2. Molar Equivalence: An alternative to ratios

3. In a balanced chemical equation, the balancing numbers reflect the ratio in which the chemicals react. This controls not only the amount of products created, but also the amount of reactants remaining after the reaction has been completed.

4. Being able to calculate these amounts is a very important unit in chemistry, but many students have difficulty in using ratios consistently to calculate the desired answers. Molar Equivalence represents a systematic process which has proven to be an effective teaching method for most students. The following slides will take students through a typical example, with the steps and thought processes needed to successfully solve the problem.

5. For example: Calculate the moles of water produced when 12 moles of butane are burned in excess oxygen.

6. STEP 1 Write the chemical equation. C 4 H 10 + O 2  CO 2 + H 2 O

7. STEP 2 Balance the equation.The balancing numbers represent the number of UNITS of each molecule present in the balanced equation. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O

8. STEP 3 Using the equation as a data table, record all values given underneath the appropriate molecule. If only mass is provided, calculate the number of moles of that molecule present. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 12 moles

9. STEP 4 Determine the MOLAR EQUIVALENCE.The molar equivalence (M.E.) represents the number of moles in one unit.M.E. = n / balancing number 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 12 moles bal # = 2 units M.E. = n / bal # = (12 moles) / (2 units) = 6 moles / unit

10. STEP 5 Determine the moles of the desired products and reactants by rearranging the equation in step 4: n = M.E. * balancing number. Notice how the units cancel out, and that we can solve all of the products simultaneously. In the case of reactant, the moles calculated is the number of moles which will be used as a result of the reaction. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 12 molesn = M.E. * bal # n = M.E. * bal #n = M.E. * bal # bal # = 2 units= (6 moles/unit) (13 units)= (6 moles/unit) (8 units)= (6 moles/unit) (10 units) M.E. = n / bal #= 78 moles= 48 moles= 60 moles = (12 moles) / (2 units) = 6 moles / unit

11. This concept can also be expanded to include the concept of limiting reagents. eg. Calculate the moles of each molecule when 4.0 moles of butane are reacted with 20. moles of oxygen.

12. Follow steps 1-4 as before, except that now there will two molar equivalences being calculated. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 4.0 molesn = 20. moles bal # = 2 unitsbal # = 13 units M.E. = n / bal #M.E. = n / bal # = (4.0 moles) / (2 units)= (20. moles) / (13 units) = 2.0 moles / unit= 1.538 mol/unit

13. STEP 5 Compare the molar equivalences. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 4.0 molesn = 20. moles bal # = 2 unitsbal # = 13 units M.E. = n / bal #M.E. = n / bal # = (4.0 moles) / (2 units)= (20. moles) / (13 units) = 2.0 moles / unit= 1.538 mol/unit

14. Whichever molecule has the LOWER molar equivalence value is the LIMITING REAGENT. The molecule with the HIGHER value is the EXCESS REAGENT – this value is invalid and should be ignored as if it was never calculated in the first place.

15. STEP 5 Compare the molar equivalences. Cancel out the higher molar equivalence. 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 4.0 molesn = 20. moles bal # = 2 unitsbal # = 13 units M.E. = n / bal #M.E. = n / bal # = (4.0 moles) / (2 units)= (20. moles) / (13 units) = 2.0 moles / unit= 1.538 mol/unit

16. STEP 6 Use the molar equivalence of the limiting reagent to calculate the moles of product created and the moles of reactant used in the reaction. n initial = 4.0 molesn = M.E. * bal #n = M.E. * bal # n used = M.E. * bal #= (1.538 mol/unit) (8 units)= (1.538 mol/unit) (10 units) = (1.538 mol/unit) (2 units)= 12.3 moles= 15.38 moles = 3.077 moles n remaining = n initial – n used = 4.0 moles – 3.077 moles = 0.923 moles 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O n = 20. moles bal # = 13 units M.E. = n / bal # = (20. moles) / (13 units) = 1.538 mol/unit

17. This systematic process will work regardless of the number of molecules involved in the balanced equation.