Determining Chemical Determining Chemical FormulasFormulas

empirical formula- empirical formula- consists of the consists of the symbols for the elements combined in symbols for the elements combined in a compound, with subscripts showing a compound, with subscripts showing the smallest whole number mole ratio the smallest whole number mole ratio of the different atoms in the compoundof the different atoms in the compound

CHCH33 = empirical formula (does not = empirical formula (does not exist)exist)

CC22HH66 = molecular formula (ethene) = molecular formula (ethene)

Empirical FormulasEmpirical Formulas The formulas of ionic The formulas of ionic

compounds are compounds are empirical formulasempirical formulas by the definition of by the definition of ionic formulas.ionic formulas.

The formulas of The formulas of molecular molecular compounds compounds may or may or may notmay not be the same be the same as its empirical as its empirical formula.formula.

Calculating an Empirical Calculating an Empirical FormulaFormula

1)1) If the elements are in % composition by If the elements are in % composition by mass form, covert the percentages to mass form, covert the percentages to grams.grams.

2)2) Convert the masses of each element to Convert the masses of each element to moles by dividing the mass of the element moles by dividing the mass of the element by its molar mass.by its molar mass.

3)3) Select the element with the smallest Select the element with the smallest number of moles and divide the number number of moles and divide the number of moles of each element by that number of moles of each element by that number which will give you a 1:---:--- ratio.which will give you a 1:---:--- ratio.

4)4) IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to each number ratio, apply the numbers to each element. If one of the number is not close element. If one of the number is not close to a whole number, use a multiplier to to a whole number, use a multiplier to convert the ratio to a whole number ratio.convert the ratio to a whole number ratio.

1-1- If the elements are in % If the elements are in % composition by mass form, covert composition by mass form, covert the percentages to grams.the percentages to grams.

e.g.e.g. C = 40.0% C = 40.0% 40.0 g 40.0 gH = 6.67% H = 6.67% 6.67 g 6.67 gO = 53.3% O = 53.3% 53.3 53.3

gg

2-2- Convert the masses of each Convert the masses of each element to moles by dividing the element to moles by dividing the mass of the element by its molar mass of the element by its molar mass.mass.

e.g.e.g. C = 40.0/12 = 3.33 molC = 40.0/12 = 3.33 molH = 6.67/1 = 6.67 molH = 6.67/1 = 6.67 molO = 53.3/16 = 3.33 molO = 53.3/16 = 3.33 mol

3-3- Select the element with the Select the element with the smallest number of moles and smallest number of moles and divide the number of moles of each divide the number of moles of each element by that number which will element by that number which will give you a 1:---:--- ratio.give you a 1:---:--- ratio.

e.g.e.g. C = 3.33/3.33 = 1C = 3.33/3.33 = 1H = 6.67/3.33 = 2H = 6.67/3.33 = 2O = 3.33/3.33 = 1O = 3.33/3.33 = 1

4-4- IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to number ratio, apply the numbers to each element. If one of the number each element. If one of the number is not close to a whole number, use is not close to a whole number, use a multiplier to convert the ratio to a a multiplier to convert the ratio to a whole number ratio.whole number ratio.

e.g.e.g. 1:2:1 ratio 1:2:1 ratio CH CH22OO

Calculating an Empirical FormulaCalculating an Empirical Formula Sample Problem Sample Problem 32.38% Na, 22.65% S, & 44.99% O.32.38% Na, 22.65% S, & 44.99% O.1-1- convert to 32.38 g Na, 22.65 g S, & 44.99 g Oconvert to 32.38 g Na, 22.65 g S, & 44.99 g O 2-2- 32.38 32.38 ÷ 22.99 = 1.408 mol Na÷ 22.99 = 1.408 mol Na

22.65 ÷ 32.07 = 0.7063 mol S22.65 ÷ 32.07 = 0.7063 mol S44.99 ÷ 16.00 = 2.812 mol O44.99 ÷ 16.00 = 2.812 mol O

3-3- 1.408 ÷ 0.7063 = 1.993 mol Na 1.408 ÷ 0.7063 = 1.993 mol Na 2 2

0.7063 ÷ 0.7063 = 1 mol S0.7063 ÷ 0.7063 = 1 mol S2.812 ÷ 0.7063 = 3.981 mol O 2.812 ÷ 0.7063 = 3.981 mol O 4 4

4-4- Rounding Rounding 2:1:4 2:1:4 Na Na22SOSO44

Calculating an Empirical Calculating an Empirical FormulaFormula

Review sample problem M on page Review sample problem M on page 247.247.

Do practice problems #1, 2, & 3 on Do practice problems #1, 2, & 3 on page 247.page 247.

Practice problem #1 page 247Practice problem #1 page 247

63.52% iron (Fe)63.52% iron (Fe) 36.48% sulfur (S)36.48% sulfur (S)

Convert % to grams:Convert % to grams: Fe = 63.52g S = 36.48gFe = 63.52g S = 36.48g

Divide each element by its molar mass:Divide each element by its molar mass:Fe = 63.52/55.8 = 1.14 molFe = 63.52/55.8 = 1.14 molS = 36.48/32.1 = 1.14 molS = 36.48/32.1 = 1.14 mol

Divide each number of moles by the smallest Divide each number of moles by the smallest number:number:

Fe = 1.14/1.14 = 1Fe = 1.14/1.14 = 1 S = 1.14/1.14 = 1S = 1.14/1.14 = 1

Ratio = 1:1 so Ratio = 1:1 so FeSFeS is the empirical formula is the empirical formula

Practice problem #2 page 247Practice problem #2 page 247

K = 26.56%K = 26.56% Cr = 35.41% Cr = 35.41% O = 38.03% O = 38.03%

K = 26.56/39.1 = 0.679 molK = 26.56/39.1 = 0.679 molCr = 35.41/52.0 = 0.681 molCr = 35.41/52.0 = 0.681 molO = 38.03/16.0 = 2.38 molO = 38.03/16.0 = 2.38 mol

K = 0.679/0.679 = 1K = 0.679/0.679 = 1Cr = 0.681/0.679 = 1.003Cr = 0.681/0.679 = 1.003O = 2.38/0.679 = 3.51O = 2.38/0.679 = 3.51

1:1:3.5 ratio1:1:3.5 ratioDouble the ratio to get whole numbers Double the ratio to get whole numbers 2:2:7 2:2:7

Empirical formula is Empirical formula is KK22CrCr22OO77

Practice problem #3 page 247.Practice problem #3 page 247.

20.0 g calcium & bromine20.0 g calcium & bromine

4.00 g Ca so 16.00 g Br4.00 g Ca so 16.00 g Br

Already in grams so divide by molar mass:Already in grams so divide by molar mass:4.00/ 40.1 = .0997 mol Ca4.00/ 40.1 = .0997 mol Ca16.00/79.9 = .2003 mol Br16.00/79.9 = .2003 mol Br

Ca = .0997/.0997 = 1Ca = .0997/.0997 = 1Br = .2003/.0997 = 2.009 Br = .2003/.0997 = 2.009 2 2

Empirical formula is Empirical formula is CaBrCaBr22

Ch 7 quiz #4Ch 7 quiz #4Empirical FormulasEmpirical Formulas

1-1- A compound is 27.3% carbon and A compound is 27.3% carbon and 72.7% 72.7% oxygen by mass. What is the oxygen by mass. What is the empirical empirical formula of the compound?formula of the compound?

2-2- A compound is 11.1% hydrogen A compound is 11.1% hydrogen and and

88.9% oxygen. What is its empirical 88.9% oxygen. What is its empirical formula?formula?

Calculating a Molecular Calculating a Molecular FormulaFormula

molecular formula-molecular formula- the actual formula of the actual formula of a a molecularmolecular compound (it may or may not compound (it may or may not be the same as the empirical formula of the be the same as the empirical formula of the compound)compound)

1)1) The molar mass of a compound is The molar mass of a compound is determined by analytical means & is given.determined by analytical means & is given.

2)2) Calculate the formula mass of the empirical Calculate the formula mass of the empirical formula. Divide the molar mass of the formula. Divide the molar mass of the compound by its empirical mass.compound by its empirical mass.

3)3) ““Multiply” the empirical formula by this Multiply” the empirical formula by this factor.factor.

Calculating a Molecular Calculating a Molecular FormulaFormula

1)1) empirical formula = Pempirical formula = P22OO55

2)2) molecular mass is 283.89molecular mass is 283.89

3)3) empirical mass is 141.94empirical mass is 141.94

4)4) Dividing the molecular mass by Dividing the molecular mass by the empirical mass gives a the empirical mass gives a multiplication factor of : multiplication factor of :

283.89 283.89 ÷ 141.94 ÷ 141.94 = 2.0001 = 2.0001 2 2

5)5) 2 x (P2 x (P22OO55) ) P P44OO1010

Chapter 7 ProblemsChapter 7 Problems

Do practice Do practice problems #1 & problems #1 & 2 on page 249.2 on page 249.

Do section Do section review review problems #1-4 problems #1-4 on on page 249.page 249.

Practice problem #1 page 249Practice problem #1 page 249

empirical formula = CHempirical formula = CH

formula mass = 78.110 amuformula mass = 78.110 amu

empirical mass = ? = 12.0 + 1.0 = 13.0 empirical mass = ? = 12.0 + 1.0 = 13.0 amuamu

molecular mass / empirical mass = molecular mass / empirical mass = 78.110/13.0 = 6.008 78.110/13.0 = 6.008 multiplication multiplication factor of 6factor of 6

molecular formula = CH x 6 molecular formula = CH x 6 C C66HH66

Practice problem #2 page 249Practice problem #2 page 249

formula mass = 34.00 amuformula mass = 34.00 amu0.44 g H & 6.92 g O0.44 g H & 6.92 g O11stst find empirical formula: find empirical formula:

H = 0.44/1.0 = 0.44H = 0.44/1.0 = 0.44O = 6.92/16.0 = 0.43O = 6.92/16.0 = 0.43

0.44/0.43 0.44/0.43 1 H & 0.43/0.43 1 H & 0.43/0.43 1 O 1 Oempirical formula = HOempirical formula = HOempirical mass = 17.0empirical mass = 17.0

formula mass / empirical mass = formula mass / empirical mass = 34.00/17.0 = 234.00/17.0 = 2

HO x 2 HO x 2 H H22OO22

Section review problem #1 page 249.Section review problem #1 page 249.

36.48% Na36.48% Na 25.41% S25.41% S 38.11% O 38.11% O

36.48/23.0 = 1.58 mol Na36.48/23.0 = 1.58 mol Na

25.41/32.1 = 0.792 mol S25.41/32.1 = 0.792 mol S

38.11/16.0 = 2.38 mol O38.11/16.0 = 2.38 mol O

1.58/0.792 = 1.995 1.58/0.792 = 1.995 2 2

0.792/0.792 = 1 0.792/0.792 = 1 1 1

2.38/0.792 = 3.005 2.38/0.792 = 3.005 3 3

2:1:3 2:1:3 Na Na22SOSO33

Section review problem #2 page 249.Section review problem #2 page 249.

53.70% Fe53.70% Fe 46.30% S46.30% S

53.70/55.8 = 0.962 mol Fe53.70/55.8 = 0.962 mol Fe46.30/32.1 = 1.44 mol S46.30/32.1 = 1.44 mol S

0.962/0.962 = 1 Fe0.962/0.962 = 1 Fe1.44/0.962 = 1.50 S1.44/0.962 = 1.50 S

1:1.5 doubled 1:1.5 doubled 2:3 2:3 Fe Fe22SS33

Section review problem #3 page 249Section review problem #3 page 249

1.04 g K1.04 g K 0.70 g Cr0.70 g Cr 0.86 g O0.86 g O

1.04/39.1 = .0266 mol K1.04/39.1 = .0266 mol K0.70/52.0 = .0135 mol Cr0.70/52.0 = .0135 mol Cr0.86/16.0 = .0538 mol O0.86/16.0 = .0538 mol O

.0266/.0135 = 1.97 .0266/.0135 = 1.97 2 2

.0135/.0135 = 1.0135/.0135 = 1

.0538/.0135 = 3.99 .0538/.0135 = 3.99 4 4

Empirical formula = KEmpirical formula = K22CrOCrO44

Section Review problem #4 page 249Section Review problem #4 page 249

4.04 g N 11.46 g O4.04 g N 11.46 g O f.m. = 108.0 amu f.m. = 108.0 amu

4.04/14.0 = .289 mol N4.04/14.0 = .289 mol N11.46/16.0 =.716 mol O11.46/16.0 =.716 mol O

.289/.289 = 1.289/.289 = 1 .716/.289 = 2.45.716/.289 = 2.45double ratio double ratio 2:5 2:5 N N22OO55

e.f.m. = 108e.f.m. = 108f.m./e.f.m. = 108/108 = 1f.m./e.f.m. = 108/108 = 1

empirical formula is same as molecular formula empirical formula is same as molecular formula NN22OO55

To find molar mass: add the masses of To find molar mass: add the masses of the elements in the formula of the the elements in the formula of the compound.compound.

To find number of grams (mass): To find number of grams (mass): multiply # of moles times the molar multiply # of moles times the molar mass of the compound.mass of the compound.

To find the number of moles: divide To find the number of moles: divide the number of grams by the molar the number of grams by the molar mass of the compound.mass of the compound.

To calculate % composition by mass: To calculate % composition by mass:

1- find the molar mass of a compound1- find the molar mass of a compound

2- divide the mass of each element by 2- divide the mass of each element by thethe

molar mass of the compoundmolar mass of the compound

3- multiply by 100 to convert each 3- multiply by 100 to convert each ratio to a ratio to a

percentpercent

Calculating an Empirical Calculating an Empirical FormulaFormula

1)1) If the elements are in % composition by If the elements are in % composition by mass form, convert the percentages to mass form, convert the percentages to grams.grams.

2)2) Convert the masses of each element to Convert the masses of each element to moles by dividing the mass of the element moles by dividing the mass of the element by its molar mass.by its molar mass.

3)3) Select the element with the smallest Select the element with the smallest number of moles and divide the number number of moles and divide the number of moles of each element by that number of moles of each element by that number which will give you a 1:---:--- ratio.which will give you a 1:---:--- ratio.

4)4) IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to each number ratio, apply the numbers to each element. If one of the number is not close element. If one of the number is not close to a whole number, use a multiplier to to a whole number, use a multiplier to convert the ratio to a whole number ratio.convert the ratio to a whole number ratio.

Calculating a Molecular Calculating a Molecular FormulaFormula

molecular formula-molecular formula- the actual formula of the actual formula of a a molecularmolecular compound (it may or may not compound (it may or may not be the same as the empirical formula of the be the same as the empirical formula of the compound)compound)

1)1) The molar mass of a compound is The molar mass of a compound is determined by analytical means & is given.determined by analytical means & is given.

2)2) Calculate the formula mass of the empirical Calculate the formula mass of the empirical formula. Divide the molar mass of the formula. Divide the molar mass of the compound by its empirical mass.compound by its empirical mass.

3)3) ““Multiply” the empirical formula by this Multiply” the empirical formula by this factor.factor.

Final Practice- chapter 7 part 2Final Practice- chapter 7 part 21-1- Determine the molar mass of the Determine the molar mass of the

compound compound NaNa33POPO4 4 ..

2-2- How many moles of COHow many moles of CO22 are in 198 g ? are in 198 g ?

3-3- What is the mass of 2.25 moles of HWhat is the mass of 2.25 moles of H22O ?O ?

4-4- What is the % composition of each element What is the % composition of each element of of the compound Pthe compound P44OO1010 ? ?

5-5- What is the empirical formulas of a What is the empirical formulas of a compound that compound that is 25.9% N and 74.1% O ? is 25.9% N and 74.1% O ? What is it molecular What is it molecular formula if its molecular formula if its molecular mass is 216 ?mass is 216 ?

Final Practice- chapter 7 part 2Final Practice- chapter 7 part 2

1-1- Determine the molar mass of the Determine the molar mass of the compound compound NaNa33POPO4 4 ..

Na = 3 x 23.0 = 69.0Na = 3 x 23.0 = 69.0

P = 1 x 31.0 = 31.0P = 1 x 31.0 = 31.0

O = 4 x 16.0 = O = 4 x 16.0 = 64.064.0

164.0 g/mol164.0 g/mol

2-2- How many moles of COHow many moles of CO22 are in 198 g ? are in 198 g ?

C = 1 x 12.0 = 12.0C = 1 x 12.0 = 12.0

O = 2 x 16.0 = O = 2 x 16.0 = 32.032.0

44.0 g/mol44.0 g/mol

198/44.0 = 4.5 mol CO198/44.0 = 4.5 mol CO22

3-3- What is the mass of 2.25 moles of HWhat is the mass of 2.25 moles of H22O ?O ?

H = 2 x 1.0 = 2.0H = 2 x 1.0 = 2.0

O = 1 x 16.0 = O = 1 x 16.0 = 16.016.0

18.0 g/mol18.0 g/mol

2.25 mol x 18.0 g/mol = 40.5 g H2.25 mol x 18.0 g/mol = 40.5 g H22OO

4-4- What is the % composition of each element What is the % composition of each element of of the compound Pthe compound P44OO1010 ? ?

P = 4 x 31.0 = 124.0P = 4 x 31.0 = 124.0

O = 10 x 16.0 = O = 10 x 16.0 = 160.0160.0

284.0 g/mol284.0 g/mol

P = 124/284(100) = 43.7%P = 124/284(100) = 43.7%

O = 160/284 (100) = 56.3%O = 160/284 (100) = 56.3%

5-5- What is the empirical formulas of a What is the empirical formulas of a compound that compound that is 25.9% N and 74.1% O ? is 25.9% N and 74.1% O ? What is it molecular What is it molecular formula if its molecular formula if its molecular mass is 216 ?mass is 216 ?

N = 25.9/14.0 = 1.85N = 25.9/14.0 = 1.85

O = 74.1/16.0 = 4.63O = 74.1/16.0 = 4.63

N = 1.85/1.85 = 1N = 1.85/1.85 = 1

O = 4.63/1.85 = 2.5O = 4.63/1.85 = 2.5

1:2.5 doubled 1:2.5 doubled 2:5 so empirical formula = N 2:5 so empirical formula = N22OO55

e.f.m. = (2 x 14) + (5 x 16) = 108e.f.m. = (2 x 14) + (5 x 16) = 108

216 (mfm)/108 (efm) = 2216 (mfm)/108 (efm) = 2 2 x 2:5 2 x 2:5 4:10 4:10 N N44OO1010