Stoichiometry…
Can You Make 2.00 Grams of a Compound?
Purpose:
Use your skills of predicting chemical equations, balancing equations, and calculating molar mass to solve a complex
stoichiometry problem. Then test your laboratory techniques by mixing the reactants and isolating exactly 2.00 g of a compound.
Background:
Stoichiometry is the branch of chemistry that deals with the numerical relationships and mathematical proportions of
reactants and products in chemical reactions. For any chemical equation, the mole, mass, and atom ratios must remain constant. The
coefficients from a balanced chemical equation provide the definite ratios between the number of moles of reactants and products
involved (equation 1).
N2(g) + 3H2(g)  2NH3(g)
Equation 1:
One mole of any chemical substance contains 6.02 x 1023 particles (molecules, atoms, etc) and has a mass equal to the sum of
atomic masses of all elements in the formula. Therefore, the mole ratio of a balanced chemical equation can be converted into a mass
ratio or number of particles ratio. These ratios can be used as conversion factors to determine the exact amounts of reactants and
products that form during a chemical change (equation 2).
Equation 1:
Balanced equation:
N2(g) + 3H2(g)  2NH3(g)
Molar mass:
28.0
Mass:
28.0 g + 6.0 g
2.0
17.0
=
34.0 g
When two ionic compounds are placed in solution, a double replacement reaction can occur. Double replacement reactions
are slao called double displacement or exchange reactions and follow the patter: A +X- + B+Y  B+X- + A+Y . Normally, the
reactants are chosen so one product will precipitate out of solution while the other ions stay in solution. In this way, one of the
products can be easily isolated by filtration. Equation 3 is an example of a double replacement reaction.
Equation 3:
Pb(NO3)2(aq) + KI(aq)  PbI2(s) + KNO3(aq)
When the reacting ionic compounds are dissolved in water, they separate into the four different ions that are free to move
around and then re-combine to form the new products. If one of the combinations of ions is insoluble in water, it will precipitate out of
solution. In equation 3, lead(II) iodide (PbI 2) precipitates out of solution. Some ionic solids also contain water molecules that are
trapped within their crystal structure. These are called hydrated compounds. An example of a hydrated compound would be copper(II)
sulfate pentahydrate, Cu(SO4)  5H2O. During the dissolving process, the water molecules are released into the solution.
In order to produce the 2.00 grams of desired product in the laboratory using the double replacement reaction, a balanced
equation is required and the product that will precipitate must be identified. The next step is to determine the molar mass of each
reactant and product, and finally use dimensional analysis to determine the quantities of reactants required. The following four steps
describe the process in detail.
Step 1: Write a balanced formula equation for the reaction.
Preparing a balanced chemical formula equation is the first step in understanding a reaction. Start the process by determining
what reaction is occurring and write out the names of all reactants and products.
Equation 4:
Reaction:
nitrogen + hydrogen = ammonia
Next, prepare an unbalanced equation by writing out the chemical formula for each reactant and product.
Equation 5:
Unbalanced equation:
N2 + H2  NH3
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Finally, the chemical equation must be balanced. The key principle in balancing equations is that the atoms are conserved in a
chemical reaction. Therefore, the same number and type of atoms must be found among the reactants and products of a
reaction. For example, if you start with two nitrogen atoms in the reactants, you must have two nitrogen atoms in the
products.
Equation 6:
Balanced equation:
N2 + 3H2  2NH3
Step 2: Predict which compound will form a precipitate using general solubility rules.
When working with aqueous solutions, it is helpful to understand a few rules concerning which substances are soluble in
water and which will form precipitates. Use the solubility table in your reference packets.
Step 3: Calculate the mass of reactants needed to produce the 2.00 g of precipitate.
To calculate the mass of reactants needed to produce 2.00 g of a precipitated product requires the determination of the molar
mass of each reactant and product. Molar mass (also called molecular weight) is the mass in grams of one mole of a
substance. Molar mass is determined by adding the atomic masses of all of the atoms in the chemical formula. For example,
the chemical formula CuSO4  5H2O shows that in 1 mole of this compound, there are 1 mole of Cu ions, 1 mole of SO 4
ions, and 5 moles of H2O which gives a total of 1 mole of Cu, 1 mole of S, 10 moles of H and 9 moles of O atoms.
The total mass of each element in the compound is equal to the number of moles of atoms of the element multiplied by its
atomic mass. The molar mass of copper (II) sulfate pentahydrate is equal to the sum of the total masses of each element.
Number of moles of atoms
1 mole of Cu
1 mole of S
9 moles of O
10 moles of H





Atomic mass
63.55 g Cu/mol
32.06 g S/mol
16.00 g O/mol
1.01 g H/mol
=
=
=
=
=
Molar Mass of CuSO4  5H2O
Total mass of elements
63.55 g Cu
32.06 g S
144.0 g O
10.1 g H
249.7 g
Once the molar mass of each reactant and product is determined, the next step is to determine the number moles in 2.00 g of
precipitate using equation 5.
Equation 5:
grams
1
*
1 mole
molar mas s
 moles
For example, if 2.00 g of copper(II) sulfate pentahydrate is desired, using dimensional analysis, the solution would be:
2.00 g Cu(SO )  5H O
4
2
*
1
1 mole Cu(SO )  5H O
4
2
 0.00801mol Cu(SO )  5H O
4
2
249.7 g Cu(SO )  5H O
4
2
Lastly, use the coefficients in the balanced equation to determine the number of moles of each of the reactants required to
produce the desired number of moles of precipitate. Once the number of moles is determined, use the following equation to
calculate the number of grams of each reactant.
Equation 6:
moles
1
*
molar mas s
1 mole
 grams
Step 6: Use the calculated masses of reactants to make and recover the 2.00 grams of precipitate.
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After the chemical equation is balanced, the precipitated product is determined and the mass of each reactant required to
produce 2.00 g of product is calculated, the reaction can be performed in the lab. Both of the reactants need to be dissolved in
separate beakers using 25 mL of distilled water for each reactant. Then the two solutions need to be mixed together to form
the precipitate. The precipitate will be recovered by filtration using a pre-massed piece of filter paper. The precipitate will be
dried and then the mass of the solid determined. The precipitate will be turned in to your teacher and the percent error
calculated for the experiment.
Student Tips:

When writing chemical formulas for the reactants and products, remember that the sum of all of the positive and negative
charges in the compound must equal zero.

Use subscripts to indicate the number of ions needed to equalize the charges. Parentheses are required to enclose
polyatomic ions when there is more than one in the formula, such as the two nitrate ions in Ba(NO3)2.

After the formulas for the individual compounds are correctly written, only the coefficients can be changed to balance
the equation and to obtain the same number of atoms of each element in both the reactants and products.

Before measuring out the reactants, do the calculated mass values seem reasonable? According to the Law of
Conservation of Mass, the total mass of all reactants must equal the total mass of all products.
Safety Information:
The compounds selected for this experiment have low to moderate toxicity, but avoid breathing the dust or getting any in
your eyes or skin. Zinc sulfate is a skin and mucous membrane irritant and is mildly toxic. Sodium carbonate and potassium carbonate
may be skin irritants. Magnesium sulfate irritates eyes and the respiratory tract. Calcium chloride is moderately toxic.
All solutions may be flushed down the drain and solids thrown in the trash. Wear chemical goggles.
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