Zumdahl’s Chapter 3
Stoichiometry
Chapter Contents
Aston’s Atomic Mass
The Mole
Molar Mass
% Composition
Molecular Formulae
Chemical Equations
Balancing Chemical
Equations
Stoichiometric
Calculations
Limiting Reactant
Calculations
Aston’s Atomic Masses
(works for molecules too!)
Mass Spectrometer
Fast atomic ions (current) bend near magnet
Deflection varies inversely with inertia!
Multiple isotopes differ in mass (inertia)
and so give multiple beam deflections.
98.892 % 12C at 12 amu by definition.
1.108 % 13C at 13.00338 amu
averageC  12.011 atomic mass units
Avogadro’s Mole
Definition: One mole of 12C atoms weighs
exactly 0.012 kg (12 g)
Thus, 1 amu  1 g / NAv
Since atoms combine by numbers, NAv has
the advantage of showing combinations by
weights.
NAv = 6.02213671023, the SI count
Molar Masses & % Composition
Trivial; weigh them in mass spectrometer?
Since atomic masses of elements never
vary, MW = sum of atomic weights times
number of atoms in molecule (subscripts).
MW(P4O10) = 4(AWP) + 10(AWO) =
4(30.97) + 10(16.00) = 283.9 g mol–1
% O is 100%  160.0/283.9 = 56.36%
Formula Weight Analysis
Atomic Absorption Spectrometry (AA)
Intensity of atomic glows in controlled flame
gives proportion of atom in molecule.
Organic combustion analysis for CxHyOz
Burn in excess oxygen, O2
Trap and weigh resultant H2O and CO2
Convert weights to moles H and C, resp.
Get mass O by difference with original mass
Scale moles to find simplest integer x, y, & z.
Molecular Formulae
Empirical formula
from elemental
composition must be
scaled by ratio of
MW:FW to obtain
molecular formula.
Standard formulae
show combinations:
Hg6(PO4)2 (why OK?)
Structural formula
give geometric
information as well:
ClH2CCH2COOH, 3chloropropanoic acid,
& digests to C3H5O2Cl
Mineral formulae
show co•crystalites
like Y2(CO3)3•3H2O
Chemical Equations
How many reactants  how many products?
Chemical equations not only codify the perfect
proportions but also note conditions:
CaCO3(s) + 2 HCl(aq)  Ca2+(aq) + 2 Cl–(aq)
.
+ CO2(g) + H2O(l)
Catalysts, photons (h), or heat () may stand
above the reaction arrow ().
The key is molecular consumption/production.
Balancing Chemical Reactions
First know all the reactants and products!
Balance first atoms appearing in only one
molecule on each side. Let’s burn TNT:
C7H5(NO2)3 + O2  7CO2 + H2O + N2
C7H5(NO2)3 + O2  7CO2 + H2O + 1.5N2
C7H5(NO2)3 + O2  7CO2 + 2.5H2O + 1.5N2
C7H5(NO2)3 + 5.25O2  7CO2 + 2.5H2O + 1.5N2
4C7H5(NO2)3 + 21O2  28CO2 + 10H2O + 6N2
Stoichiometric Calculations
While TNT is solid, the O2 and the products
are gases. 1 mol TNT is worth ? moles gas?
4C7H5(NO2)3 + 21O2  28CO2 + 10H2O + 6N2
 Moles gas = ¼(28 + 10 + 6 – 21) = 5.75
Since each gas mole is ~1000 the volume of
a solid mole, TNT is destructive by rapidly
increasing its volume by a factor of 5,750!
More Calculations
A “megaton” of TNT (measure of nuclear
bomb destructivity) would produce what
weight of CO2?
106 tons [ 2000 lbs / ton ] [ 0.4536 kg / lbs ] = 9.072108 kg
MW[C7H5(NO2)3] = 7(12)+5(1)+6(16)+3(14) = 227 g = 0.227 kg
4C7H6(NO2)3 + 21O2  28CO2 + 10H2O + 6N2
9.072  108 kg TNT  [1 mol TNT/0.227 kg TNT] 
[28 mol CO2/4 mol TNT]  [0.044 kg CO2/1 mol CO2]
1.23  109 kg CO2 = 1.23 megatons CO2
Limiting Reactant Calculations
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
What weight of molten iron is produced by
1 kg each of the reactants?
Needed weights are MW(Fe2O3) and 2AW(Al)
or 159.7 g Fe2O3 and 53.96 g Al, respectively.
Smallest ratio Available:Needed is limiting!
1000 g/159.7 g = 6.26 Fe2O3 < 1000 g/53.96 g = 18.52 Al
6.26 mol Fe2O3  [2 mol Fe/1 mol Fe2O3]  [55.85 g Fe/1 mol Fe]
= 699 g Fe