Empirical and Molecular Empirical and Molecular FormulasFormulas

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Empirical and MolecularEmpirical and Molecular Molecular Formula:Molecular Formula:

The formula for a compound that exist as The formula for a compound that exist as the nbr. of atoms.the nbr. of atoms.

Ex. HEx. H220022 Hydrogen Peroxide Hydrogen Peroxide Empirical Formula:Empirical Formula:

Indicate the lowest whole number ratio of Indicate the lowest whole number ratio of atoms of each element in a compound.atoms of each element in a compound.

Ex. HO Ex. HO Hydrogen Peroxide Hydrogen Peroxide Ho, Ho, Ho!!!Ho, Ho, Ho!!!

The molecular formula of glucose is… The molecular formula of glucose is… CC66HH1212OO6, , 6, , its empirical formula is…its empirical formula is…

CHCH22O …O … So what’s my point?So what’s my point?

Empirical formulas are lowest whole Empirical formulas are lowest whole number ratios of atoms of each type …number ratios of atoms of each type …to determine them experimentally to determine them experimentally we need numbers of atoms!we need numbers of atoms!

Consider the following:Consider the following: Analysis of a compound at CSI North Analysis of a compound at CSI North

Haven Labs indicates that it containsHaven Labs indicates that it contains 32.38 g Na32.38 g Na 22.65 g S 22.65 g S 44.99 g O44.99 g O How can we use this data to How can we use this data to

determine the formula?...Think… determine the formula?...Think… Mass Mass number of moles! number of moles!

Remember doing mole Remember doing mole conversions?conversions?

Use the molar mass of element Use the molar mass of element convert to moles.convert to moles.

you now have a ratio of moles in the you now have a ratio of moles in the compound.compound.

Divide each mole amount by the Divide each mole amount by the smallest mole amount… smallest mole amount… whole # whole # ratio???ratio???

Use ratio to write subscriptsUse ratio to write subscripts

1.408 mol Na : 0.7063 mol S : 1.408 mol Na : 0.7063 mol S : 2.812mol O2.812mol O

Divide each by smallest Divide each by smallest

1.993 mol Na : 1 mol S : 3.981 mol O1.993 mol Na : 1 mol S : 3.981 mol O

NaNa22SOSO44

Try this..Try this.. Analysis indicates that a compound Analysis indicates that a compound

contains 78.1% Boron and 21.9% contains 78.1% Boron and 21.9% hydrogen determine the empirical hydrogen determine the empirical formula.formula.

Why is this one any different from Why is this one any different from the last example?the last example?

Assume you have a 100 gram Assume you have a 100 gram sample…now how many grams of sample…now how many grams of boron do you have?boron do you have?

7.22 mol B : 21.7 mol H7.22 mol B : 21.7 mol H

Divide through by smallest Divide through by smallest 1 mol B : 3.01 mol H1 mol B : 3.01 mol H

BHBH33

Try this one …Try this one …

Analysis of a 10.150 g sample of a Analysis of a 10.150 g sample of a compound known to contain only compound known to contain only phosphorus and oxygen indicates a phosphorus and oxygen indicates a phosphorus content of 4.433 g. What phosphorus content of 4.433 g. What is the empirical formula?is the empirical formula?

Grams of P = 4.433g Grams of O = ?Grams of P = 4.433g Grams of O = ?

Now try it…Now try it…

0.1431 mol P : 0.3573 mol O0.1431 mol P : 0.3573 mol O

Divide by smallest Divide by smallest

1 mol P : 2.497 mol O …1 mol P : 2.497 mol O …

Now what?...Now what?...Multiply through by two Multiply through by two to obtain whole numbersto obtain whole numbers… … 2 : 52 : 5

PP22OO55

Find empirical formulas and the Find empirical formulas and the empirical masses…empirical masses…

1. 36.48% Na, 25.41% S, 38.11 % O…1. 36.48% Na, 25.41% S, 38.11 % O…NaNa22SOSO33 126 g/mol126 g/mol

2.2. 53.70 % iron and 46.30% sulfur…53.70 % iron and 46.30% sulfur…FeFe22SS33 208 g/mol208 g/mol

3.3. 1.04 g K, 0.70 g Cr and 0.86 g O…1.04 g K, 0.70 g Cr and 0.86 g O…KK22CrOCrO44 194 g/mol194 g/mol

4.4. 36.51% C, 6.20% H and 57.29 % O …36.51% C, 6.20% H and 57.29 % O …CHCH22OO 30 g/mol30 g/mol

Molecular FormulasMolecular Formulas You have already solved for the You have already solved for the

Empirical FormulaEmpirical Formula Now find the mass of the Empirical Now find the mass of the Empirical

FormulaFormula Ex. CHEx. CH22O O ~30 g/mol ~30 g/mol

You will be given a molecular weight You will be given a molecular weight to start with (~184.62 g/mol)to start with (~184.62 g/mol)

Now divide the molecular mass by Now divide the molecular mass by the Empirical massthe Empirical mass

This will tell you the ratio, or the This will tell you the ratio, or the amount of times that the Empricial amount of times that the Empricial Formula goes into the Molecular Formula goes into the Molecular formulaformula (184.62 g/mol)/(30 g/mol)(184.62 g/mol)/(30 g/mol)

Answer should be close to a whole Answer should be close to a whole numbernumber 6.154 6.154 6 6

Multiply the Empirical Formula by Multiply the Empirical Formula by that whole nbr. and you have the that whole nbr. and you have the molecular formulamolecular formula CC66HH1212OO66

Quantitative AnalysisQuantitative Analysis Various methods are used to Various methods are used to

determine the determine the massmass or percent or percent composition of each element in a composition of each element in a given compound…more about those given compound…more about those later…later…

Analysis by CombustionAnalysis by Combustion

HWHW Read pg 332-339Read pg 332-339

#1-2 on pg 339#1-2 on pg 339

Empirical and Molecular Formula Empirical and Molecular Formula handouthandout

One interesting note!One interesting note! A compound with a molecular mass of A compound with a molecular mass of

59.7 g is determined to be made up of 59.7 g is determined to be made up of 40% Na and 60% Cl. What is the Empirical 40% Na and 60% Cl. What is the Empirical Formula? What is the Molecular Formula?Formula? What is the Molecular Formula?

Empirical Formula – NaClEmpirical Formula – NaCl Molecular Formula – NaClMolecular Formula – NaCl What are the charges (oxidation nbrs) on What are the charges (oxidation nbrs) on

the Na cation and Cl anion?the Na cation and Cl anion? +1 and -1+1 and -1 What is the overall charge of the cmpd?What is the overall charge of the cmpd? ZEROZERO

Oxidation Numbers by GroupOxidation Numbers by Group

1+

2+

Tend to have more than one oxidation number (+1 to +7)

3+ varies 3- 2- 1-

0

3+

3+ or 4+

The sum of the oxidation nbrs in an The sum of the oxidation nbrs in an ionic cmpd must be zero.ionic cmpd must be zero.

Ex. Magnesium IodideEx. Magnesium Iodide MgMg+2+2 II-1-1

Goes to Goes to MgIMgI22 Again the charge on the cmpd is Again the charge on the cmpd is

00

Aluminum Oxide AlAluminum Oxide Al22OO33??

Percent CompositionPercent CompositionBelieve it or not you’ve Believe it or not you’ve

already done it.already done it.

% comp.% comp. Tells you the percent of the mass Tells you the percent of the mass

made up by each element in the made up by each element in the compoundcompound

The mass of each element in a The mass of each element in a compound compared to the entire compound compared to the entire mass of the compound and mass of the compound and multiplied by 100 %multiplied by 100 %

Find the percent composition of Find the percent composition of copper (I) sulfide, Cucopper (I) sulfide, Cu22SS

A sample of an unkn. Compound with A sample of an unkn. Compound with a mass of 0.2370 g is extracted from a mass of 0.2370 g is extracted from the roots of a plant. Decomposition the roots of a plant. Decomposition of the sample produces 0.09480 g of of the sample produces 0.09480 g of carbon, 0.1264 g of oxgyen, and carbon, 0.1264 g of oxgyen, and 0.0158 g of hydrogen. What is the 0.0158 g of hydrogen. What is the percentage comp. of the compound?percentage comp. of the compound?

HydratesHydrates inorganic salts which contain a inorganic salts which contain a

specific number of water molecules specific number of water molecules that are loosely attached that are loosely attached

Written as: CuSOWritten as: CuSO44 • 5H• 5H22OO Copper Sulfate with 5 water Copper Sulfate with 5 water

moleculesmolecules

Formula for a hydrateFormula for a hydrate Finding the empirical formula for a Finding the empirical formula for a

hydrate involves finding the mass hydrate involves finding the mass difference of the hydrate and difference of the hydrate and anhydrous salt by driving off the anhydrous salt by driving off the excess water:excess water:

DEMO:DEMO:http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.html

Example you conduct an experiment using Example you conduct an experiment using CuWOCuWO44 • xH• xH22OO Mass of hydrate = 2.286 gMass of hydrate = 2.286 g Mass of anhydrous salt = 2.050 gMass of anhydrous salt = 2.050 g Mass of HMass of H22O driven off = 0.236 gO driven off = 0.236 g % H% H22O = 0.236/2.286 = 10.3%O = 0.236/2.286 = 10.3% % salt = 2.050/2.286 = 89.68%% salt = 2.050/2.286 = 89.68% CuWOCuWO44 = 311.4 g/mol = 311.4 g/mol Mol salt = 2.050g (1mol/311.4g) = 0.00658 molMol salt = 2.050g (1mol/311.4g) = 0.00658 mol Mol HMol H22O = 0.236g (1mol/18g) = 0.0131 molO = 0.236g (1mol/18g) = 0.0131 mol Divide by the smallest mol amountDivide by the smallest mol amount 0.0131 mol/0.00658 mol =~2 ; 2 water molecules0.0131 mol/0.00658 mol =~2 ; 2 water molecules CuWOCuWO44 • 2H • 2H22OO