The object of this Chemistry Project Report is to relate some aspect of chemistry to my major. I am a Secondary Mathematics Major, so I decided to do my paper on the mathematics of Chemistry; stoichiometry. Enjoy!

Jaclyn Bloemendaal

Dr. Fred King

Chemistry 115, Section I0I

December 2, 2002

Reaction Stoichiometry

Chemistry, the branch of science concerned with the study of matter and the changes that matter can undergo. So, when a person thinks about writing a chemistry paper, there are hundreds if not thousands of different aspects they could use for a topic. Narrow the choices down by looking for an aspect of chemistry that is related to mathematics. Bingo! That gives you stoichiometry. But what exactly is stoichiometry and how is it used in chemistry and everyday life?

Stoichiometry can be defined in many different ways. According to Jones and Atkins, it is the quantitative relation between the amounts of reactants consumed and products formed in chemical reactions as expressed by the balanced chemical equation for the reaction. A website entitled “Stoichiometry Review” defines it as the study of the quantitative relationships that exist in chemical formulas and chemical reactions. Another definition is the branch of chemistry and chemical engineering that deals with the quantities of substances that enter into, and are produced by, chemical reactions. The website titled “From Caveman to Chemist: Stoichiometry” gives a definition that anyone can understand, not just people who know a lot about chemistry; it is a big long word which describes the general process of figuring out how much stuff you need to make something.

Before a person can begin using stoichiometry, there is a key concept in chemistry they must understand. This concept is the mole, abbreviated mol. The mole is basically the unit of chemical amount and it represents Avogadro’s number of atoms, ions, or molecules; 6.0221x10²³. There is no direct way of measuring this quantity, so chemists use conversion factors.

The most basic stoichiometry is Mole-to-Mole calculation. This is when a balanced chemical equation is used to determine the moles of a substance needed to produce a certain amount of another substance. To do this, the conversion factor called the mole ratio is used. The mole ratio is the substance required over the substance given, and this ratio is obtained by using the coefficients of the materials as they show up in the balanced equation.

Mole-to-Mole calculations can be applied to the synthesis of ammonia. Using this information, take the chemical equation

N_{2 (g)} + 3H_{2 (g)} à 2NH_{3 (l)}

and determine the moles of ammonia that 5 moles of nitrogen would yield. So, it would be set this up like this:

Moles of NH₃ = (5 mol N₂) x (2 mol NH₃/ 1 mol N₂) = 10 mol NH₃

The next step up in stoichiometry is Mass-to-Mass calculation. These calculations are used to calculate the mass of a product that can be obtained from a certain amount of reactant. Mass-to-Mass calculations use Mole-to-Mole calculations in them, but they are much more complex because another conversion factor is used in these calculations, the formula weight. The formula weight is the grams of the substance in one mole over a mole of the substance.

Taking an example from our chemistry book by Jones and Atkins, we want to find the number of kilograms of iron that can be obtained from 10.0 kg of iron (III) oxide. The chemical equation that represents this reaction is

Fe₂O_{3 (s)} + 3CO _(g) à 2Fe _(s) + 3CO_{2 (g)}

We have to use the mole ratio 2 mole Fe/1 mol Fe₂O₃. The calculation is done in four steps: (1) find the molar mass of Fe₂O₃, (2) calculate the moles of Fe₂O₃, (3) calculate the moles of Fe, and (4) find the mass of Fe.

(1) Molar Mass = 2(55.85) + 3(16.00) g/mol = 159.70 g/mol

(2) Moles Fe₂O₃ = (10.0 kg Fe₂O₃) x (1000 g/1 kg) x (1 mol Fe₂O₃/159.70 g Fe₂O₃)

(3) Moles Fe = (10.0 kg Fe₂O₃) x (1000 g/1 kg) x (1 mol Fe₂O₃/159.70 g Fe₂O₃) x

(2 mol Fe/1 mol Fe₂O₃)

(4) Mass Fe = (10.0 kg Fe₂O₃) x (1000 g/1 kg) x (1 mol Fe₂O₃/159.70 g Fe₂O₃) x

(2 mol Fe/1 mol Fe₂O₃) x (55.85 g Fe/1 mol Fe) x (1 kg/1000 g) = 6.99 kg

Mass-to-Mass calculations like these are used in a process called gravimetric analysis. Gravimetric analysis is the use of mass measurements to determine an amount of substance that is present. Usually an insoluble compound would be precipitated out of a solution. Then, the precipitate is filtered and weighted. From the weighed mass, the amount of the original substances can be calculated. A real life application would be gravimetric analysis of water to monitor levels of lead.

Mole-to-Mole and Mass-to-Mass calculations both assume that all reactants going into a chemical reaction are used up and the reaction has gone to completion. In a large majority of reactions this is not the case because the reactants are not always present in the exact mole ratio required. This is due to something called a limiting reactant; a reactant that governs the maximum yield of product in a given reaction. The limiting reactant is supplied in a smaller amount than what is required to react fully with the given amount of the other reactant. This means that it will be fully consumed in the reaction.

The following example of limiting reaction stoichiometry is also taken from the chemistry book by Atkins and Jones. Calcium carbide reacts with water to form calcium hydroxide and the flammable gas ethyne. Which is the limiting reactant when 100. g reacts with 100. g of calcium carbide? There are two ways to do this calculation, and the chemical equation for this reaction is

CaC_{2 (s)} + 2H₂O à Ca(OH)_{2 (aq)} + C₂H₂

The first method begins by calculating the molar mass of both calcium carbide and water, and then converting the given masses into moles. Then, find the mole ratio and find the amount of water needed to react with 100. g of calcium carbide. The molar mass for calcium carbide is 64.10 g/mol, and the molar mass of water is 18.02 g/mol. The mole ratio is 1 mol CaC₂/2 mol H₂O.

(1) Moles of CaC₂ = (100. g CaC₂) x (1 mol CaC₂/64.10 g CaC₂) = 1.56 mol CaC₂

Moles of H₂O = (100. g H₂O) x (1 mol H₂O/18.02 g H₂O) = 5.55 mol H₂O

(2) Moles H₂O to react = (1.56 mol CaC₂) x (2 mol H₂O/ 1 mol CaC₂)= 3.12 mol H₂O

Because 3.12 moles of water are needed to react with the calcium carbide and 5.55 moles are supplied, the calcium carbide is the limiting reactant.

The second method takes the number of moles of each reactant and uses the mole ratios from the chemical equation to determine how many moles of C₂H₂ can be formed. The mole ratios required for this method are 1 mol C₂H₂/1 mol CaC₂ and 1 mol C₂H₂/2 mol H₂O.

(1) Mol C₂H₂ from CaC₂ = (1.56 mol CaC₂) x (1 mol C₂H₂/1 mol CaC₂) = 1.56 mol C₂H₂

(2) Mol C₂H₂ from H₂O = (5.55 mol H₂O) x (1 mol C₂H₂/2 mol H₂O) = 2.78 mol C₂H₂

Because the number of moles formed by calcium carbide is smaller than that formed by water, it is the limiting reactant.

Stoichiometry is very important when it is involved with solutions. Large numbers of medications are given to patients in a solution, and it is imperative that they are administered in the correct dosages. Many medications, when given in large doses, can make a patient sick or even kill them. Stoichiometry in solutions becomes important in chemistry when mixing solutions together.

When performing stoichiometry in solution, molar concentration, or molarity for short, must be known or calculated. Molarity is the number of moles of solute divided by the total volume of the solution measured in liters. The units of molarity are moles/liter, and they are abbreviated M. Here is an example of taken from the chemistry book by Ebbing and Gammon: A sample of NaNO₃ weighing 0.38 g is placed in a 50.0-mL volumetric flask. The flask is then filled with water to the mark on the neck. What is the molarity of the resulting solution?

In order to find the molarity, calculating the moles of solute is vital. Convert the grams of NaNO₃ to moles, and then divide that by the liters of solution to find molarity. 0.38 g NaNO₃ is 4.47x10^-3 mol NaNO₃, and 50.0 mL is 50.0x10^-3 L. So, the calculation would look like this:

Molarity = 4.47x10^-3 mol/50.0x10^-3 L = 0.089 M NaNO₃

Stoichiometry is very important in cases where chemists need to study a substance that is available in limited quantity. Solutions play an extremely important role here because they can be used to transfer small amounts of a substance from one container to another. To do this, simply dilute, or reduce the concentration, of a solution by adding more solvent. This way, the same amount of solute is present in a much larger quantity. For this purpose, a person must know that there is a relationship between the molarity of the solution before dilution and the molarity of the solution after dilution.

Rearrange the molarity equation so that is looks like this:

Moles of solute = molarity x liters of solution

This equation is true for both the initial molarity (M_i) and volume (V_i) and the final molarity (M_f) and volume (V_f). Since the moles of solute stay the same, the relationship can be expressed in this form

M_i x V_i = M_f x V_f

Here is an example. After diluting a 5.0 M solution of NH₄OH, you get 50 mL of 0.225 M solution of NH₄OH. How many milliliters of the original solution were diluted? So apply the information in the problem to the formula above.

(5.0 M) x Vi = (0.225M) x (50.0 mL), Vi = 2.25 mL

Finally, there is stoichiometry that can be done with reacting gases. This is comes into play because quite frequently it is important to now the volume of gases produced in a reaction. This is important because one of the most common causes of explosion in industries is the increasing pressure in a closed container filled by reacting gases. By performing the stoichiometry of reacting gases, we can predict the volume of gas involved in the reaction and prevent explosions.

An important conversion factor used in these stoichiometric equations is the normal gas volume; 24.4 L gas/1 mol gas. That value occurs when the temperature of the gas is 25^oC, but at 0^oC the volume is 22.4 L gas/1 mol gas. An example from the website titled “From Caveman to Chemist: Stoichiometry” explains this process. In the fermentation of glucose, what volume of carbon dioxide is produced when 100 grams of glucose is fermented to ethanol? The balanced chemical equation for this reaction is

C₆H₁₂O_{6 (s)} à 2CO_{2 (g)} + 2C₂H₅OH _(l)

This is how the problem would be set up:

Liters CO₂ = 100 g C₆H₁₂O₆ x (1 mol C₆H₁₂O₆/180 g C₆H₁₂O₆) x (2 mol CO₂/1 mol C₆H₁₂O₆) = 27 L

So, it is evident that reaction stoichiometry is important in every aspect of chemistry. For every reaction there is a balanced chemical equation, and from these equations you can calculate many things using stoichiometry. It can be simply described as the mathematics of chemistry, but there is nothing simple about it. Stoichiometry is very important when it comes to medicine because incorrect doses can have adverse affects on the patient or even cause death. Also, if too much reactant is used in a chemical reaction, it will be left over and wasted or can cause a reaction other than the intended one. This makes a difference when performing reactions with substances that are very expensive or available in small quantities. Unfortunately, this has only begun to scratch the surface of what stoichiometry can do.

Resources:

Ebbing, Darrell D., and Steven D. Gammon. General Chemistry. 6^th ed. New York: Houghton Mifflin Co., 1999.

“From Caveman to Chemist: Stoichiometry.” 20 November 2002. http://cator.hsc.edu/~kmd/caveman/projects/stoich/.

Jones, Loretta, and Peter Atkins. Chemistry: Molecules, Matter, and Change. 4^th ed. New York: W.H. Freeman and Co., 2000.

“Mols, Percents, and Stoichiometry.” 20 November 2002. http://www.chemtutor.com/mols.htm#stoi.

“Stoichiometry.” 20 November 2002. http://www.shodor.org/UNChem/basic/stoic/.

“Stoichiometry Review.” 20 November 2002. http://www.bhs.berkeley.k12.ca.us/departments/science/chemistry/Glimme%20Chem/Stoichiometry/Stoichiometry.html.

I would like to thank the above mentioned resources because I used many of their examples and definitions. If you would like to view the web pages, click on the web address and it will take you there.

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