Determining Formulae (OCR AS Chemistry A): Revision Note

Deducing molecular & empirical formulae

Deduce the molecular and empirical formula of the following compounds:

Answers:

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:

• So, the empirical formula is CO₂

Calculating empirical formula from percentages

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

Answer:

• So, the empirical formula is C₃H₄ 

Calculating molecular formula

The empirical formula of X is C₄H₁₀S and the relative molecular mass of X is 180.2

What is the molecular formula of X?

(A_r data: C = 12.0, H = 1.0, S = 32.1)

Answer:

1. Calculate relative mass of the empirical formula

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

     • Relative empirical mass = (12.0 x 4) + (1.0 x 10) + (32.1 x 1)

     • Relative empirical mass = 90.1

2. Divide relative molecular mass of X by relative empirical mass

   • Ratio between M_r of X and the M_r of the empirical formula = 180.2 / 90.1

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

3. Multiply each number of elements by the ratio

   • (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.

1. Calculate the empirical formula of X.

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

Answer:

Answer 1:

• So, the empirical formula of compound X is CH₂Cl

Answer 2:

1. Calculate relative mass of the empirical formula:

   • Relative empirical mass = (1 x C) + (2 x H) + (1 x Cl)

     • Relative empirical mass = (1 x 12.0) + (2 x 1.0) + (2 x 35.5)

     • Relative empirical mass = 49.5

2. Divide relative molecular mass of X by relative empirical mass

   • Ratio between M_r of X and the M_r of the empirical formula = 99.0/49.5

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

3. Multiply each number of elements by the ratio

   • (C₄ x 2) + (H₁₀ x 2) + (S x 2)

     • (C_{1 x 2}) + (H_{2 x 2}) + (Cl_{1 x 2}) = (C₂) + (H₄) + (Cl₂)

     • The molecular formula of X is C₂H₄Cl₂

Calculating water of crystallisation

10.0 g of hydrated copper sulfate are heated to a constant mass of 5.59 g.

Calculate the formula of the original hydrated copper sulfate.

(M_r data: CuSO₄ = 159.6, H₂O = 18.0) 

Answer:

Water of crystallisation calculations follow the same principles as empirical formula calculations:

• Start with the masses of the anhydrous salt and water lost

• Divide each mass by the Mr of the corresponding component to get moles

• Divide by the smaller number to get a whole number ratio

• Write the formula as: Salt•xH₂O, where x is the number of moles of water per mole of salt

This method gives the formula of the hydrated compound.