Empirical & Molecular Formula (AQA AS Chemistry)

Calculating empirical formula from mass

Determine the empirical formula of a compound that contains 2.72 g of carbon and 7.28 g of oxygen.

Answer:

Elements	Carbon	Oxygen
Mass of each element (g)	2.72	7.28
Atomic mass	12.0	16.0
Moles = mass / Ar	= 0.227	= 0.455
Ratio (divide by smallest value)	= 1	= 2

• So, the empirical formula of the compound is CO₂

Calculating empirical formula from percentage

Determine the empirical formula of a hydrocarbon that contains 90.0% carbon and 10.0% hydrogen.

Answer:

Elements	Carbon	Hydrogen
Mass of each element (g)	90.0	10.0
Atomic mass	12.0	1.0
Moles = mass / Ar	= 7.5	= 10.0
Ratio (divide by smallest value)	= 1	= 1.33
Convert to whole number ratio (x3 for this example)	1 x 3 = 3	1.33 x 3 = 4

• So, the empirical formula of the compound is  C₃H₄ 

Calculating molecular formula

The empirical formula of X is C₄H₁₀S and the relative molecular mass of X is 180

What is the molecular formula of X?

(A_r data: C = 12, H = 1, S = 32)

Answer:

Step 1: Calculate relative mass of the empirical formula

• Relative empirical mass = (C x 4) + (H x 10) + (S x 1)
• Relative empirical mass = (12 x 4) + (1 x 10) + (32 x 1)
• Relative empirical mass = 90

Step 2: Divide relative formula mass of X by relative empirical mass

• Ratio between M_r of X and the M_r of the empirical formula = 180/90
• Ratio between M_r of X and the M_r of the empirical formula = 2

Step 3: Multiply each number of elements by 2

• (C₄ x 2) + (H₁₀ x 2) + (S x 2)     =    (C₈) + (H₂₀) + (S₂)
• Molecular Formula of X is C₈H₂₀S₂

Calculating empirical formula and molecular formula

Analysis of a compound X shows that it contains 24.2 % by mass of carbon, 4.1 % by mass of hydrogen and 71.7% by mass of chlorine.

Calculate the empirical formula of X.

Use this empirical formula and the relative molecular mass of X (M_r = 99.0) to calculate the molecular formula of X.

Answer:

Elements	Carbon	Hydrogen	Chlorine
Value (g or %)	24.2	4.1	71.7
Atomic mass	12.0	1.0	35.5
Moles = mass / Ar	= 2.02	= 4.1	= 2.02
Ratio (divide by smallest)	= 1	= 2	= 1

• So, the empirical formula of compound X is CH₂Cl
• The relative formula mass of the empirical formula is:
  • Relative formula mass = (1 x C) + (2 x H) + (1 x Cl)
  • Relative formula mass = (1 x 12.0) + (2 x 1.0) + (2 x 35.5)
  • Relative formula mass = 49.5
• Divide the relative formula mass of X by the relative formula mass of the empirical formula
  • Ratio between M_r of X and the M_r of the empirical formula = 99.0/45.9
  • Ratio between M_r of X and the M_r of the empirical formula = 2
• Multiply each number of elements by 2
  • (C_{1 x 2}) + (H_{2 x 2}) + (Cl_{1 x 2}) = (C₂) + (H₄) + (Cl₂)
  • The molecular formula of X is C₂H₄Cl₂