Lesson

Stoichiometry Problems

Putting the steps into practice

To show how this method is used to convert the given unit to the unit asked for in the problem, I will solve a typical stoichiometric problem using this mechanism. Here is a two-part sample problem:

If a 2.8 g sample of mercuric oxide is decomposed by heating:

a) How many grams of mercury will be produced?
b) How many moles of oxygen will be produced?

Corresponding calculation for the first arrow:

The second arrow goes from under HgO to under Hg. Therefore, it asks the student to convert moles of HgO to moles of Hg.

Corresponding calculation for the second arrow:

The last arrow goes from under Hg (from moles of Hg) to above Hg (to grams of Hg).

Corresponding calculation for the third arrow:

Corresponding calculation for the first arrow:

The second arrow shows the student to go from under the HgO (moles of HgO) to under the O₂
(moles of O₂).

Corresponding calculation for the second arrow:

Notice that the number of arrows corresponds to the number of conversion factors in the dimensional analysis.

Gas Laws

This method can also be used to solve stoichiometric problems involving gases, when the conditions are not at STP.If the vertical arrow passing through a gas points down (pointing to moles), the student uses the ideal gas law to solve for moles represented as n in the ideal gas law (n=PV/RT), where n is the number of moles of the gas, P is the gases pressure, V is the volume of gas, R is the ideal gas law constant, and T is the Kelvin temperature of the gas.If the vertical arrow passing through a gas points up (pointing to L), the student uses the ideal gas law to solve for V (V=nRT/P).The following sample problem illustrates this:

What volume of carbon dioxide will be collected at 25^oC and .965atm when 25.6g of calcium carbonate undergoes decomposition?

A common mistake that students make is to use the ideal gas law with materials that are not gases.If the arrow passes through a gas, they should use the ideal gas law.If the arrow does not pass through a gas, the ideal gas law should not be used.In this example, because the first arrow passes through solid CaCO₃ (not a gas), the student does not use the ideal gas law equation for the first step.

Corresponding calculation for the first arrow:

The second arrow goes from under CaCO₃ (moles of CaCO₃) to under CO₂ (moles of CO₂).

Corresponding calculation for the second arrow:

The third arrow passes only through the gas CO₂ so it is the only place where a gas law should be used. Because the arrow points to the X L the ideal gas law should be solved for the variable measured in liters which is V (V = nRT/P). However, to do this calculation, the moles of CO₂ (n) first needs to be calculated using the stoichiometry from step 1 and step 2.

Now that the number of moles of CO₂ is known the ideal gas law can be used.It should also be noted that since we are in the middle of series of multiplication steps, the number of significant figures recorded in the answer for n should not yet be rounded.

Corresponding calculation for the third arrow:

Notice that I have used parentheses instead of multiplication signs in the solution of the ideal gas law.Counting the number of multiplication signs used in the student’s attempted solution will not match with the number of arrows on their drawn path if times signs are used in the gas law.A general rule that may be helpful is to use a different symbol to represent multiplication signs in the gas law (parentheses or asterisks) than is used in the stoichiometry.

Molarity Problems

Another type of problem that benefits from this process is using a downward arrow through the data for a given gas and using stoichiometry to solve for the molarity of a solution.

If .033 L of hydrogen is collected at 0 ^oC and 1.00 atm from the reaction of solid aluminum and .500-M hydrochloric acid, what volume of .500-M HCl reacted?

The first arrow goes only through the gas H₂ and points to moles of (underneath) H₂, so the ideal gas law has to be solved for “moles of H₂^” (n = PV/RT).Since this is a gas law, I will use the * to represent the multiplication in the gas law and the typical “x” as the stoichiometric multiplication.

The second arrow goes from under H₂ (mol H₂) to under HCl (mol HCl).

The third arrow goes through an aqueous solution of HCl. When a vertical arrow goes through an aqueous solution, its molarity needs to be used. Since the solution is .500-M HCl, 1 L of this solution contains .500 moles of HCl. (.500 mol HCl = 1-L HCl solution)

This problem also illustrates a way to use the molar volume of any gas at STP as 22.4 L/mol. Since H₂ is .033 L of a gas at STP, the student could also change liters of H₂ to moles of H₂ in the following manner and then finish the problem as previously shown:

Electrolysis applications

A final way to use this technique is with an electrolysis problem. The following problem will show the way to use time to measure the flow of electrons and then the standard stoichiometry method to answer an electrolysis problem.

If a 10.0-amp current flows into a solution of AlCl₃ for 2.00 minutes, what mass of Al will be plated on the cathode?

Summary of the set up

After the students have learned all of the various molar conversions, I condense them into a final table that shows the possible units above and below the chemical equations used in stoichiometric problems and the conversion factor that will get them from above the equation to below and vice versa:

Using this table as a guide, the student can place a given number and its unit in the stoichiometric problem and X with its unit above or below the final chemical formula. The type of molar conversion that needs to be used to convert the given unit to moles is given on the downward arrow, while the mole-to-mole ratio is found from the chemical equation’s coefficients and is written on the horizontal arrow on the bottom of the table. The needed conversion factor for moles to the final unit asked for in the problem is indicated by the upward arrow in the table.