients by putting suitable coefficients.
Never change the subscripts while balancing the equation.
CH4 + 2O2
CO2 + 2H2O
2C2H6(g) +
7O2(g)
4CO2(g) + 6H2O(g)
Chemical equation must have an equal # of atoms of each element on either side.
Stoichiometry in chemical reactions.
This is the quantitative study of reactants and products in a chemical reaction.
Chemical equation gives a description of chemical reactions. There are two parts of a chemical
equation: reactants (left of the arrow) and products (right of the arrow):
2H2 + O2
H2O
Read the + sign as “reacts with” and the arrow as “produces”.
Numbers in front of the formulas are called Stoichiometric coefficients; give number of
moles( or molecules) of reactants and products.
Numbers in the formulas (they appear as a subscripts); give number of atoms in a molecule.
Example: Consider the reaction of methane with oxygen:
CH4 + O2
CO2 + H2O
Atoms in the reactants: 1 C, 4 H, and 2 O.
Atoms in the products: 1 C, 2 H, and 3 O.
Quantitative information from balanced chemical Equations
Consider the balanced reaction:
2H2(g)
+
O2(g)
2H2O(l)
2 moles
1 mole
2 moles
23
23
Molecules 2(6.02 x 10 )
6.02 x 10
2(6.02 x 1023)
4 x 1.008g
2 x 16.00g
2 x 18.02g
Limiting reactant.
The one or more reactants that are completely used up in a chemical reaction
are called limiting reactants.
Consider 10 H2 molecules mixed with 7 O2 molecules
According to balanced chemical equation, the stoichiometric ratio of H2 to O2 is 2 to 1:
2H2(g) + O2(g)
2H2O(l)
i.e. 2 molecules of hydrogen react completely with 1 molecule of oxygen
This means that 10 H2 molecules require 5 O2 Molecules. Since we have 7 O2
molecules, the limiting reactant is Hydrogen as it is completely used up. The O2 is
present in excess by 2 molecules.
2CO + O2
2CO2
We know that 2 moles of CO( 2 x28.01g) react with 1 mole of Oxygen (2 x 16.00g) to
form 2 moles of Carbon dioxide( 2 x 44.01g). If we mix 1 mole of CO with 2 moles of O2,
CO is completely used up in the reaction leaving O2 excess. Thus CO is the limiting
reagent.
Reaction yield
Actual yield
Percentage yield
x 100
=
Theoretical yield
Theoretical yield is the amount of product that would be formed if all the limiting
reactant were reacted.