Stoichiometry
1790's Lavoisier proposed the Law of Conservation of Mass: Matter can be neither
created nor
destroyed and there is no detectable change in mass during chemical reactions.
During chemical reactions, atoms are simply rearranged.
Stoichiometry: study of the quantitative relationships of chemical processes.
3.1 Counting by Weighing
As you know, it is far more practical to count many small objects by weighing them, than
by counting them individually. This is especially true of atoms. We could not count
them individually because of their extremely small size. So we are forced to do so by
a different method. That is where Avogadro’s mole comes into play.
3.2 Atomic Masses
Because atoms are so tiny, it is impractical to report their masses in grams, so an arbitrary
unit called the atomic mass unit was developed. It is symbolized: amu or sometimes µ.
• 1 amu = 1.66 x 10-24 g or
1 g = 6.02 x 1023 amu
• The relative standard of atomic mass is the isotope carbon-12. A 12C atom = exactly
12.0000 amu.
• Other atoms have nearly, but not exactly whole number masses. The reasons they are
not whole numbers is something we can investigate later if you wish.
The term atomic mass usually refers to the mass of a single atom or isotope of an element.
The term atomic weight refers to a weighted average of all the isotopes of an element, and
therefore, atomic weight as reported on the periodic table is most often not a whole
number.
The Mass Spectrometer
One of the most important tools of the modern chemist is the mass spectrometer. This
instrument allows highly accurate analysis of chemical samples with regard to the
weights of the various particles making up the sample. A picture of the instrument is
shown on p. 78 in your text.
A gaseous sample of material is subjected to a stream of high-energy electrons. The
electrons produce (+) charged ionic particles.
• The (+) ions are accelerated down an evacuated tube by a (-) charged electric grid.
• The ions are shot through 2 focusing slits which produce a narrow beam of (+) ions.
• The (+) ions are then deflected by a magnetic field. Ions are thus separated based on
their mass (m) to charge (e) ratio (m/e).
• Most of the ions are +1 charge, so deflection depends largely on mass. Lighter atoms
are deflected more than heavier ones.
•
By varying the acceleration voltage on the grid, and/or the strength of the magnetic
field, particles of varying mass can be separated with great accuracy. Furthermore, the
relative number of ions striking the detection grid, can also be determined with great
accuracy, giving precise information on the relative abundance of each type of particle
by mass.
The mass spec provided the first incontrovertible evidence for atomic isotopes and highly
accurate information on atomic masses, relative abundance of each isotope and
therefore atomic weights.
3.3 The Mole
1811: Amadeo Avogadro suggested that equal volumes of all gases under equal
conditions of temperature and pressure contained the same number of molecules.
Big deal, right? Actually, this may have been the single most important contribution to
chemistry and perhaps all science up to that point in man's history. This discovery
eventually lead to the concept of the mole, and the entire branch of chemistry known
as analytical and quantitative chemistry.
A mole is 6.0221367 x 1023 objects. Those objects obviously have a mass. And since
different objects have differing masses, a mole of one type of object has a different
mass than a mole of another type of object, in much the same way that a dozen
marbles has a different mass than a dozen bowling balls. There are 12 of each item,
but their weights and masses are vastly different.
6.02 x 1023 is Avogadro's number. This number was chosen so that atoms and
molecules could be counted by simply using their atomic, molecular or formula
weights.
• The atomic wt. of an element or the molecular/formula wt. of a compound
expressed in grams contains exactly 6.02 x 1023 particles of that substance (give or
take a few million).
• 12.0000 g of 12C atoms contains 6.02 x 1023 C atoms = 1 mol of 12C.
• 24.3 g of Mg atoms contains 6.02 x 1023Mg atoms = 1 mol of Mg.
• 180.0 g of C6H12O6 contains 6.02 x 1023 molecules of C6H12O6 = 1 mol of C6H12O6.
• 58.5 g of NaCl contains 6.02 x 1023 NaCl formula units = 1 mol of NaCl.
3.4 Molar Masses and Molecular Weights
Molecular weight usually refers to the mass of a single particle (atom, ionic formula unit
or molecule).
We even use different terms depending on the particle: atomic weight, formula weight or
molecular weight. The unit will be the amu (atomic mass unit.)
The term “molar mass” refers to the mass in grams of one mole of these particles.
For substances represented by a molecular formula, molecular weight is easily calculated
by simply adding the atomic weights of all the atoms in the formula together.
• Ex. C6H12O6
M. W. =
6 C atoms = 6(12.0 amu) = 72.0 amu
12 H atoms = 12(1.0 amu) = 12.0 amu
6 O atoms = 6(16.0 amu) = 96.0 amu
Total = 180.0 amu
For ionic substances which are represented best by an empirical formula, a formula
weight is calculated in the same way.
• Ex. NaCl
F. W. = 1 Na ion = 23.0 amu
1 Cl ion = 35.5 amu
Total = 58.5 amu
Pay attention to the situation and the problem, but for most situations, 1 decimal
place atomic, molecular and formula weights are sufficient. If greater accuracy
is required, use more decimal places from the periodic table, but pay attention to
sig. figs when doing so.
Never let the numbers from the periodic table be the thing that limits the significant
figures in a calculation problem.
3.5 Percentage Composition from Formulas
These problems are closely related to calculating formula wts. To calculate the % by
mass (weight) of an element in a compound, simply divide the total weight of that
element in the compound formula by the total formula weight.
• Ex. What is the % by weight of C in C6H12O6? Ans: 72.0 amu/180.0 amu x 100% =
40.0%
3.6 Empirical Formulas from Analyses
Problems of this type are basically working % composition problems backwards, with one
twist.
•
•
•
•
•
Using the chemical formula for a compound, calculate the molecular/formula weight.
Assume that you have 100.0 g of the compound.
Divide the total mass of each element in the compound by its atomic wt.
You now have the number of moles of each element in 100 g of the cpd.
To find the empirical formula, simply convert the ratio of moles from your
calculations to their lowest whole number ratio.
Example: Phosgene gas was used as a chemical weapon in WW I. It is highly toxic and
very effective, however, if the wind changed, you were hosed. It still has industrial
uses today. Phosgene is composed of C, O and Cl. Chemical analysis of phosgene
produces the following results: 12.1% C, 16.2% O and 71.7% Cl by Mass. What is
the empirical formula of phosgene?
Moles C = 12.1 g (1 mol C/12.0 g) = 1.01 mol C
Moles O = 16.2 g (1 mol O/16.0 g) = 1.01 mol O
Moles Cl = 71.7 g (1 mol Cl/35.5 g) = 2.02 mol Cl
It is now obvious that the C:O:Cl ratio is 1:1:2, so the empirical formula for phosgene
is COCl2.
Sometimes the ratios are no so neat. In these cases, take the lowest number in the group,
and divide all the numbers in the set by this value. This will set the lowest value at 1,
and hopefully a simple relationship between it and the other values will emerge.
3.7 Chemical Equations
Chem. rxns (reactions) are represented by balanced equations, which are both qualitative
and quantitative.
Equations must be balanced using coefficients: CH4 + 2O2  CO2 + 2H2O
•
C8H18 + O2  CO2
+
H2O becomes 2C8H18 + 25O2  16CO2 + 18H2O
Symbols which frequently are found in chemical equations that give additional
information about the components or conditions include: g, l, s, aq, up and down arrows,
 , Ni
3.8 Balancing Equations
The basic consideration here is the conservation of mass. All atoms present at the start of
a chemical process must also be there at the end.
1. Determine what type if reaction is occurring. Write the correct formulas for the
reactants and products, and the physical state of each.
2. Write the unbalanced equation that summarizes the reaction in Step 1.
3. Balance by inspection starting with the most complicated molecules. Write in the
coefficients necessary to get the atoms on the reactant side to balance with those
on the reactant side.
3.9 Stoichiometric Calculations: Amounts of Reactants and Products.
Coefficients in a balanced chemical equation represent:
• relative numbers of molecules and
• relative number of moles.
Since moles can't be measured on a balance, these relationships can be used to calculate
measurable quantities (e.g.. grams, liters, etc.)
Ex. 2 CuFeS2(s) + 5 O2(g)  2 Cu(s) + 2 FeO(s) + 4 SO2(g)
chalcopyrite
How much copper metal can be produced from 1.00 g of chalcopyrite ore?
1.00 g CuFeS2 x 1 mole CuFeS2 x 2 mol Cu
x 63.5 g Cu = 0.347 g Cu
183 g CuFeS2
2 mol CuFeS2
1 mol Cu
Sequence: g reactant  mol reactant  mol product  g product
Each arrow represents a calculation. Any time you go from moles to grams or vice versa,
use the molar mass of the substance. Any time you must go from moles of one
substance to moles of another, use the mole ratio from the balanced equation.
The Sample Exercises on Pp. 105-6 may be helpful.
3.10 Limiting Reactant (or limiting reagent)
Under given conditions, the mol ratios in which substances react is usually fixed.
Therefore, the amount of 1 material may limit the amount of another that can react.
• 3 molecules of H2 + 1 molecule of O2  2 molecules of H2O + 1 molecule H2
The O2 limits how much water can form and therefore how much H2 can react.
Problem: A solution contains 3.50 g of dissolved Na3PO4. A second solution contains
6.40 g of dissolved Ba(NO3)2. How many grams of Ba3(PO4)2 can be prepared?
Calculating Yield. It is possible to calculate the amount of product that should be created
by a reaction. This calculated value is the theoretical yield. However, this amount is
rarely produced, for a variety of reasons. Dividing the actual yield by the theoretical
yield produces the percent yield. There are sample exercises on pp. 95-6.