Stoichiometry

Stoichiometry

What is the stoichiometry?

Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

What You Should Expect?

Given : Amount of reactants 

Question: how much of products can be formed.

 Example  : 2 A + 2B ====>   3C

What do you need?

            You will need to use

1. molar ratios,
2. molar masses,
3. balancing and interpreting equations, and
4.  conversions between grams and moles.

Note: This type of problem is often called "mass-mass."

Steps Involved in Solving Mass-Mass Stoichiometry Problems

l  Balance the chemical equation correctly

l  Using the molar mass of the given substance, convert the mass given to moles.

l  Construct a molar proportion (two molar ratios set equal to each other)

l  Using the molar mass of the unknown substance, convert the moles just calculated to mass.

 Stoichiometry is simply the math behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation.

What is a Chemical Equation?

In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C"represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You should have reviewed naming schemes, or nomenclature, in earlier readings.

A chemical equation is an expression of a chemical process. For example:

AgNO₃(aq) + NaCl(aq) ---> AgCl(s) + NaNO₃(aq) 

In this equation, AgNO₃ is mixed with NaCl. The equation shows that the reactants (AgNO₃ and NaCl) react through some process (--->) to form the products (AgCl and NaNO₃). Since they undergo a chemical process, they are changed fundamentally.

Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means the compound is dissolved in water. Finally, the (g) sign means that the compound is a gas.

Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of molecules, or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed.

On some occasions, a variety of information will be written above or below the arrows. This information, such as a value for temperature, shows what conditions need to be present for a reaction to occur. For example, in the graphic below, the notation above and below the arrows shows that we need a chemical Fe₂O₃, a temperature of 1000° C, and a pressure of 500 atmospheres for this reaction to occur.

The Mole

Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.022 x 10²³) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 10²³ carrots. In the equation above there are no numbers in front of the terms, so each coefficient is assumed to be one (1). Thus, you have the same number of moles of AgNO₃, NaCl, AgCl, NaNO₃.

Converting between moles and grams of a substance is often important. This conversion can be easily done when the atomic and/or molecular mass of the substance(s) are known. Given the atomic or molecular mass of a substance, that mass in grams makes a mole of the substance. For example, calcium has an atomic mass of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.

Balancing Chemical Equations

Sometimes, however, we have to do some work before using the coefficients of the termselement in the reactants equals the number of atoms of that same element in the products. To do this we have to figure out the relative number of molecules of each term expressed by the term's coefficient.

Balancing a simple chemical equation is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term.

2Fe₃O₄

This term expresses two (2) molecules of Fe₃O₄. In each molecule of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen atoms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.

Now let's try balancing the equation mentioned earlier:

Al + Fe₃O₄---> Al₂O₃+ Fe

Developing a strategy can be difficult, but here is one way of approaching a problem like this.

1. Count the number of each atom on the reactant and on the product side.
2. Determine a term to balance first. When looking at this problem, it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is:

Al + 3 Fe₃O₄---> 4 Al₂O₃+ Fe

Be sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.

3. Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:

Al +3 Fe₃O₄---> 4Al₂O₃+ 9Fe

4. Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side. Now, we're done, and the balanced equation is:

8Al + 3Fe₃O₄ ---> 4Al₂O₃ + 9 Fe

Density

Density refers to the mass per unit volume of a substance. It is a very common term in chemistry.

Concentrations of Solutions

The concentration of a solution is the "strength" of a solution. A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water. It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration.

The concentration of a solution is typically given in molarity. Molarity is defined as the number of moles of solute (what is actually dissolved in the solution) divided by the volume in liters of solution (the total volume of what is dissolved and what it has been dissolved in).

Molality =  moles  of  solute 

                  Liters of solution

Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do.

Example: If 5.00 g of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?
One of our first steps is to convert the amount of NaOH given in grams into moles:

5,00 g NaOH X               1 mole              = 0,125 moles

            1             (22,9 + 16,00 + 1,008)g

Now we simply use the definition of molarity: moles/liters to get the answer

Molality =  0,125    moles = 0,025 mol/L   

                  5,00 L of soln

So the molarity (M) of the solution is 0.025 mol/L.

Molality is another common measurement of concentration. Molality is defined as moles of solute divided by kilograms of solvent (the substance in which it is dissolved, like water).

Molality = moles of solute

                  Kg of solvent

Molality is sometimes used in place of molarity at extreme temperatures because the volume can contract or expand.

Example: If the molality of a solution of C₂H₅OH dissolved in water is 1.5 and the mass of the water is 11.7 kg, figure out how much C₂H₅OH must have been added in grams to the solution.

Our first step is to substitute what we know into the equation. Then we try to solve for what we don't know: moles of solute. Once we know the moles of solute we can look at the periodic table and figure out the conversion from moles to grams.

Molality = moles solute

                  Kg solvent

Now we simply use the definition of molarity: moles/liters to get the answer

Molality = moles solute

                  Kg solvent

1,5 mols = moles solute

       Kg       11,7 kg

1,5 moles X 11,7 kg = 17,55 moles

        Kg

17,55 moles X (2*12,01) + (6*1,008) + 16 = 808,5 gr C₂H₅OH

            1                                  1 moles           

It is possible to convert between molarity and molality. The only information needed is density.

Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution knowing that the density is 3.25 g/mL. To do this problem we can assume one (1) liter of solution to make the numbers easier. We need to get from the molarity units of mol/L to the molality units of mol/kg. We work the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams in a kg. This conversion will only be accurate at small molarities and molalities.

0,3 mol X  1 ml   X    1 L    X 1000 g  = 0,09 mols/kg

    1 L        3,25 g    1000 mL    1 kg

It is also possible to calculate colligative properties, such as boiling point depression, using molality. The equation for temperature depression or expansion is

ΔT= K_f × m

Where:

ΔT is temperature depression (for freezing point) or temperature expansion (for boiling point) (°C). K_f is the freezing point constant (kg°C/mol) m is molality in mol/kg.

Example: If the freezing point of the salt water put on roads is -5.2° C, what is the molality of the solution? (The K_f for water is 1.86 °C/m.)

This is a simple problem where we just plug in numbers into the equation. One piece of information we do have to know is that water usually freezes at 0° C.

ΔT       = K_f * m

ΔT/K_f    = m

m          = 5.2/1.86

m         = 2.8 mols/kg