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A compound is found to be 38.43% c, 4.85%H and 56.72%Cl with a molar mass of 250 g/mol. Determine the molecular formula

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Answer:

∴molecular formula => C₁₂H₁₈O₆

Step-by-step explanation:

A good calculation sequence to follow for empirical formula is ...

% => grams => moles => mole ratio => reduce mole ratio => empirical ratio

then determine whole number multiple for molecular formula using ...

molecular weight = (Whole Number Multiple)(empirical formula weight)

Applying to your posted problem ...

C: 38.43% => 38.43g => (38.43g/12g·mol⁻¹) => 3.2025 moles C

H: 4.85% => 4.85g => (4.85g/1g·mol⁻¹) => 4.85 moles H

Cl: 56.72% => 56.72g => (56.72g/35g·mol⁻¹) => 1.6206 moles Cl

mole ratio C:H:Cl => 3.2025 : 4.85 : 1.6206

reduced mole ratio => divide mole ratio values by the smaller mole value in the set.

That is, reduced mole ratio = empirical formula ratio ...

=> (3.2025/1.6206):(4.85/1.6206):(1.6206/1.6206) => 2 : 3 : 1

∴ empirical formula => C₂H₃O

empirical formula weight = 2(12) + 3(1) + 1(16) = 43g/mol

Whole no multiple = mol wt / emp wt = 250g/mol / 43g/mol ≅ 6

∴molecular formula => C₁₂H₁₈O₆

User Mariska
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