The empirical formula for a compound is CH2O

The empirical formula for a compound is CH2O. The compound has a molar mass of 90.1 g/mol. Determine the molecular formula for the compound.

The Correct Answer and Explanation is :

Solution: Determining the Molecular Formula

Step 1: Determine the Empirical Formula Mass

The empirical formula of the compound is CH₂O. To find its empirical formula mass, we sum the atomic masses of its constituent elements:

  • Carbon (C): 12.01 g/mol
  • Hydrogen (H): 2 × 1.008 g/mol = 2.016 g/mol
  • Oxygen (O): 16.00 g/mol

Empirical formula mass = 12.01 + 2.016 + 16.00 = 30.03 g/mol

Step 2: Find the Molecular Formula Factor

To determine the molecular formula, we use the given molar mass of the compound (90.1 g/mol) and divide it by the empirical formula mass:

[
\frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{90.1}{30.03} \approx 3
]

This means that the molecular formula is 3 times the empirical formula.

Step 3: Determine the Molecular Formula

Now, we multiply each subscript in CH₂O by 3:

  • Carbon: C × 3 = C₃
  • Hydrogen: H₂ × 3 = H₆
  • Oxygen: O × 3 = O₃

Thus, the molecular formula of the compound is C₃H₆O₃.


Explanation

The empirical formula represents the simplest whole-number ratio of atoms in a compound. However, the molecular formula provides the actual number of atoms in a molecule. The molecular formula is a multiple of the empirical formula. In this case, the empirical formula CH₂O has a mass of 30.03 g/mol, and the given molecular mass is 90.1 g/mol. Since 90.1 / 30.03 ≈ 3, we scale up the empirical formula by a factor of 3, leading to the molecular formula C₃H₆O₃. This formula corresponds to molecules like glyceraldehyde or lactic acid, which are biologically significant compounds.

Now, let’s generate an image representing this molecular structure.

Here is a 3D molecular model of C₃H₆O₃, illustrating the atomic arrangement with Carbon (black), Hydrogen (white), and Oxygen (red). The structure represents a common molecular form like lactic acid. Let me know if you need any modifications or explanations!

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