Empirical Formulae
Amount of Substance Worksheet
Easy Empirical Formulae Questions
1. A substance was analysed and found to contain the following percentages of 271.2g of carbon, 54.2g of hydrogen and 144.6g of oxygen. The substance has an Mr of 104.
a) What is the molecular formula of the substance?
Step 1: Calculate Moles
C: 271.2 / 12.0 = 22.6
H: 54.2 / 1.0 = 54.2
O: 144.6 / 16.0 = 9.0375
H: 54.2 / 1.0 = 54.2
O: 144.6 / 16.0 = 9.0375
Step 2: Find Ratio (divide by smallest: 9.0375)
C: 22.6 / 9.0375 = 2.5
H: 54.2 / 9.0375 = 6.0
O: 1.0
Ratio = 2.5 : 6 : 1 → Multiply by 2 → 5 : 12 : 2
H: 54.2 / 9.0375 = 6.0
O: 1.0
Ratio = 2.5 : 6 : 1 → Multiply by 2 → 5 : 12 : 2
Step 3: Determine Molecular Formula
Empirical Formula = C5H12O2
Empirical Mass = (5×12) + 12 + (2×16) = 104
Since Empirical Mass (104) matches Mr (104):
Formula = C5H12O2
Empirical Mass = (5×12) + 12 + (2×16) = 104
Since Empirical Mass (104) matches Mr (104):
Formula = C5H12O2
b) Suggest a display formula for the substance.
Any pentanediol isomer is acceptable, for example: Pentane-1,5-diol
2. A substance was analysed and found to contain 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. The substance has an Mr of 88.
a) What is the molecular formula of the substance?
Step 1: Calculate Moles
C: 54.5 / 12 = 4.54
H: 9.1 / 1 = 9.1
O: 36.4 / 16 = 2.275
H: 9.1 / 1 = 9.1
O: 36.4 / 16 = 2.275
Step 2: Ratio (divide by 2.275)
C: 2
H: 4
O: 1
Empirical Formula: C2H4O (Mass = 24+4+16 = 44)
H: 4
O: 1
Empirical Formula: C2H4O (Mass = 24+4+16 = 44)
Step 3: Molecular Formula
Mr = 88. Ratio = 88 / 44 = 2.
Multiply empirical by 2.
Formula = C4H8O2
Multiply empirical by 2.
Formula = C4H8O2
b) Suggest a display formula for the substance.
Could be a carboxylic acid or an ester.
Example: Butanoic Acid or Ethyl Ethanoate.
Example: Butanoic Acid or Ethyl Ethanoate.
3. A substance was analysed and found to contain 68.64g of carbon, 10.01g of hydrogen, 22.88g of oxygen and 20.02g of nitrogen. The substance has an Mr of 85.
a) What is the molecular formula of the substance?
Step 1: Calculate Moles
C: 68.64 / 12 = 5.72
H: 10.01 / 1 = 10.01
O: 22.88 / 16 = 1.43
N: 20.02 / 14 = 1.43
H: 10.01 / 1 = 10.01
O: 22.88 / 16 = 1.43
N: 20.02 / 14 = 1.43
Step 2: Ratio (divide by 1.43)
C: 4
H: 7
O: 1
N: 1
Empirical: C4H7ON (Mass = 48+7+16+14 = 85)
H: 7
O: 1
N: 1
Empirical: C4H7ON (Mass = 48+7+16+14 = 85)
Step 3: Molecular Formula
Since empirical mass matches Mr (85):
Formula = C4H7ON
Formula = C4H7ON
b) Suggest a display formula for the substance.
A likely candidate is 2-hydroxybutane nitrile or cyclobutanamide. The images for the compounds are:
4. A substance was analysed and was found to contain 36.1% carbon, 5.3% hydrogen, 48.1% oxygen and 10.5% nitrogen. It has an Mr of 133.
a) What is the molecular formula of the substance?
Step 1: Calculate Moles
C: 36.1 / 12 = 3.008
H: 5.3 / 1 = 5.3
O: 48.1 / 16 = 3.006
N: 10.5 / 14 = 0.75
H: 5.3 / 1 = 5.3
O: 48.1 / 16 = 3.006
N: 10.5 / 14 = 0.75
Step 2: Ratio (divide by 0.75)
C: 4
H: 7
O: 4
N: 1
Empirical: C4H7O4N (Mass = 48+7+64+14 = 133)
H: 7
O: 4
N: 1
Empirical: C4H7O4N (Mass = 48+7+64+14 = 133)
Step 3: Molecular Formula
Matches Mr (133):
Formula = C4H7NO4
Formula = C4H7NO4
b) Which amino acid is this?
Aspartic Acid
5. Fumaric acid has the formula HO2CCH=CHCO2H.
a) What is the empirical formula for fumaric acid?
Molecular Formula: C4H4O4
Divide by common factor (4):
CHO
Divide by common factor (4):
CHO
b) If you analysed a sample of fumaric acid, what percentage mass of each element would you have?
Mr = (4×12) + 4 + (4×16) = 116
% C = (48 / 116) × 100 = 41.4%
% H = (4 / 116) × 100 = 3.4%
% O = (64 / 116) × 100 = 55.2%
% C = (48 / 116) × 100 = 41.4%
% H = (4 / 116) × 100 = 3.4%
% O = (64 / 116) × 100 = 55.2%
Hard Empirical Formula Questions
1. An organic compound was found to contain 37.5g of carbon, 4.5g of hydrogen, and 25.0g of oxygen. The molecular ion peak in mass spectrographic analysis was found to be 86. The compound was tested, and was found not to react with Tollens’ reagent, acidified dichromate, or sodium carbonate. However, when it was treated with NaBH4 it could then react with acidified dichromate. Identify the compound, and draw its skeletal formula.
Working Out
Step 1: Determine Formula
Moles: C(3.125), H(4.5), O(1.56)
Ratio: C(2), H(2.88 ≈ 3), O(1) → Empirical C2H3O (Mass 43)
Mr = 86 (Molecular Ion Peak) → Ratio = 2
Molecular Formula: C4H6O2
Ratio: C(2), H(2.88 ≈ 3), O(1) → Empirical C2H3O (Mass 43)
Mr = 86 (Molecular Ion Peak) → Ratio = 2
Molecular Formula: C4H6O2
Step 2: Deduce Structure
– No Tollens: Not an aldehyde.
– No Dichromate: Not a primary/secondary alcohol.
– No Na2CO3: Not a carboxylic acid.
– Reacts with dichromate AFTER NaBH4 reduction: Contains a Ketone.
Identity: Butane-2,3-dione (CH3COCOCH3)
– No Dichromate: Not a primary/secondary alcohol.
– No Na2CO3: Not a carboxylic acid.
– Reacts with dichromate AFTER NaBH4 reduction: Contains a Ketone.
Identity: Butane-2,3-dione (CH3COCOCH3)
2. An organic compound was found to contain 124.2g of Carbon, 17.7g of hydrogen and 28.1g of oxygen. The Mr was found to be 98. When mixed with bromine water it did not react. It also did not react with acidified potassium dichromate, even under reflux. Name the compound and draw its formula.
Working Out
Step 1: Determine Formula
Moles: C(10.35), H(17.7), O(1.76)
Ratio (div by 1.76): C(5.9 ≈ 6), H(10), O(1)
Formula: C6H10O (Mass 98 matches Mr)
Ratio (div by 1.76): C(5.9 ≈ 6), H(10), O(1)
Formula: C6H10O (Mass 98 matches Mr)
Step 2: Deduce Structure
– No C=C double bond (Bromine water).
– Not an alcohol/aldehyde (Dichromate).
– Cyclic Ketone fits.
Identity: Cyclohexanone
– Not an alcohol/aldehyde (Dichromate).
– Cyclic Ketone fits.
Identity: Cyclohexanone