Download Empirical and Molecular Formulas Empirical Formula: The smallest

Empirical and Molecular Formulas
Empirical Formula: The smallest whole number mole ratio of elements in a compound
 Assume that each percent by mass represents the mass of the element in a 100.00-g sample
How to Calculate Empirical Formula
1 : Calculate the % composition of each element (If not given) and change % into grams
2nd: Calculate Molar Mass of each element
3rd: Determine simplest whole # ratio
4th: Write Empirical Formula
st
Example #1
The mass of C is 48.64g, the mass of H is 8.16g, and the mass of O is 43.20g. Find the Empirical Formula
(EF).
Step 1: Find Molar mass of each element
48.64 g of C X 1 mol of C/12.01g of C (atomic mass) =4.050 mol of C
8.16 g of H X 1 mol of H/1.008g of H (atomic mass) =8.10 mol of H
43.20 g of O X 1 mol of O/16.00g of O (atomic mass) =2.700 mol of O
Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1
4.050/2.7 = 1.5 mol of C
8.10/2.7 = 3 mol of H
2.7/2.7 = 1 mol of O
Then look at the three numbers of moles and determine the lowest number they can be multiplied by to
get all whole numbers. In this case the number is 2.
4.050/2.7 = 1.5 mol of C x2=3 mol of C
8.10/2.7 = 3 mol of H X 2=6 mol of H
2.7/2.7 = 1 mol of O X2=2 mol of O
Step 3: Create Empirical Formula from moles in Step two
C3H6O2
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Example #2
Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of
40.68% C, 5.08 H, and 54.24% oxygen and has a molar mass of 118.1g/mol. Determine the
empirical formula for succinic acid.
Step 1: Determine molar mass.
1st: Convert percentages into grams of elements.
2nd: Use molar mass formula to find moles.
40.68 g of C X 1mol C/12.01g (atomic mass) of C= 3.390 mol of C
5.08 g of H X 1mol H/1.008g (atomic mass) of H= 5.04 mol of H
54.24g of O X 1mol O/16.00g (atomic mass) of O= 3.390 mol of O
Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1
4.068/3.390 = 1 mol of C
5.08/3.390 = 1.5 mol of H
3.390/3.390 = 1 mol of O
Then look at the three numbers of moles and determine the lowest number they can be multiplied by to
get all whole numbers. In this case the number is 2.
4.068/3.390 = 1 mol of C X 2 =2 mol of C
5.08/3.390 = 1.5 mol of H X 2=3 mol of H
3.390/3.390 = 1 mol of O X 2 =2 mol of O
Step 3: Create Empirical Formula from moles in Step two
C2H3O2
Practice Problems for Empirical Formula
Directions: What is the empirical formula of the compounds below?
1. 75% C, 25% H
2. 52.7% K, 47.3% Cl
3. 22.1% Al, 25.4% P, and 52.5% O
4. 13% Mg, 87 Br
5. 32.4% Na, 22.5% S, and 45.1% O
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Molecular Formula
Molecular Formula: A formula that states that actual number of atoms of each element in one
compound or formula unit of the substance.
How to Calculate Molecular Formula:
1st:
2nd:
n=
Experimentally determined Molar Mass of the compound
Mass of Empirical formula of the compound
Multiple the subscripts in the empirical formula by n
Example #1
Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of
40.68% C, 5.08 % H, and 54.24% oxygen and has a molar mass of 118.1g/mol. Determine the
Molecular formula for succinic acid. Empirical Formula for Succinic Acid is: C2H3O2 and the molar
mass of Succinic acid is 59.04 g/mol
Step 1: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical
formula to determine n
n=
Experimental Molar mass of succinic acid
Molar mass of C2H3O2
n= 118.1g/mol
59.04 g/mol
n= 2.000
Step 2: Multiple subscripts by 2 (n)
C4H6O4
Example #2
The empirical formula of a compound is NO2. Its molecular mass is 92g/mol. What is the
molecular formula?
Step 1: Determine the empirical formula molar mass
N has 1 atom 1 X 14.0067= 14.0067g/mol
O has 2 atoms 2X 16.00=
32.00g/mol
46.0067 g/mol
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Step 2: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical
formula to determine n
n=
Experimental Molar mass of NO2
Molar mass of NO2
n= 92g/mol
46.0067 g/mol
n= 2.000
Step 2: Multiple subscripts by 2 (n)
N2O4
Practice problems for Molecular Formula
1) The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is the
molecular formula?
2) A compound is found to be 40% C, 6.7 H and 53.5%O. Its Molecular mass is 60 g/mol. What is
its molecular formula?
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