Unit 4 Notes.notebook
October 21, 2015
Empirical and Molecular Formulas
Empirical Formula
Empirical Formula
A formula that gives the simplest whole­
number ratio of the atoms of each element in a
compound.
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Molecular Formula
Empirical Formula
H2O2
HO
C6H12O6
CH3O
C2H4O2
CH2O
CH3O
CH2O
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Ex: Determine the empirical formula for a
compound containing 2.128 g Cl and 1.203 g
Ca.
1. Find mole amounts.
2.128 g Cl x 1 mol Cl = 0.0600 mol Cl
35.45 g Cl
Steps
1. Find mole amounts.
2. Divide each mole by the smallest moles.
1.203 g Ca x 1 mol Ca = 0.0300 mol Ca
40.08 g Ca
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2. Divide each mole by the smallest mole.
Cl = 0.0600 mol Cl = 2.00 mol Cl
0.0300 mols
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A compound weighing 298.12g consists of 72.2% Magnesium & 27.8% Nitrogen by mass. What is the empirical formula?
Hint
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
Ca = 0.0300 mol Ca = 1.00 mol Ca
0.0300 mols
Ratio = 1 Ca: 2 Cl
Empirical Formula = CaCl2
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Unit 4 Notes.notebook
October 21, 2015
A compound weighing 298.12 g consists
of 72.2% magnesium and 27.8% nitrogen
by mass. What is the empirical formula?
Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g
N – (27.8%/100)*298.12 g = 82.88 g
Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole
24.3 g
N – 82.88 g * ( 1 mole ) = 5.92 mole
14.01 g
Divide by small: Mg ­ 8.86 mole/5.92 mole = 1.50
N ­ 5.92 mole/5.92 mole = 1.00
Multiply ‘til whole: Mg – 1.50 x 2 = 3.00
N – 1.00 x 2 = 2.00
Molecular Formula
The molecular formula gives the actual number of
atoms of each element in a molecular compound.
1.
2.
3.
4.
Steps
Find the empirical formula.
Calculate the Empirical Formula Mass.
Divide the molar mass by the “EFM”.
Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH2O3.
2. “EFM” = 62.03 g
3. 124.06/62.03 = 2
4. 2(CH2O3) = C2H4O6
Mg3N2
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Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The molar
mass of the compound is 92.0 g/mol.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.
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B. Divide each mole by the smallest mole.
N = 0.350 mols = 1.00 mol N
0.350 mols
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Empirical formula.
A. Find mole amounts.
4.90 g N x 1 mol N = 0.350 mol N
14.01 g N
11.2 g O x 1 mol O = 0.700 mol O
16.00 g O
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Molecular formula
= 92.0 g/mol = 2.00
Molar Mass
Emp. Formula Mass 46.01 g/mol
Molecular Formula = 2 x Emp. Formula = N2O4
O = 0.700 mols = 2.00 mol O
0.350 mols
Empirical Formula = NO2
Empirical Formula Mass = 46.01 g/mol
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Unit 4 Notes.notebook
October 21, 2015
A 528.39 g compound containing only
carbon, hydrogen, and oxygen is found to
be 48.38% carbon and 8.12% hydrogen by
mass. The molar mass of this compound is
known to be ~222.25 g/mol. What is its
molecular formula?
A 528.39 g compound containing only carbon, hydrogen,
and oxygen is found to be 48.38% carbon and 8.12%
hydrogen by mass. The molar mass of this compound is
known to be ~222.25 g/mol. What is its molecular
formula?
g C – (48.38/100)*528.39 g = 255.64 g
g H – (8.12/100)*528.39 g = 42.91 g
g O – (43.5/100)*528.39 g = 229.85 g
mole C ­ 255.64 g * ( 1 mole ) = 21.29 mol
12.01 g
mole H – 42.91 g * ( 1 mole ) = 42.49 mol
1.01 g
mole O – 229.85 g * ( 1 mole ) = 14.37 mol
16.00 g
Oct 13­8:15 AM
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?
From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O C – 21.29/14.27 = 1.49
H – 42.49/14.27 = 2.98 (esentially 3)
O – 14.27/14.27 = 1.00
C – 1.49 x 2 = 3
H – 3 x 2 = 6
O – 1 x 2 = 2
C 3 H6 O 2
Oct 13­8:15 AM
Oct 13­8:15 AM
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?
From last slide: Empirical formula = C3H6O2
“EFM” = 74.09
Molar mass = 222.25 = ~3
EFM 74.09
3(C3H6O2) = C9H18O6
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