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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

1st: 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

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

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

3.390/3.390 = 1 mol of C

5.04/3.390 = 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.

3.390/3.390 = 1 mol of C X 2 =2 mol of C

5.04/3.390 = 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% B

5. 32.4% Na, 22.5% S, and 45.1% O

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: n= Experimentally determined Molar Mass of the compound

Mass of Empirical formula of the compound

2nd: 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

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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