April 3, 2014

•Stoichiometry

Stoichiometry • Stoichiometry is the study of quantities of materials

consumed and produced in chemical reactions

• Stoikheion (Greek, “element”) + • -metry (English, measurement)

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!!

2 H2 + O2 2 H2O

• 2 molecules of H2 react with 1 molecule of O2 to form 2 molecules of water

IT ALSO MEANS• 2 moles of H2 react with 1 mole of O2 to form 2

moles of water

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 H2 + O2 2 H2O

• Hydrogen to oxygen: 2 mol H2

1 mol O2

• Hydrogen to water: • Oxygen to water: • Oxygen to hydrogen: • Water to oxygen:

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 H2 + O2 2 H2O

• Hydrogen to oxygen: 2 mol H2

1 mol O2

• Hydrogen to water: 2 mol H2

2 mol H2O• Oxygen to water: • Oxygen to hydrogen: • Water to oxygen:

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 H2 + O2 2 H2O

• Hydrogen to oxygen: 2 mol H2

1 mol O2

• Hydrogen to water: 2 mol H2

2 mol H2O• Oxygen to water: 1 mol O2

2 mol H2O

• Oxygen to hydrogen: • Water to oxygen:

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 H2 + O2 2 H2O

• Hydrogen to oxygen: 2 mol H2

1 mol O2

• Hydrogen to water: 2 mol H2

2 mol H2O• Oxygen to water: 1 mol O2

2 mol H2O

• Oxygen to hydrogen: 1 mol O2

2 mol H2

• Water to oxygen:

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 H2 + O2 2 H2O

• Hydrogen to oxygen: 2 mol H2

1 mol O2

• Hydrogen to water: 2 mol H2

2 mol H2O• Oxygen to water: 1 mol O2

2 mol H2O

• Oxygen to hydrogen: 1 mol O2

2 mol H2

• Water to oxygen: 2 mol H2O 1 mol O2

Calculating Masses of Reactants and Products1. 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.

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

MOL (GIVEN) MOL (UNKNOWN)• 1. 2H2 + O2 2H2O • A. How many moles of water are produced from 1 mole of H2?

Mole ratio H2 to H2O : 2 mol H2 = 2 mol H2O

1 mol H2 | 2 mol H2O = 1 x 2 = 1 mol H2O 2 mol H2 2

MOL (GIVEN) MOL (UNKNOWN)• 1. 2H2 + O2 2H2O • B. How many moles of water are produced from 3 moles of O2?

Mole ratio O2 to H2O : 1 mol O2 = 2 mol H2O

3 mol O2 | 2 mol H2O = 3 x 2 = 6 mol H2O 1 mol O2 1

MOL (GIVEN) MOL (UNKNOWN) MASS (UNKNOWN)• 2. CH4 + 2O2 CO2 + 2H2O• A. How many grams of water are produced from one mole of CH4?

(molar mass H2O = 18.02 g/mol)

Mole ratio CH4 to H2O : 1 mol CH4 = 2 mol H2O

1 mol CH4 | 2 mol H2O | 18.02 g = 1 x 2 x 18.02 = 36.04 g H2O 1 mol CH4 1 mol H2O 1 x 1

MOL (GIVEN) MOL (UNKNOWN) MASS (UNKNOWN)• 2. CH4 + 2O2 CO2 + 2H2O• B. How many grams of O2 are required to produce 5 mol CO2? (molar

mass O2 = 32.00 g/mol)Mole ratio CO2 to O2 : 1 mol CO2 = 2 mol O2

5 mol CO2 | 2 mol O2 | 32.00 g = 5 x 2 x 32.00 = 320 g O2

1 mol CO2 1 mol O2 1 x 1

mass, g (given) mol (given) mol (unknown)• 3. CH4 + 2O2 CO2 + 2H2O• A. How many moles of CO2 can be produced from 4.0 g CH4? (molar

mass CH4 = 16.05 g/mol)Mole ratio CH4 to CO2 : 1 mol CH4 = 1 mol CO2

4.0 g CH4 | 1 mol CH4 | 1 mol CO2 = 4.0 x 1 x 1 = 0.25 mol CO2

16.05 g CH4 1 mol CH4 16.05 x 1

mass, g (given) mol (given) mol (unknown)• 3. CH4 + 2O2 CO2 + 2H2O• B. How many moles of CH4 are required to react with 2.0 g O2? (molar

mass O2 = 32.00 g/mol)Mole ratio O2 to CH4: 2 mol O2 = 1 mol CH4

2.0 g O2 | 1 mol O2 | 1 mol CH4 = 2.0 x 1 x 1 = 0.031 mol CH4

32.00 g O2 2 mol O2 32.00 x 2

mass, g (given) mol (given) mol (unknown) mass, g (unknown)• 4. ____NH3 + ____O2 ____NO + ____H2O

First you must balance the equation. N: 1 = N: 1 H: 3 ≠ H: 2 O: 2 = O: 2

mass, g (given) mol (given) mol (unknown) mass, g (unknown)• 4. __2__NH3 + ____O2 ____NO + __3__H2O

N: 1 2 ≠ N: 1H: 3 6 ≠ H: 2 6O: 2 ≠ O: 3 4

mass, g (given) mol (given) mol (unknown) mass, g (unknown)• 4. __2__NH3 + ____O2 _2__NO + __3__H2O

N: 1 2 = N: 1 2 H: 3 6 = H: 2 6O: 2 ≠ O: 3 4 5 Leave oxygen for last!

mass, g (given) mol (given) mol (unknown) mass, g (unknown)• 4. __4__NH3 + __5__O2 _4__NO + __6__H2O

N: 1 2 4 = N: 1 2 4H: 3 6 12 = H: 2 6 12O: 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.

mass, g (given) mol (given) mol (unknown) mass, g (unknown)

• 4. 4 NH3 + 5 O2 4 NO + 6 H2O• A. How many grams of H2O can be formed from 1.0 g NH3?

(molar mass NH3 = 17.04 g/mol, molar mass H2O = 18.02 g/mol)Mole ratio NH3 to H2O: 4 mol NH3 = 6 mol H2O

1.0 g NH3 | 1 mol NH3 | 6 mol H2O | 18.02 g = 1.0 x 1 x 6 x 18.02 = 1.59 g H2O 17.04 g NH3 4 mol NH3 1 mol H2O 17.04 x 4

mass, g (given) mol (given) mol (unknown) mass, g (unknown)

• 4. 4 NH3 + 5 O2 4 NO + 6 H2O• B. If 5.0 g NO are formed, how many grams of H2O are also formed? (molar mass

NO = 30.01 g/mol)Mole ratio NO to H2O: 4 mol NO = 6 mol H2O

5.0 g NO| 1 mol NO | 6 mol H2O | 18.02 g = 5.0 x 1 x 6 x 18.02 = 4.50 g H2O 30.01 g NO 4 mol NO 1 mol H2O 30.01 x 4