Stoichiometry

Stoichiometry

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

The MoleThe Mole

1 dozen =1 gross = 1 ream =

1 mole =

121445006.022 x 1023

There are exactly 12 grams of carbon-12 in one mole of carbon-12.

Avogadro’s NumberAvogadro’s Number6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).

Amadeo Avogadro

I didn’t discover it. Its just named after

me!

Calculations with Moles:Calculations with Moles:Converting moles to gramsConverting moles to grams

How many grams of lithium are in 3.50 moles of lithium?

3.50 mol Li= g Li

1 mol Li

6.94 g Li 45.1

Calculations with Moles:Calculations with Moles:Converting grams to molesConverting grams to moles

How many moles of lithium are in 18.2 grams of lithium?

18.2 g Li= mol Li

6.94 g Li

1 mol Li 2.62

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 3.50 moles of lithium?

3.50 mol Li = atoms Li

1 mol Li

6.022 x 1023 atoms Li 2.11 x 1024

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.022 x 1023 atoms Li

1.58 x 1024

6.94 g Li 1 mol Li

(18.2)(6.022 x 1023)/6.94

Calculating Formula MassCalculating Formula MassCalculate the formula mass of magnesium Calculate the formula mass of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) ==

84.32 g84.32 g

Calculating Percentage Calculating Percentage CompositionComposition

Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..From previous slide:From previous slide:24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g 24.31 100 28.83%

84.32Mg

12.01 100 14.24%84.32

C 48.00 100 56.93%84.32

O 100.00

FormulasFormulas

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

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical 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.

FormulasFormulas (continued)(continued)

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

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued)(continued)

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

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Empirical Formula DeterminationDetermination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

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

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

4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.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)(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 Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:


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

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

33 55 22Empirical formula:C3H5O

2

Finding the Molecular Finding the Molecular FormulaFormulaThe empirical formula for adipic acid The empirical formula for adipic acid

is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g



3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

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

146 273



3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146 273

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

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44

ReviewReview: Chemical : Chemical EquationsEquations

Chemical change involves a Chemical change involves a reorganization of reorganization of

the atoms in one or more substances.the atoms in one or more substances.CC22HH55OH + 3OOH + 3O22 2CO2CO22 + 3H + 3H22OO

reactantsreactants productsproducts

11 mole of ethanol mole of ethanol reacts with reacts with 33 moles of moles of oxygenoxygento produce to produce 22 moles of carbon dioxide moles of carbon dioxide and and 33 moles of watermoles of water

When the equation is balanced it has When the equation is balanced it has quantitative significance:quantitative significance:

Solving a Stoichiometry Solving a Stoichiometry ProblemProblem

1.1. Balance the equation.Balance the equation.2.2. Convert masses to moles.Convert masses to moles.3.3. Determine which reactant is limiting.Determine which reactant is limiting.4.4. Use moles of limiting reactant and Use moles of limiting reactant and

mole ratios to find moles of desired mole ratios to find moles of desired product.product.

5.5. Convert from moles to grams.Convert from moles to grams.

Working a Stoichiometry Working a Stoichiometry ProblemProblem

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 Working a Stoichiometry ProblemProblem

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

Limiting Limiting ReactantReactant The The limiting reactantlimiting reactant is the reactant is the reactant

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

Documents

Stoichiometry