STOICHIOMETRY

Study of the amount of substances consumed and produced in a

chemical reaction.

The Mole and Stoichiometry

1. The Mole

a. SI unit for a given amount of substance

b. abbreviation – mol

c. standard mole = the amount of a substance that contains as many particles as there are found in exactly 12 g of carbon-12

The Mole and Stoichiometry

2. Avogadro’s number –

a. the number of particles in exactly one mole of a pure substance

b. 6.022 x 1023

c. 12 g Carbon-12 = one mol = 6.022 x 1023 atoms of Carbon-12

The Mole and Stoichiometry 3. Molar mass – a. mass of one mole of a pure substance b. unit g/mol c. for elements found on the periodic table d. Examples:

nitrogen: 14.0 g/mol (1 mol of nitrogen = 14.0 grams = 6.02 x 1023 atoms

nitrogen) silver: 107.9 g/mol (1 mol of silver= 107.9 grams = 6.02 x 1023 atoms

silver)

The Mole and Stoichiometry

e. Calculating Molar masses for compounds

1. Identify the name and number of each atom found in the compound

2. multiply the number of atoms by atomic mass for that element (found on the Periodic table).

3. add all the masses together

The Mole and Stoichiometry

f. Example:

H2O :

Al2 (PO4) 3

The Mole and Stoichiometry 4. Gram/Mole Conversions

a. Mass to moles

Given: grams / molar mass = moles

Example: How many moles are there in 12 g of MgS?

The Mole and Stoichiometry b. Moles to mass

Given moles X molar mass = grams

Example: What is the mass of 2.5 moles of FeCl3?

The Mole and Stoichiometry5. Mass/Mass Conversions

(Grams A)( 1 mol A) (coefficient B)( mass B)

( 1 )( mass A )(Coefficient A)(1 mol B)

Example

2H2 + O2 2H2O

How many grams of water is produced from 5.00 g hydrogen?

STOICHIOMETRY• Setting the problem up:

(mass-mass)

• 1. Write and balance the equation..

• 2. Identify the given and unknown.

• 3. Identify the mole ratio (coefficients from the balanced equation for the given and unknown).

• 4. Calculate the molar mass for the given and unknown.

SOLVING THE PROBLEM:

1. Convert grams of given to moles of given (Divide by molar mass of given)

2. Convert moles of given to moles of unknown (Multiply by the mole ratio- coefficient of unknown/coefficient of given)

3. Convert moles of unknown to grams of unknown(Multiply by the molar mass of unknown)

Limiting Reactants

                    

A.    Limiting Reactant – the reactant that limits the amount of the reactant that can combine, and the amount of product formed in a chemical reaction

B.    Excess Reactant – the substance that is not

used up completely in a reaction

•

Limiting Reactants  C    Problem-solving steps for determining

limiting/excess reactant

1.      Write and balance the equation2.      Identify the amounts of the two reactants

and unknown3.      Identify mole ratio 4.      Calculate the molar mass for the reactants

and unknown

Limiting Reactants•       • 5.       Determine moles of both reactants by

dividing mass (given in problem) by molar mass (step 4)

• 6.     Choose one of the reactants as reactant A and the other as reactant B

• 7.     Multiply the moles of reactant A by the mole ratio of reactant B/reactant A.

Limiting Reactants•     

• 8. Compare the moles of reactant B needed (answer from step 7) to actual number of moles present (answer from step 5).

• If amount is less than what you have then reactant B is in excess and reactant A is the limiting reactant

      If amount needed is more than what you have then reactant B is the limiting reactant and reactant A is in excess.

Determining the amount produced

• 1.      Find the limiting reactant 2.      __________ the moles of reactant (calculated in Step C5 above) by the mole ratio of product/reactant

• 3. Multiply moles of product (answer from step D2) by the molar mass of the _________  

IV.  Percent Yield•  

• 1.      Theoretical yield – the amount of product predicted to be produced on the

basis of the balanced chemical equation

• 2.      Actual yield – the amount of product actually obtained in a reaction

• 3. actual yield

• % yield = theoretical yield X 100

Percent Composition

• Steps in finding Percent Composition:• #1 Calculate the molar mass of the

compound• #2 Divide the mass of the element by the

Total• #3 Multiply by 100.• #4 Check answer: the Parts should add

up to 100.

Percent Composition

• Example: Find the percent composition of each element in carbon dioxide.

Empirical Formulas

• Empirical Formula – Lowest whole number formula for a compound

Empirical FormulasSteps in solving:

#1 Convert grams of given to moles (divide by atomic mass)

# 2 Find mole ratio – divide all parts by the smallest

# 3 IF the ratio is a whole number – then place these in the formula as subscripts

HOWEVER, if you have fractions- MUST multiply all by a common factor to arrive at a small whole number

Empirical Formulas

• EXAMPLE 1: A certain compound contains 52% nickel, 9.6% carbon and 38.4%oxygen. What is its empirical formula?

Empirical Formulas

• EXAMPLE 2: A compound is found to contain 43.7% phosphorus and 56.9% oxygen. Calculate its simplest formula.

Molecular Formulas

• Molecular formula – whole number multiples of an empirical formula:

• The molecular formula is the true formula

• Example: HO (empirical)

• H2O2 (Molecular)

Molecular Formulas

Steps in solving:

#1 Molecular mass = x Empirical mass

# 2 (Empirical formula) x

Molecular Formulas

• Example : The molecular mass for a hydrocarbon was found to be 64 g/mol. If the empirical formula is

• CH4, what is the molecular formula?