Section 6.2 Page 268 Empirical and Molecular Formulas

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Section 6.2 Page 268Empirical and
Molecular Formulas
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Empirical vs. Molecular Formula

• Empirical Formula (Ef)
• A formula that gives the simplest whole-number
  ratio of the atoms of each element in a compound.

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Empirical vs. Molecular Formula

• Empirical Formula (Ef)
• A formula that gives the simplest whole-number
  ratio of the atoms of each element in a compound.
• Molecular Formula (Mf)
• A formula that gives the actual number of atoms
  of each element in a compound.

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Molecular Formula (actual) Empirical Formula (simplest)
H2O2 HO
C6H12O6 CH2O
CH3O
CH3O
CH2O
CH3OOCH C2H4O2
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• Steps to determining the empirical formula.
• Step 1. Find mole amounts.

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• Steps to determining the empirical formula.
• Step 1. Find mole amounts.
• Step 2. Divide each mole amount by the smallest
  mole amount.

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• Example 1
• Determine the empirical formula for a compound
  containing 2.128 g Cl and 1.203 g Ca.

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• 1. Find mole amounts.

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• 1. Find mole amounts.
• nCl 2.128 g 0.0600 mol Cl
• 35.45 g/mol

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• 1. Find mole amounts.
• nCl 2.128 g 0.0600 mol Cl
• 35.45 g/mol
• nCa 1.203 g 0.0300 mol Ca
• 40.08 g/mol

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• 1. Find mole amounts.
• nCl 2.128 g 0.0600 mol Cl
• 35.45 g/mol
• nCa 1.203 g 0.0300 mol Ca
• 40.08 g/mol

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• 2. Divide each mole by the smallest mole.

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• 2. Divide each mole by the smallest mole.
• Cl 0.0600 mol Cl 2.00 mol Cl
• 0.0300
• Ca 0.0300 mol Ca 1.00 mol Ca
• 0.0300

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• 2. Divide each mole by the smallest mole.
• Cl 0.0600 mol Cl 2.00 mol Cl
• 0.0300
• Ca 0.0300 mol Ca 1.00 mol Ca
• 0.0300
• Ratio 1 Ca 2 Cl
• Empirical Formula CaCl2

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Example 2A compound consists of 50.81 zinc,
and 16.04 phosphorus, and 33.15 oxygen. What is
the empirical formula?
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Example 2A compound consists of 50.81 zinc,
and 16.04 phosphorus, and 33.15 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 16.04
phosphorus, and 33.15 oxygen. What 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 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g
P30.97g/mol 16.04g
O16g/mol 33.15g
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol
P30.97g/mol 16.04g
O16g/mol 33.15g
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol
O16g/mol 33.15g
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol 2.072 0.518 4
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5 1.5 x 2 3
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol 2.072 0.518 4
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5 1.5 x 2 3
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1 1 x 2 2
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol 2.072 0.518 4
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5 1.5 x 2 3
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1 1 x 2 2
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol 2.072 0.518 4 4 x 2 8
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A compound consists of 50.81 zinc, and 16.04
phosphorus, and 33.15 oxygen. What is the
empirical formula?
Atom Mass Mole Mole Ratio Whole Ratio
Zn65.38g/mol 50.81g nZn 50.81g 65.38g/mol 0.777 mol 0.777 0.518 1.5 1.5 x 2 3
P30.97g/mol 16.04g nP 16.04g 30.97g/mol 0.518 mol 0.518 0.518 1 1 x 2 2
O16g/mol 33.15g nO 33.15g 16g/mol 2.072 mol 2.072 0.158 4 4 x 2 8
Therefore empirical formula would be Zn3P2O8
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Example 3The percentage composition of a
compound is 48.63 carbon, 21.59 oxygen, 18.90
nitrogen, and the rest hydrogen. Find the
empirical formula for the compound.
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Molecular Formula

• The molecular formula gives the actual number
  of atoms of each element in a molecular compound.

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Molecular Formula

• The molecular formula gives the actual number
  of atoms of each element in a molecular compound.
• Steps
• 1. Find the empirical formula.
• 2. Calculate the Empirical Formula Molar Mass.
  (MEf)
• 3. Divide the Molecular Formula Molar Mass
  (MMf) by the MEf.
• 4. Multiply empirical formula by factor.

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Molecular Formula

• The molecular formula gives the actual number
  of atoms of each element in a molecular compound.
• Steps
• 1. Find the empirical formula.
• 2. Calculate the Empirical Formula Molar Mass.
  (MEf)
• 3. Divide the Molecular Formula Molar Mass
  (MMf) by the MEf.
• 4. Multiply empirical formula by factor.
• Note The molecular formula molar mass (MMf) is
  always given 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 2A compound containing carbon,
hydrogen, iodine and oxygen is found to be 18.0
carbon, 2.5 hydrogen, 63.5 iodine, and 16.0 O.
The molar mass of this compound is known to be
400 g/mol. What is its molecular formula?

1. Find the empirical formula.
2. Then you can find the molecular formula.

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Mass Spectrometry Demo!
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Poison Identification.
Acetone C3H6O Hydrogen cyanide HCN Ethylene
Glycol C2H6O2 Isopropanol C3H8O Methanol
CH4O Sodium hypochlorite NaOCl Acetaminophen
C8H9NO2
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