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The molecular mass of a compound is '60'. If it contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen, what is the molecular formula?
C ₂ H ₄ O ₂
C ₃ H ₆ O ₃
CH ₂ O
C ₄ H ₈ O ₄
C ₂ H ₄ O ₂
To determine the molecular formula, we first calculate the empirical formula by dividing the mass percentages of each element by their respective atomic masses. The empirical formula of this compound is CH2O (ratio 1:2:1), which has a mass of 30. Since the given molecular mass is 60, we multiply the empirical formula by n=3060=2, yielding C2H4O2.
To determine the molecular formula, we first calculate the empirical formula by dividing the mass percentages of each element by their respective atomic masses. The empirical formula of this compound is CH2O (ratio 1:2:1), which has a mass of 30. Since the given molecular mass is 60, we multiply the empirical formula by n=3060=2, yielding C2H4O2.
Think of the empirical formula as the simplest recipe for a batch of cookies, while the molecular formula tells you exactly how many batches (n) you need to make to reach the target total weight of the final product.
Remember 'EM-N-MO': Empirical Mass first, then N factor, then Molecular formula.
n=EmpiricalFormulaMassMolecularMass — Scaling factor
Molecular Formula=n×EmpiricalFormula — Conversion formula
The molecular formula is obtained by multiplying the empirical formula by an integer n, where n=EmpiricalFormulaMassMolecularMass. First, we find the molar ratio of atoms: C (40/12=3.33), H (6.7/1=6.7), and O (53.3/16=3.33). Dividing these by the smallest value (3.33) gives a whole number ratio of 1:2:1, resulting in the empirical unit CH2O.
The empirical formula represents the simplest whole-number ratio of atoms in a compound.
The molecular formula represents the actual number of atoms of each element in one molecule.
For C2H4O2, the molar mass calculation is (2×12)+(4×1)+(2×16)=60 g/mol.
Different compounds can share the same empirical formula but have different molecular masses.
Allows identification of chemical composition from percentage data.
Distinguishes between substances with identical empirical ratios but different molar masses.
Requires accurate experimental percentage composition data.
Empirical formulas provide no information about molecular structure or isomerism.
Analytical chemistry for identifying unknown organic compounds.
Forensic analysis in determining the nature of substances.
Atomic masses used: C=12, H=1, O=16.
Option B (C3H6O3) has a mass of 90; Option C (CH2O) is the empirical formula with mass 30; Option D (C4H8O4) has a mass of 120.
A is correct — The compound with an empirical formula of CH2O and a molar mass of 60 must be C2H4O2 (Acetic acid).
Always verify if the calculated empirical formula mass divides the molecular mass into a clean integer; if not, re-check your percentage-to-mole conversions.
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