Stoichiometry
Readings for this section
Petrucci: Chapter 2 and Chapter sections 3.1 - 3.3 and Chapter 4
Stoichiometry
Stoichiometry refers to all quantitative aspects of chemical composition and reactions.
The word comes from the Greek words for "element" and "measure"
Atomic Structure
Consider Hydrogen. A hydrogen atom can be thought of as containing two particles,
Proton p
mp = 1.6726 x 10-27 kg
Charge = + 1.6022 x 10-19 C
Electron e-
m e = 9.1096 x 10-31 kg
Charge = - 1.6022 x 10-19
C
One model of the hydrogen atom has the two types of particle orbit each other, much like the moon and earth do. Using newtonian physics to describe their paths, we see that the larger mass object moves very little while the smaller mass object moves much more. Since the mass of the proton is about 1836 times that of an electron, it will move 1836 times less than the electron (since F = ma).
Thus, we can picture the atom as a central "fixed" nucleus with an electron orbiting around it. This is in line with the Rutherford model of the atom, which was later developed further by Bohr.
Unfortunately, this model doesn't explain why the electrons don't simply give up their energy and decay their orbit until they contact the nucleus and stay there. Neutonian physics predicts that the moon and earth (and all orbiting bodies) will eventually suffer this same fate.
A
more modern treatment of the atom using quantum mechanics predicts that the atom
can be thought of more like a fixed nucleus with electron position being undeterminable,
but a "likelihood" of locating the electron around the nucleus would look like
a probability cloud (e- cloud).
e- density depends on
distance r (more later)
and on angular position
There are 3 types of Forces in Nature:
-
gravitational
-
electrostatic ¬ important in Chemistry
-
nuclear
All electrostatic forces can be described using Coulomb's
Law
-1C = charge of 6.241 x 1018 electrons
We define e as equal to the charge of one proton
e = 1.602 x 10-19 C
So therefore,
-e = charge of one electron
we often use e as a unit of charge (more convenient)
e.g. H+ ® proton ® charge = +1e
e- ® electron ® charge = -1e
He2+ ® Helium Nucleus ® charge = 2e
But mHe = 4 mH
\ He
Þ 2 protons + 2 Neutrons
+e charge neutral
Nucleus
Mass of Atoms
As a first approximation, we can add up the mass of the individual particles within an atom to calculate it's mass. This method yields calculated masses which are not very accurate for various reasons but we can still use the concept.
When we write atomic symbols, we often include the mass number and atomic number as superscripted and subscripted prefixed numbers, respectively where
Mass Number = Total # of Nucleons
Atomic number Z = Number of Protons in the nucleus.
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subscript 1 is Not needed
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all have eight protons.
Chemical reactions
® all nuclei remain unchanged only the electrons are redistributed.
Nuclear reactions
® Nuclei are changed
(high energy bombardment by proton, He2+, etc.
Chemicals with different isotopes have very similar chemical & physical properties
| Freezing | Boiling | |
| H2O | Oº C | 100ºC |
| (heavy water) D2O | 3.82 ºC | 101.42 ºC |
In a naturally occurring sample of hydrogen, we find mostly H, but some D
relative abundance
0.99985 H
0.00015 D
We can use the relative abundances along with the atomic mass of each isotope to determine an average mass for a sample of hydrogen.
Example: Calculate the mass of an "average" atom of hydrogen found in nature if the isotope masses of H = 1.00783 u and of D = 2.01410 u.
Mass of H (= 1.00783)* 0.99985 = 1.00768 u
Mass of D (= 2.01410)* 0.00015 = 0.00030 u
Average Mass of Hydrogen = 1.00798
u
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NOTE: Symbol u stands for atomic
mass unit |
We can use average measured mass of an element to determine the relative abundances of the isotopes that comprise it.
Example Chlorine has two isotopes 
, with masses of 34.96885 u and 36.96590 u respectively.
The average atomic mass of Chlorine is 35.453 u. What are the relative abundances (in percent) of the two isotopes?
Let fractional abundance (mass fraction) of 35Cl be x
Therefore, the fractional abundance of 37Cl is 1 - x .
we thus have
(34.96885 u) (x) + (36.96590 u) (1-x) = 35.453 u
often some algebra: x = 0.7576 1-x = 0.2424
\ percent abundance of 35Cl = 75.76%
percent abundance of 37Cl = 24.24%
We often use atomic mass or Relative Atomic Mass as a short form for average atomic mass.
See Table on the inside Front cover of Petrucci (current text book).
The accuracy to which the atomic masses are known vary, depending on stability of isotopes and on constancy of isotopic ratios. The uncertainty in the last decimal place in the table is indicated by the number in parentheses listed after each of the masses from the table. We see that the number of figures of accuracy differs for each element.
Conservation of Mass & Energy
Much of the work of measuring and calculating amounts in chemistry depends on the law of conservation of mass, which can be stated as
Mass can neither be created, not destroyed.
Let's try a simple test of this law.
If we add the mass of
2 protons = 2 x 1.00728 u = 2.01456 u
2 neutrons = 2 x 1.00866 u = 2.01732 u
2 electrons = 2 x 0.00055 u = 0.00110 u
4.03298 u
but mass of 4He is 4.00260 u
difference is 0.03038 u
This is because of energy released as the particles combine to form the atom.
E = mc2
So we see that mass is not conserved. Energy and mass can be inter-converted
Thus: original Law
Mass can neither be created, nor destroyed, is false.
We now have Two Laws:
- In a chemical reaction, atoms (and hence mass) are neither created nor destroyed.
- Energy can be neither created nor destroyed but can be changed from one form to another.
The Mole
Formula mass & Molecular Mass
Formula Mass
H2O2 ==> 2(1.008 u) + 2(16.00 u) = 34.02 u
Since H2O2 is a molecule, this is a Molecular Formula
Molecular Mass of H2O2 is 34.02 u
(empirical formula for hydrogen peroxide is HO)
NaCl Þ 22.99 u + 35.45 u = 58.44 u
NaCl does not form Molecules \ NaCl is not a molecular formula, it is the Empirical formula
Emperical Formula Mass of NaCl is 58.44 u
Distinction between Empirical Formula and Molecular Formula.
Substance , Empirical Formula, Molecular Formula
Water H2O H2O
Hydrogen Peroxide HO H2O2
Table Salt
NaCl
O (Sodium Chloride)
Ethane
CH3
C2H6
Benzene
CH
C6H6
Iron (metal)
Fe
O
The atomic mass unit (u) is too small for real-world experiments, where mass is commonly measured in grams (g). We define a way of relating u to g.
Define N = # of u in one g
i.e. 1 g = Nu = 6.022 x 1023u
we call N Þ Avagadro's constant (or number)
N = 6.022045 x 1023
This number is conveniently used as a unit of measuring the amount (number) of elementary particles (atom, molecule, ion, etc…) in a sample .
The unit is called the Mole.
Mole Þ the amount of any substance which contains as many elementary entities as there are atoms in exactly 0.012 kg (12g) of 12C.
When the unit mole is used, the elementary entities must be stated, e.g., Atoms, Molecules, Ions, etc…
Average Atomic Mass , Mass of one mole of Atoms
C atom 12.01 u 12.01 g
O atom 16.00 u 16.00 g
Al atom 26.98 u 26.98 g
Molecular Mass , Mass of one mole of Molecules
CO = 28.01 u 28.01 g
CH4 = 16.04 u 16.04 g
Example # moles of Al in 50.00 g of aluminium
50 g Al = ? mol Al atoms
we know Atomic mass Al = 26.98 u
\ 1 mol Al = 26.98 g Al
1 = 1 mol Al atoms
26.98 g Al
50.00 gAl
= 1.853 mol Al atoms
The mass of 1 mole of a substance is its molar mass (M).
For example, the mass of 1 mole of H = 1.008 g
\ Molar Mass H (MH) = 1.008 g/mole
= 1.008 g mol-1
Mass of 1 mole of H2?
1 mol H2 = 2 mol H
mass = # moles x molar mass
= 2 mol x 1.008
g mol-1
= 2.016 g
We must be careful when talking of moles. If we say: "what is the mass of one mole of hydrogen?" we have not distinguished between H atoms and H2 molecules.
\ always specify the elementary particle of interest when discussing moles.
Chemical Calculations: Composition & Formulas
We are often interested in understanding the composition of a substance.
One way of expressing this is in % composition. We can express the percentage of a substance found within a given sample using several measurement techniques. For example, we often use volumes to compare liquids mixed together, c.f. the amount alcohol in beer is about 5%v/v. In other words, 5% of the volume of a bottle of beer is alcohol.
Another very comon measurement technique is measure the mass of the material.
In a given chemical compound, we can use mass measurements to determine the percent by mass of each element found within the compound.
% mass of element mass of element in 1 molecular formula
in compound = or empirical formula of compound x 100%
Molecular Mass or Empirical formula
Mass of Compound or = mass of element in 1 mole of Cpd x 100%
Molar Mass of Cpd
for example:
Mass of one mole of Fe2O3
Þ (55.85 g) (2) + (16.00) (3) = 159.70 g
\ Molar Mass = 159.70 g mol-1
One mole of Fe2O3 contains 2 mol Fe atoms
Mass Fe = 55.85g × 2 mol = 111.70 g
mol
% Fe = 111.70g × 100% = 69.94 %
159.7 g
Similarily for Oxygen:
% O - 3 * (16.00) * 100% = 30.06%
159.7 g
or
100% - 69.94% = 30.06%
Example: how much Nitrogen is needed to make 10.00 kg of ammonia?
Ammonia Þ NH3 ® 1 atom N for every Molecule NH3
% N = Mass of 1 mol N
x 100%
Mass of 1 mol NH3
= 14.01
g x 100% = 82.25%
14.01 g + 3(1.008) g
Mass nitrogen needed to make 10.00 g NH3?
Empirical Formula determination.
e.g. water is 11.19% Hydrogen and 88.81 % oxygen by mass.
Assume 100
g H2O
® 11.19 g H
88.81 g O
Ratio # mol
H atoms = 11.10 mol = 2
® H
# mol O atoms = 5.55 mol 1
® O
Hence H2O
Example: A compound containing only Phosphorous and oxygen was found to have the following composition: 43.7% P 56.3% O by mass .
What is empirical Formula?
Assume 100 g of compound ® 43.7 g P
56.3 g O
Mass measurements to determine the amount of each element in an unknown compound can often be made using analytical techniques whereby the component elements of the compounds are separated and are collected to be weighed. For example, in the combustion of a compound which contains carbon and hydrogen and oxygen like ascorbic acid, we can use the device pictured below to measure the amount of hydrogen (collected as water) and the amount of carbon (collected as carbon dioxide) and with some work, we can calculate the amount of oxygen and hence determine the empirical formula.
Example. A sample of 6.49 mg of ascorbic acid is burned: The mass of the H2O absorber increased by 2.64 mg while that of the CO2 absorber increased by 9.74 mg. What is the empirical formula for Ascorbic acid?
Molar mass CO2 = 12.01 + 2(16.00) = 44.01 g mol-1
Molar mass H2O = 2(1.008) + 16.00 = 18.02 g mol-1
Mass O in Ascorbic Acid = 6.49 mg - (2.66 mg + 0.295 mg) = 3.53 x 10-3 g O
Set up as a table to show data:
| C | H | O | |
| mass | 2.66 x 10-3 g | 0.295 x 10-3 g | 3.53 x 10-3 g |
| moles (of atoms) |
 =2.21 ×10-4 |
 =2.92 ×10-4 |
 =2.21 ×10-4 |
| mole ratio of atoms |
 |
 |
 |
| =1.00 | =1.32 | =1.00 | |
| Whole # ratio of atoms | 3 | 3.96 @ 4 | 3 |
\ empirical Formula = C3H4O3
Chemical Equations
Many chemical reactions can be represented, at least in summary, by listing the reactants and products, separated by an arrow to represent the chemical process involved. Although it is not common that reactions would occur in a single step as might be implied by such a construct, this method is still very useful. This method represents, essentially, the initial and final states of the chemical system involved, not the actual chemical process involved.
Consider the reaction of carbon and oxygen to produce carbon dioxide. We can represent this reaction easily as described above with the following construction.
C + O2 ® CO2
In most reactions, we must do a bit more work to properly identify the proper initial and final state of the chemical system. For thermodynamics, it is very important to exactly identify the state of each chemical.
C(s) + O2(g) ® CO2(g)
Here, we can easily see that carbon is in the solid state while oxygen and carbon dioxide are gases. Here, we have a completed chemical reaction. It properly identifies the states of the chemical system before and after the reaction takes place. Yet, this reaction is still far from completely characterized. We know, for example, that carbon exists as graphite, buckyballs or as other aggregate particles, but rarely as individual atoms as this formula seems to imply.
Other reactions lead to further complications. We need to be concerned with the law of conservation of matter. "Matter can neither be created, nor destroyed. Thus, there must be the same number of atoms of each element on the reactant side of a chemical reaction as on the product side. We'll use this idea to help balance reactions where simply listing the compounds is not sufficient as it was above.
Let's consider the reaction between water and oxygen to produce water. We can write this reaction as a gas-phase reaction and eliminate complications like phase changes, etc.
H2(g) + O2(g) ® H2O(g)
Simply listing the compounds as we have done here does not satisfy the law of conservation of matter. We could easily decide what to do in the above case and rewrite this equation so that is does. By inspection, we have:
H2(g) + 1/2 O2(g) ® H2O(g)
This equation now follows the law of conservation of matter. There are exactly the same numbers of atoms of each element on both sides of the equation.
Let's redo this balancing process using a more rigorous method. This method (with a little modification later) will be useful to you to balance even the hardest of chemical reactions for the rest of your chemistry career. Here are the steps involved:
| 1) Write the unbalanced reaction by simply listing the reactants and products. |
H2(g) + O2(g) ® H2O(g) |
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| 2) Use the chemicals as column headers and add totals for reactants and products. | total | H2 | aO2 | bH2O | total | |
| H | 2 | 2 | 0 | 2b | 2b | |
| 3) Assign letter variables
(red) to indicate the amount of each compound,
starting with the second one. 4) Count each element and put totals in the respective columns |
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| O | 2a | 0 | 2a | b | b | |
| 5) solve for the variables | H: 2 = 2b so, b = 1 O: 2a = b so, a = ½ |
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| 6) Finish the equation by putting the proper coefficients in place (if a coefficient turns out to be negative, move that chemical to the other side of the reaction). | H2(g) + 1/2 O2(g) ® H2O(g) | |||||
That seemed fairly simple. What if the chemicals involved have charges? We must be able to account for the fact that we can't produce or destroy electrons and therefore, the charge should remain constant too.
In this case, we can simply add a row to tally the charge just as we did for each element and still continue as normal.
Let's try another example.
Balance the following reaction.
NO3–(aq) + Cu(s) + H2O(l) ® Cu2+(aq) + NO(aq) + H+(aq)
| Elements | | | | | | | | | | | | |
totals | | | | | | | | | | | | |
NO3– | aCu | bH2O | | | | | | | | | | | | |
cCu2+ | dNO | eH+ | | | | | | | | | | | | |
Totals |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| N | 1 | 1 | 0 | 0 | 0 | 1 | 0 | d | ||||
| O | 3+b | 3 | o | b | 0 | d | 0 | d | ||||
| Cu | a | 0 | a | 0 | c | 0 | 0 | c | ||||
| H | 2b | 0 | 0 | 2b | 0 | 0 | e | e | ||||
| charges | -1 | –1 | 0 | 0 | 2c | 0 | e | 2c+e |
Now, according to the law of conservation of matter, we now have the following equations.
1 = d
3+b = d ... so, b=–2 (this must mean water is on the product side, not the reactant side as we have it here.)
a = c
2b = e so 2×(–2) = e = –4 (this means the H+ is on the wrong side in the table above too. It is a reactant)
–1 = 2c+e = 2c – 4 so 2c = 3 or c = 3/2 (= a)
We've now solved for all the coefficients and we can write the balanced chemical equation.
NO3– + 3/2 Cu + 4 H+ ––––> 3/2 Cu2+ + NO + 2 H2O
Note that this reaction involves the oxidation of copper and reduction of the nitrate ion. This type of reactions is normally considered the most difficult of reactions to balance yet it was straightforward here. We'll study REDOX reactions later in the course in more detail.
The coefficients we determined here are called stoichiometric coefficients and they describe the stoichiometry of the reaction. In other words, they describe the amounts or measure of each compound involved in the reaction in terms of molecules (or moles of molecules).
Calculations using balanced chemical equations.
How many grams of chlorine are needed to react with 0.245 g of hydrogen to give HCl? How much HCl is formed?
1) Write the balanced chemical reaction: H2 + Cl2 ––––> 2 HCl
2) Use what we know to calculate stepwise towards our answer. Never try to figure out a single equation of a single method to solve these equations. Always take it a step at a time and look for common methodologies in the steps.
According to the stoichiometry, #mol Cl2 = #mol H2 = 0.122 mol.
0.122 mol Cl2 * 70.90 g/mol = 8.65 g Cl2
Also, accordingly, #mol HCl = 2*0.122 mol = 0.244 mol.
0.244 mol HCl * 36.46 g/mol = 8.90 g HCl.
Limiting reagent is the reagent that is used up first in a reaction. We can often determine from the stoichiometry whether a given initial amount of a chemical is in sufficient quantity to react or whether it will be used up before the other chemicals are finished reacting. If all the chemicals are completely used up at exactly the same time the initial reaction mixture was said to me composed of a stoichiometric mixture of the reactants. This rarely occurs unless carefully planned. It is not even a desirable situation in many case.
Example: Consider 3 moles of SO2 reacting with 2 moles of O2 to give SO3
- What is the limiting reagent?
- What is the maximum amount of SO3 that can form?
- How much of the remaining reactant is left after the reaction is completed?
1) Balance the reaction: 2 SO2 + O2 ––––> 2 SO3.
The stoichiometric coefficients tell us the ratio of each reactant needed. In this case for every two moles of SO2, we need 1 mole of O2. Our initial conditions are given as 3 moles of SO2 and 2 moles of O2
Trial and error is often the best way in this type of situation.
a) Let's assume that O2 is the limiting reagent.
| 2 moles of O2 means we need 2×2 = 4 moles of SO2. But we
only have 3 moles of SO2 so O2 must not be the limiting reagent.
The only other choice is the SO2. # moles of O2 used in the reaction is given using the ratios of the stoichiometric coefficients as follows:
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b) Maximum amount of product is if all SO2 is used up. Again, we use ratios of coefficients to determine this:
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c) Since, from a) we see that 1.5 mol of O2 is use up, we can easily determine that 0.5 mol of O2 is remaining after the reaction is complete.
The example given here is simplified in that all measurements are given in moles. We always must use moles when working with the stoichiometric coefficients. If our question gives us grams or volumes and concentrations, we must convert to moles first before proceeding. Remember that Dalton's Law says that we can use partial pressures in place of moles for these ratio type calculations since pressure is proportional to moles (for an ideal gas).
Here is another example of the type of calculations you may need to do in your chemistry problems. In this case, we are not interested in a reaction but in calculating concentrations in solutions. This turns out to be more an exercise in algebra than in chemistry. Note below that the units of concentration are requested in moles/dm3. We can determine easily that 1 dm3 = 1 L, so we are actually looking for the commonly used concentration term Molarity (mol L–1) but expressed as SI units.
20.36 g of NaCl is dissolved in sufficient water to form 250. mL of solution. What is the concentration in moles per dm3?
Molar Mass of NaCl = 58.44 g mol–1
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20.36g NaCl |
= 0.3484 mol NaCl |
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58.44 g/mol |
This is dissolved to form 250 mL of solution so:
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0.3484 mol NaCl |
= 1.39 M NaCl |
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0.250 L solution |
where M represents the units mol/L or mol dm–3.
What volume of Conc. H2SO4 (98% by mass, density=1.84 g cm–3) is required to make 10.0 L of 0.200 M H2SO4 solution.
In this case, we work backwards since we only really know the final solution.
# Mol H2SO4: 10.0 L × 0.200 mol L–1 = 2.00 mol H2SO4
Mass H2SO4: 2.00 mol H2SO4 × 98.08 g/mol = 196 g H2SO4. (NOTE: 98.08 is not the 98% mentioned in the question. It's the molar mass of pure H2SO4)
Mass Conc. H2SO4:
Volume of Conc. Acid needed:
