Dougherty Valley High School Chemistry

Stoichiometry Online Help


Practice Problems Here:

Click for Stoichiometry Practice Problems


What is Stoichiometry?

The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. It usually determines how much of a substance is expected to react or be produced in a chemical reaction. It is a very mathematical part of chemistry, so be prepared for lots of calculator use.

 

Jeremias Benjaim Richter (1762-1807) was the first to describe the principles of stoichiometry. In 1792 he wrote:

 

"Die stöchyometrie (Stöchyometria) ist die Wissenschaft die quantitativen oder Massenverhältnisse zu messen, in welchen die chymischen Elemente gegen einander stehen." (German translated into English: Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another.)


The key steps (in order) to doing stoichiometry:

1. Balance your equation first.

2. Convert from grams to moles using molar mass.

3. Determine the limiting reagent [if necessary] (Use mole ratios to figure out.)

4. Use ratios to find the moles of the reactant or product you need to find.

5. Convert from moles back to grams of the new substance using that substance's molar mass.

6. Calculate percent yield [if necessary] (Divide actual yield by theoretical yield X 100%)

 

 

Sample Example Problem [to show you how it's done]:

Problem: How many grams of silicon carbide are produced when 50.0 g of silicon dioxide is heated with 32.0 g of carbon?

 

            SiO2 (s)  + C (s) ā SiC (s)  + CO (g)                                 [unbalanced]

 

 

1. Balance your equation first.

            SiO2 (s)  + 3C (s) ā  SiC (s)  +  2CO (g)                           [balanced]

 

 

2. Convert from grams to moles using molar mass.

 

            To get moles from grams of silicon dioxide (SiO2):          [molar mass = 60 g/mol]

 

                        50.0 g SiO2  X  ( 1 mol  /  60 g ) = 0.833 mol SiO2

 

            To get moles from grams of carbon (C):    [molar mass = 12 g/mol]

 

                        32.0 g C  X  ( 1 mol  /  12 g ) = 2.667 mol C

 

3. Determine the limiting reagent [if necessary] (Use mole ratios to figure out.)

 

            Figure out how many moles of C would need to react with 0.833 moles SiO2:

                        * From your balanced equation, the ratio is 3 mol C: 1 mol SiO2.

 

                        0.833 mol SiO2  X  ( 3 mol C / 1 mol SiO2 )  = 2.5 mol C needed

 

                        * But you have 2.667 mol C to start with and need 2.5 mol C to react, so you have extra C and limiting reagent is SiO2

                        * To calculate excess: 2.667 - 2.500 = 0.167 mol C in excess

 

            OR

 

            Figure out how many moles of SiO2 would need to react with 0.833 moles C:

                        * From your balanced equation, the ratio is 1 mol SiO2 : 3 mol C.

 

                        2.667 mol C  X  ( 1 mol SiO2 / 3 mol C )  = 0.889 mol SiO2 needed

 

* But you have 0.833 mol SiO2 to start with and need 0.889 mol SiO2 to react, so you have SiO2 limiting reagent because you don't have enough SiO2 needed to react

* Compare 0.889 mol SiO2 needed vs. 0.833 mol SiO2 you started with

 

 

4. Use ratios to find the moles of the reactant or product you need to find.

Since SiO2 is the limiting reagent, you use the moles of SiO2 if you have to calculate how many grams of silicon carbide (the product you're trying to find) are being produced. Use your mole ratios from your balanced equation:

 

                        0.833 mol SiO2  X  ( 1 mol SiC / 1 mol SiO2 )  = 0.833 mol SiC produced

 

 

5. Convert from moles back to grams of the new substance using that substance's molar mass.

 

            To get grams from moles of carbon (SiC):            [molar mass = 40 g/mol]

 

                        0.833 mol SiC  X  ( 40 g  /  1 mol ) = 33.33 g SiC produced

 

 

Answer to Problem: 33.33 g SiC produced = the theoretical yield