Stoichiometry and the mole

Stoichiometry and the mole

What is stoichiometry? Quantitative aspects of chemistry Stoicheon Greek root (element) Metron Greek root( to measure) Calculate how much of a reactant

needed to produce a product or how much product could be expected

Stoichiometry How much do I want? How much do I have? How much will I get?

The Mole•Defined as the number of carbon atoms in exactly 12 grams of carbon-12.

•1 mole is 6.02 x 1023 particles. •Treat it like a very large dozen •6.02 x 1023 is called Avogadro's number.

The Mole One mole = 6.022 x 1023 (Avogadro’s

number)

Avogadro’s hypothesis

When all of the following are the same: Volumes of containers Temperature of the gases The pressure exerted by and on each

gasThen, the number of molecules in each

container will be the same, too. However, the masses are not equal.

N2(g) + 3H2(g)2NH3(g)

1 3 2

1 dozen(12)

3 dozen(36)

2 dozen(24)

1 gross(144)

3 gross(432)

2 gross(288)

1mole(6.02x1023)

3 moles(18.06x1023)

2 moles(12.04 x1023)

Representative particles

• The smallest pieces of a substance.

• For an element it is an atom. –Unless it is diatomic

• For a molecular compound it is a molecule.

• For an ionic compound it is a formula unit.

Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per moleH2O = 18.02 grams per mole

Ca(OH)2 = 74.10 grams per mole

Chemical Equations

Chemical change involves a reorganization of

the atoms in one or more substances.C2H5OH + 3O2 ® 2CO2 + 3H2O

reactants products

1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

When the equation is balanced it has quantitative significance:

Mole Relations

Calculating Masses of Reactants and Products

1. Balance the equation. 2. Convert mass to moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles

of desired substance. 5. Convert moles to grams, if

necessary.

Working a Stoichiometry Problem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.1. Identify reactants and products and write the balanced equation.

Al + O2 Al2O3

b. What are the reactants?a. Every reaction needs a yield sign!

c. What are the products?d. What are the balanced coefficients?

4 3 2

Working a Stoichiometry Problem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O31 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =

12.3 g Al2O3

Percent yield The % yield is calculated from: (actual yield / theoretical yield) x 100 The theoretical yield is how much

product is predicted from balanced chemical equation.

The actual yield is how much is recovered when actual experiment is conducted.

Limiting Reactant The limiting reactant is the reactant

that is consumed first, limiting the amounts of products formed.

How to solve stoichiometry problems?

Write the balanced chemical equation

Convert to moles information on reactants and products

Use mole ratios from equation to determine number of moles of unknown

Convert moles to unit desired g xmoles xmoles yg y

Limiting reactant Reactant present in short supply

Excess reactant• Reactant in excess relative to limiting

reactant

Formulas

molecular formula = (empirical formula)n [n = integer]

molecular formula = C6H6 = (CH)6

empirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued)

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued)

Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).

Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Determination

1. Base calculation on 100 grams of compound.

2. Determine moles of each element in 100 grams of compound.

3. Divide each value of moles by the smallest of the values.

4. Multiply each number by an integer to obtain all whole numbers.


Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O


(part 2)

4.107 1.502.74

mol Cmol O

6.78 2.472.74

mol Hmol O

2.74 1.002.74

mol Omol O

Divide each value of moles by the smallest of the values.

Carbon:

Hydrogen:

Oxygen:


(part 3)Multiply each number by an integer to obtain all whole numbers.

Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00x 2 x 2 x 2

3 5 2Empirical formula:C3H5O

2

Finding the Molecular FormulaThe empirical formula for adipic acid

is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

1. Find the formula mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g



3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

2. Divide the molecular mass by the mass given by the emipirical formula.

146 273



3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g146 273

3. Multiply the empirical formula by this number to get the molecular formula.

(C3H5O2) x 2 =

C6H10O4

Combustion Analysis Technique that requires burning of an unknown substance and trap

the gases from burning.

Used to determine molecular formula of an unknown

Combustion= burning Using oxygen

Points about combustion Element that makes unknown almost always contain carbon

and hydrogen. Oxygen is often involved and nitrogen is involved sometimes.

Must know mass of the unknown substance before burning it

Unknown will be burnt in pure oxygen, present in excess Carbon dioxide and water are the products All the carbon winds up as carbon dioxide and all the

hydrogen winds up as water The end result will be to determine the empirical formula of

the substance

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Stoichiometry and the mole