section 6.2 page 268 empirical and molecular formulas
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Section 6.2 Page 268 Empirical and Molecular Formulas. Empirical vs. Molecular Formula. Empirical Formula (E f ) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Empirical vs. Molecular Formula. Empirical Formula (E f ) - PowerPoint PPT PresentationTRANSCRIPT
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Section 6.2Section 6.2 Page Page 268268
Empirical and Molecular Empirical and Molecular FormulasFormulas
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Empirical vs. Molecular Empirical vs. Molecular FormulaFormula
Empirical FormulaEmpirical Formula (E (Eff))
A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the atoms of each element in a of the atoms of each element in a compound.compound.
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Empirical vs. Molecular Empirical vs. Molecular FormulaFormula
Empirical FormulaEmpirical Formula (E (Eff))
A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the atoms of each element in a of the atoms of each element in a compound.compound.
Molecular FormulaMolecular Formula (M (Mff))
A formula that gives the A formula that gives the actualactual number of atoms number of atoms of each element in a compound.of each element in a compound.
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CHCH22OOCHCH33OOCH = COOCH = C22HH44OO22
CHCH33OOCHCH33OO
Molecular Molecular FormulaFormula
(actual)(actual)
Empirical FormulaEmpirical Formula
(simplest)(simplest)
HH22OO22 HOHO
CC66HH1212OO66 CHCH22OO
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Steps to determining the Steps to determining the empirical formula.empirical formula.
Step 1.Step 1. Find mole amounts. Find mole amounts.
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Steps to determining the Steps to determining the empirical formula.empirical formula.
Step 1.Step 1. Find mole amounts. Find mole amounts.
Step 2.Step 2. Divide each mole amount Divide each mole amount by the smallest mole amount.by the smallest mole amount.
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Example 1:Example 1:
Determine the empirical formula for a Determine the empirical formula for a compound containing 2.128 g Cl and compound containing 2.128 g Cl and 1.203 g Ca.1.203 g Ca.
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1.1. Find mole amounts.Find mole amounts.
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1.1. Find mole amounts.Find mole amounts.
nnClCl = = 2.128 g 2.128 g = 0.0600 mol Cl = 0.0600 mol Cl
35.45 g/mol35.45 g/mol
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1.1. Find mole amounts.Find mole amounts.
nnClCl = = 2.128 g 2.128 g = 0.0600 mol Cl = 0.0600 mol Cl
35.45 g/mol35.45 g/mol
nnCaCa = = 1.203 g 1.203 g = 0.0300 mol = 0.0300 mol CaCa
40.08 g/mol40.08 g/mol
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1.1. Find mole amounts.Find mole amounts.
nnClCl = = 2.128 g 2.128 g = 0.0600 mol Cl = 0.0600 mol Cl
35.45 g/mol35.45 g/mol
nnCaCa = = 1.203 g 1.203 g = = 0.03000.0300 mol mol CaCa
40.08 g/mol40.08 g/mol
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2.2. Divide each mole by the Divide each mole by the smallest mole.smallest mole.
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2.2. Divide each mole by the Divide each mole by the smallest mole.smallest mole.
Cl = Cl = 0.0600 mol Cl0.0600 mol Cl = 2.00 mol Cl = 2.00 mol Cl 0.03000.0300
Ca = Ca = 0.0300 mol Ca0.0300 mol Ca = 1.00 mol = 1.00 mol CaCa
0.03000.0300
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2.2. Divide each mole by the Divide each mole by the smallest mole.smallest mole.
Cl = Cl = 0.0600 mol Cl0.0600 mol Cl = 2.00 mol Cl = 2.00 mol Cl 0.03000.0300
Ca = Ca = 0.0300 mol Ca0.0300 mol Ca = 1.00 mol Ca = 1.00 mol Ca 0.03000.0300
Ratio – 1 Ca : 2 ClRatio – 1 Ca : 2 Cl
Empirical Formula = CaClEmpirical Formula = CaCl22
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Example 2:Example 2:A compound consists of 50.81% zinc, A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? oxygen. What is the empirical formula?
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Example 2:Example 2:A compound consists of 50.81% zinc, A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? oxygen. What is the empirical formula?
• Always assume that you have a 100g sample.
• That way you can convert percent directly to grams...
• In this case you would have 50.81g of Zn, 16.04g of P, and 33.15g of O.
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Rhyme
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol
P30.97g/mol
O16g/mol
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g
P30.97g/mol16.04g
O16g/mol33.15g
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
P30.97g/mol16.04g
O16g/mol33.15g
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
O16g/mol33.15g
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
2.0720.518= 4
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
1.5 x 2 = 3
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
2.0720.518= 4
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
1.5 x 2 = 3
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
1 x 2 = 2
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
2.0720.518= 4
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
1.5 x 2 = 3
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
1 x 2 = 2
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
2.0720.518= 4
4 x 2 = 8
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A compound consists of 50.81% zinc, and A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? is the empirical formula?
Atom Mass Mole Mole Ratio
Whole # Ratio
Zn65.38g/mol50.81g nZn = 50.81g
65.38g/mol
= 0.777 mol
0.7770.518= 1.5
1.5 x 2 = 3
P30.97g/mol16.04g nP = 16.04g
30.97g/mol
= 0.518 mol
0.5180.518= 1
1 x 2 = 2
O16g/mol33.15g nO = 33.15g
16g/mol = 2.072
mol
2.0720.158= 4
4 x 2 = 8
Therefore empirical formula would be Zn3P2O8
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Example 3:Example 3:The percentage composition of a The percentage composition of a compound is 48.63% carbon, 21.59% compound is 48.63% carbon, 21.59% oxygen, 18.90% nitrogen, and the rest oxygen, 18.90% nitrogen, and the rest hydrogen. Find the empirical formula hydrogen. Find the empirical formula for the compound.for the compound.
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Molecular FormulaMolecular Formula
The molecular formula gives the The molecular formula gives the actualactual number of number of atoms of each element in a molecular compound.atoms of each element in a molecular compound.
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Molecular FormulaMolecular Formula
The molecular formula gives the The molecular formula gives the actualactual number of number of atoms of each element in a molecular compound.atoms of each element in a molecular compound.
StepsSteps
1. Find the empirical formula.1. Find the empirical formula.
2. Calculate the Empirical Formula Molar Mass. (M2. Calculate the Empirical Formula Molar Mass. (MEfEf))
3. Divide the Molecular Formula Molar Mass (M3. Divide the Molecular Formula Molar Mass (MMfMf) by ) by the “Mthe “MEfEf”.”.
4. Multiply empirical formula by factor.4. Multiply empirical formula by factor.
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Molecular FormulaMolecular Formula
The molecular formula gives the The molecular formula gives the actualactual number of number of atoms of each element in a molecular compound.atoms of each element in a molecular compound.
StepsSteps
1. Find the empirical formula.1. Find the empirical formula.
2. Calculate the Empirical Formula Molar Mass. (M2. Calculate the Empirical Formula Molar Mass. (MEfEf))
3. Divide the Molecular Formula Molar Mass (M3. Divide the Molecular Formula Molar Mass (MMfMf) by ) by the “Mthe “MEfEf”.”.
4. Multiply empirical formula by factor.4. Multiply empirical formula by factor.
Note: The molecular formula molar mass (MNote: The molecular formula molar mass (MMfMf) is always given ) is always given in the question. in the question.
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Example 1:
Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.
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Example 1:
Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.
Step 2. “MEf” = 62.03 g
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Example 1:
Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.
Step 2. “MEf” = 62.03 g
Step 3. MMf/MEf = 124.06/62.03 = 2
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Example 1:
Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.
Step 2. “MEf” = 62.03 g
Step 3. MMf/MEf = 124.06/62.03 = 2
Step 4. 2(CH2O3) = C2H4O6
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Example 1:
Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH2O3.
Step 2. “MEf” = 62.03 g
Step 3. MMf/MEf = 124.06/62.03 = 2
Step 4. 2(CH2O3) = C2H4O6
Therefore the Molecular formula (Mf) is C2H4O6
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Example 2:Example 2:
A compound containing carbon, A compound containing carbon, hydrogen, iodine and oxygen is found hydrogen, iodine and oxygen is found to be 18.0% carbon, 2.5% hydrogen, to be 18.0% carbon, 2.5% hydrogen, 63.5% iodine, and 16.0% O. The molar 63.5% iodine, and 16.0% O. The molar mass of this compound is known to be mass of this compound is known to be 400 g/mol. What is its molecular 400 g/mol. What is its molecular formula?formula?
1. Find the empirical formula.
2. Then you can find the molecular formula.
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Mass Spectrometry Demo!Mass Spectrometry Demo!
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Poison Identification.Poison Identification.
Acetone: C3H6O
Hydrogen cyanide: HCN
Ethylene Glycol: C2H6O2
Isopropanol: C3H8O
Methanol: CH4O
Sodium hypochlorite: NaOCl
Acetaminophen: C8H9NO2
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