1. Stoichiometry in Chemistry Stoichiometry – The analysis of chemical reactions by mass, moles and balanced formulas Introduction Basic Analysis Analysis with Volume

2. Introduction Solving problems with chemical reactions involves applying the concepts of a balanced chemical formula, mole ratios, and molar mass of each chemical to a given situation. Invariably, one quantity (moles, grams, volume, etc. ) will be given, and one must solve for other quantities. To keep track of all the relationships, I encourage the use of a “spreadsheet”, separating all the different components so that they can be analyzed one step at a time.

3. The Spreadsheet Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 O 2 + 2 H 2  2 H 2 O 1 2 2 First, create a balanced formula, for example: Then add a speadsheet around the formula, with at least five rows Then list the ratio of moles in the balanced reaction, using the coefficients of the balanced formula Finally, add “Molar Mass”, “Grams”, and “Moles” to the left column

4. Basic Analysis Given a problem, one be able to set up an analysis and solve it Almost always, you will be given one piece of information in a reaction, and then given a chemical formula, you will need to figure out some other part of the chemical reaction.

5. How many grams of Oxygen are needed to combine with 16 grams of Hydrogen to make Water? First, Identify the Unknown and the Known X grams Oxygen 16 Grams Hydrogen Second, Give the Formulas for each compound in the reaction O 2 + H 2  H 2 O Third, Balance the Equation O 2 + 2 H 2  2 H 2 O

6. Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 X Now insert the information into a spreadsheet 16 grams

7. Now solve, first by solving for hydrogens’s moles. Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 X 16 grams 8 moles Moles = grams/molar mass

8. Using the mole ratio, solve for oxygen’s moles Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 X 16 grams 8 moles 4 moles

9. Now, finish the problem by solving for grams of oxygen. Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 X 16 grams 8 moles 4 moles Remember: moles = grams/molar mass 128 grams

10. Analysis with Volume To extend this technique to calculate the volume of reactants or products, one simply adds another row, Labled “Volume” at the end. Remember, under “standard” (usual) conditions, 1 mole of any gas will have a volume of 22.4 Liters

11. New Problem How many Liters of Oxygen are needed to combine with 16 grams of Hydrogen to make Water? First, Identify the Unknown and the Known X Liters Oxygen 16 Grams Hydrogen Second, Give the Formulas for each compound in the reaction O 2 + H 2  H 2 O Third, Balance the Equation O 2 + 2 H 2  2 H 2 O

12. Now insert the information into a spreadsheet. Notice that a new row has been added. Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 16 grams Volume X Liters

13. Once again, we first solve by solving for moles of the known: Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 16 grams Volume X Liters Remember: moles = grams/molar mass 8 moles

14. Now we solve for moles of oxygen: Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 16 grams Volume X Liters Remember: Use the mole ratio to solve mole to mole problems 8 moles4 moles

15. Finally, we solve for Liters of oxygen: Balanced Formula Mole Ratio Molar Mass Grams Moles O 2 + 2 H 2  2 H 2 O 1 2 2 32 2 18 16 grams Volume X Liters Remember: One mole = 22.4 Liters, so 4 moles = ? 8 moles4 moles 89.6 Liters

16. A final note… The only way to become confident (and competent) in solving stoichiometry problems is to….. Practice