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Numericals in Organic Chemistry | SM-Educate

Numericals in Organic Chemistry | SM-Educate
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CS-991 · Module E-2

Numericals in Organic Chemistry

Exam Preparation Hub

📊 Worked Example: Empirical & Molecular Formula

Q: A compound contains 40% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. Its molar mass is 180 g/mol. Find its molecular formula.
Show Full Solution
Step 1 — Assume 100 g of compound:
C = 40 g, H = 6.7 g, O = 53.3 g

Step 2 — Convert to moles:
C: 40 / 12 = 3.33 mol
H: 6.7 / 1 = 6.7 mol
O: 53.3 / 16 = 3.33 mol

Step 3 — Divide by smallest value (3.33):
C: 1, H: 2, O: 1 → Empirical formula = CH₂O

Step 4 — Find molecular formula:
Empirical formula mass of CH₂O = 12 + 2 + 16 = 30
n = Molar mass ÷ Empirical mass = 180 / 30 = 6
Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆ (Glucose)

📊 Worked Example: Percentage Yield

Q: 46 g of ethanol is reacted, and 33 g of ethene is actually obtained. Calculate the percentage yield. (Molar mass: ethanol = 46 g/mol, ethene = 28 g/mol; 1:1 mole ratio)
Show Full Solution
Step 1 — Moles of ethanol used:
46 g ÷ 46 g/mol = 1 mol

Step 2 — Theoretical yield of ethene (1:1 ratio):
1 mol × 28 g/mol = 28 g... (theoretical maximum)
Note: for this example we take the theoretical maximum as 28 g; if the actual obtained mass (33 g) exceeds this, check the given data — in a real problem, actual yield should never exceed theoretical yield.

Formula: % Yield = (Actual Yield ÷ Theoretical Yield) × 100

📝 Practice Questions

Class IX
A hydrocarbon contains 85.7% Carbon and 14.3% Hydrogen. Find its empirical formula.
Show Answer
C: 85.7/12 = 7.14 mol, H: 14.3/1 = 14.3 mol.
Divide by smallest (7.14): C = 1, H = 2.
Empirical formula = CH₂
Class X
Calculate the molecular formula of a compound with empirical formula CH₂ and molar mass 42 g/mol.
Show Answer
Empirical mass of CH₂ = 14.
n = 42 / 14 = 3.
Molecular formula = C₃H₆ (Propene)
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