FORMULAS, EQUATIONS AND MOLES

Mole Calculation

Chapter 3

Names associated with an amount

The mole (mol) is the amount of a substance that

contains as many elementary entities as there

are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

The Mole…..

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

What is the mass of one mole of:

C S

Cu Fe

Hg

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K 1 mol K39.10 g K

x x 6.022 x 1023 atoms K1 mol K

=

8.49 x 1021 atoms K

Molar Mass

The molar mass of a compound is found by adding Together the molar masses of all of its elements, takingInto account the number of moles of each element present.

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

STOICHIOMETRYQuantitative relationship among reactants and products are called stoichiometry.

Stoichiometry problems are easily solved when amounts of substances are converted from mass (in units of g, kg etc), volume (L), into moles.

Amounts in moles depend on how the chemical formula or the chemical reaction equations are written.

Chemical reaction equations are the basis for reaction stoichiometry, but when the reactants are not stoichiometric mixtures, some reactants will be in excess whereas others will be in limited supply.

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the REQUIRED quantity

4. Convert moles of REQUIRED quantity into desired units

Mass Changes in Chemical Reactions

3.8

Stoichiometric Ratio

Coefficient B Coefficient A

Moles B = X Moles A

The stoichiometric ratio is used to determine amounts of compounds consumed or produced in a balanced chemical reaction

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100

Limiting Reagent

The number of bicycles that can be assembled is limited by whichever part runs out first. In the inventory shown in this figure, wheels are that part.

The reactant that is completely consumed by the reaction

A molecular view of a Limiting reactant situation for the ammonia Synthesis. To make 4 molecules of NH3 requires 2 molecules of N2 and6 molecules of H2. If we start with 4 molecules of N2 and 6 molecules ofH2, H2 is the limiting reactant.

Limiting Reagent: It is the reactant whose number of moles divided By it’s stoichiometric coefficient is the smallest

Example:Given:

3 Fe + 4 H2O Fe3O4 + 4 H2

If 1.00 Kg of each of the reactant is allowed to react. How many Kg of POCl3 will be formed?

1.Find the limiting reagenta.Find moles of eachb.Divide by stoichiometric coefficientc.The compound with the smallest number of moles is the LR

PCl3:

1000g PCl3 X1 mol PCl3

137.3 g PCl3= 7.28 moles / 6 = 1.21 moles

Cl2:

1000g Cl2 X1 mol Cl2

70.9g Cl2= 14.1 moles / 6 = 2.35 moles

P4O10:

1000g P4O10X1 mol P4O10

283.9 g P4O10

= 3.52 moles / 1 = 3.52 moles

2. Determine the amount of POCl3 based on the limiting reagent PCl3.

7.28 moles PCl3 X10 mol POCl3

6 mol PCl3X

1 mol POCl3

153.3 g POCl3= 1.85 Kg POCl3

Note that the conversions of grams to Kg is not shown.

Solution Concentrations

Molarity: The most useful way of expressing the amount of a substance dissolved in a solution is with molarity

Molarity= Moles of soluteVolume of solution in L

Making Molar

SolutionsFrom Solids

What are molar solutions?What are molar solutions?A molar solution is one that expresses

“concentration” in moles per volumeUsually the units are in mol/Lmol/L can be abbreviated as M or [ ]Molar solutions are prepared using:

a balance to weigh moles (as grams)a volumetric flask to measure litres

L refers to entire volume, not water!Because the units are mol/L,

we can use the equation M = n/LAlternatively, we can use the factor label method

Solution Concentrations

Practice making molar solutionsPractice making molar solutions1. Calculate # of grams required to make 100 mL

of a 0.10 M solution of NaOH (see above).2. Get volumetric flask, plastic bottle, 100 mL

beaker, eyedropper. Rinse all with tap water.3. Fill a beaker with distilled water.4. Pour 20 - 30 mL of H2O from beaker into flask.5. Weigh NaOH. Add it to flask. Do step 5 quickly. 6. Mix (by swirling) until the NaOH is dissolved.7. Add distilled H2O to just below the colored line.8. Add distilled H2O to the line using eyedropper.

Making Molar

SolutionsFrom

Liquids

Dilution FormulaDilution Formula1. Calculate volume of solution of known molarity

required to make 100 mL of a 0.10 M solution of HCl.

MiVi =MfVf

2. Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water.

3. Fill a beaker with distilled water.4. Pour 20 - 30 mL of H2O from beaker into flask.5. Measure HCl (Vi). Add it to flask and mix6. Add distilled H2O to just below the colored line

(100mL).7. Add distilled H2O to the line using eyedropper.