1. April 3, 2014 Stoichiometry

2. Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”) + -metry (English, measurement)

3. Stoichiometry Stoichiometry is used to make predictions about how much product will be formed without actually doing the reaction. What you need: A balanced equation!!

4. 2 H 2 + O 2  2 H 2 O 2 molecules of H 2 react with 1 molecule of O 2 to form 2 molecules of water IT ALSO MEANS 2 moles of H 2 react with 1 mole of O 2 to form 2 moles of water

5. Mole Ratios Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H 2 + O 2  2 H 2 O Hydrogen to oxygen: 2 mol H 2 1 mol O 2 Hydrogen to water: Oxygen to water: Oxygen to hydrogen: Water to oxygen:

6. Mole Ratios Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H 2 + O 2  2 H 2 O Hydrogen to oxygen: 2 mol H 2 1 mol O 2 Hydrogen to water: 2 mol H 2 2 mol H 2 O Oxygen to water: Oxygen to hydrogen: Water to oxygen:

7. Mole Ratios Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H 2 + O 2  2 H 2 O Hydrogen to oxygen: 2 mol H 2 1 mol O 2 Hydrogen to water: 2 mol H 2 2 mol H 2 O Oxygen to water: 1 mol O 2 2 mol H 2 O Oxygen to hydrogen: Water to oxygen:

8. Mole Ratios Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H 2 + O 2  2 H 2 O Hydrogen to oxygen: 2 mol H 2 1 mol O 2 Hydrogen to water: 2 mol H 2 2 mol H 2 O Oxygen to water: 1 mol O 2 2 mol H 2 O Oxygen to hydrogen: 1 mol O 2 2 mol H 2 Water to oxygen:

9. Mole Ratios Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H 2 + O 2  2 H 2 O Hydrogen to oxygen: 2 mol H 2 1 mol O 2 Hydrogen to water: 2 mol H 2 2 mol H 2 O Oxygen to water: 1 mol O 2 2 mol H 2 O Oxygen to hydrogen: 1 mol O 2 2 mol H 2 Water to oxygen: 2 mol H 2 O 1 mol O 2

10. Calculating Masses of Reactants and Products 1.Make sure the equation is balanced. 2.Identify your given and your unknown. 3.Convert mass to moles, if necessary. 4.Set up mole ratios. 5.Use mole ratios to calculate moles of the product/reactant desired. 6.Convert moles to mass, if necessary. 1.Make sure the equation is balanced. 2.Identify your given and your unknown. 3.Convert mass to moles, if necessary. 4.Set up mole ratios. 5.Use mole ratios to calculate moles of the product/reactant desired. 6.Convert moles to mass, if necessary.

11. Important The “given” and “unknown” can both be reactants, both be products, or one reactant and one product Reaction stoichiometry will ALWAYS have a mol (given)  mol (unknown) conversion!!! To calculate mol (given)  mol (unknown), use mole ratio, which comes directly from the coefficients of the balanced chemical equation Other conversion factors may (or may not) be necessary before or after this conversion

12. MOL (GIVEN)  MOL (UNKNOWN) 1. 2H 2 + O 2  2H 2 O A. How many moles of water are produced from 1 mole of H 2 ? Mole ratio H 2 to H 2 O : 2 mol H 2 = 2 mol H 2 O 1 mol H 2 | 2 mol H 2 O = 1 x 2 = 1 mol H 2 O 2 mol H 2 2

13. MOL (GIVEN)  MOL (UNKNOWN) 1. 2H 2 + O 2  2H 2 O B. How many moles of water are produced from 3 moles of O 2 ? Mole ratio O 2 to H 2 O : 1 mol O 2 = 2 mol H 2 O 3 mol O 2 | 2 mol H 2 O = 3 x 2 = 6 mol H 2 O 1 mol O 2 1

14. MOL (GIVEN)  MOL (UNKNOWN)  MASS (UNKNOWN) 2. CH 4 + 2O 2  CO 2 + 2H 2 O A. How many grams of water are produced from one mole of CH 4 ? (molar mass H 2 O = 18.02 g/mol) Mole ratio CH 4 to H 2 O : 1 mol CH 4 = 2 mol H 2 O 1 mol CH 4 | 2 mol H 2 O | 18.02 g = 1 x 2 x 18.02 = 36.04 g H 2 O 1 mol CH 4 1 mol H 2 O 1 x 1

15. MOL (GIVEN)  MOL (UNKNOWN)  MASS (UNKNOWN) 2. CH 4 + 2O 2  CO 2 + 2H 2 O B. How many grams of O 2 are required to produce 5 mol CO 2 ? (molar mass O 2 = 32.00 g/mol) Mole ratio CO 2 to O 2 : 1 mol CO 2 = 2 mol O 2 5 mol CO 2 | 2 mol O 2 | 32.00 g = 5 x 2 x 32.00 = 320 g O 2 1 mol CO 2 1 mol O 2 1 x 1

16. mass, g (given)  mol (given)  mol (unknown) 3. CH 4 + 2O 2  CO 2 + 2H 2 O A. How many moles of CO 2 can be produced from 4.0 g CH 4 ? (molar mass CH 4 = 16.05 g/mol) Mole ratio CH 4 to CO 2 : 1 mol CH 4 = 1 mol CO 2 4.0 g CH 4 | 1 mol CH 4 | 1 mol CO 2 = 4.0 x 1 x 1 = 0.25 mol CO 2 16.05 g CH 4 1 mol CH 4 16.05 x 1

17. mass, g (given)  mol (given)  mol (unknown) 3. CH 4 + 2O 2  CO 2 + 2H 2 O B. How many moles of CH 4 are required to react with 2.0 g O 2 ? (molar mass O 2 = 32.00 g/mol) Mole ratio O 2 to CH 4 : 2 mol O 2 = 1 mol CH 4 2.0 g O 2 | 1 mol O 2 | 1 mol CH 4 = 2.0 x 1 x 1 = 0.031 mol CH 4 32.00 g O 2 2 mol O 2 32.00 x 2

18. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. ____NH 3 + ____O 2  ____NO + ____H 2 O First you must balance the equation. N: 1 = N: 1 H: 3 ≠ H: 2 O: 2 = O: 2

19. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. __2__NH 3 + ____O 2  ____NO + __3__H 2 O N: 1 2 ≠ N: 1 H: 3 6 ≠ H: 2 6 O: 2 ≠ O: 3 4

20. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. __2__NH 3 + ____O 2  _2__NO + __3__H 2 O N: 1 2 = N: 1 2 H: 3 6 = H: 2 6 O: 2 ≠ O: 3 4 5Leave oxygen for last!

21. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. __4__NH 3 + __5__O 2  _4__NO + __6__H 2 O N: 1 2 4 = N: 1 2 4 H: 3 612 = H: 2 6 12 O: 2 10 ≠ O: 3 4 5 10 There is no way to fit 2 evenly into 5. Try doubling everything and then see if you can make oxygen balance.

22. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. 4 NH 3 + 5 O 2  4 NO + 6 H 2 O A. How many grams of H 2 O can be formed from 1.0 g NH 3 ? (molar mass NH 3 = 17.04 g/mol, molar mass H 2 O = 18.02 g/mol) Mole ratio NH 3 to H 2 O: 4 mol NH 3 = 6 mol H 2 O 1.0 g NH 3 | 1 mol NH 3 | 6 mol H 2 O | 18.02 g = 1.0 x 1 x 6 x 18.02 = 1.59 g H 2 O 17.04 g NH 3 4 mol NH 3 1 mol H 2 O 17.04 x 4

23. mass, g (given)  mol (given)  mol (unknown)  mass, g (unknown) 4. 4 NH 3 + 5 O 2  4 NO + 6 H 2 O B. If 5.0 g NO are formed, how many grams of H 2 O are also formed? (molar mass NO = 30.01 g/mol) Mole ratio NO to H 2 O: 4 mol NO = 6 mol H 2 O 5.0 g NO| 1 mol NO | 6 mol H 2 O | 18.02 g = 5.0 x 1 x 6 x 18.02 = 4.50 g H 2 O 30.01 g NO 4 mol NO 1 mol H 2 O 30.01 x 4