Understanding Molecular Formulas from Molar Mass and Empirical Formulas

When working with chemical compounds, it's essential to understand the relationship between the empirical formula and the molecular formula to accurately determine the structure and composition of the compounds. In this article, we'll walk through the process of finding the molecular formula for a compound given its molar mass and empirical formula. We'll specifically focus on a compound with a molar mass of 30.0 g/mol and an empirical formula of CH3.

Cracking the Code: Decoding the Empirical Formula and Molar Mass

To determine the molecular formula from the empirical formula and the molar mass, we need to follow a systematic approach. Let's break down the steps:

Calculate the Molar Mass of the Empirical Formula CH3: Determine the Ratio of the Molar Mass of the Compound to the Molar Mass of the Empirical Formula: Use the Ratio to Determine the Molecular Formula:

In the given example, the empirical formula is CH3. Let's go through each step in detail.

Step 1: Calculate the Molar Mass of the Empirical Formula CH3

Molar Mass of CH3:

nCarbon (C) has a molar mass of about 12.01 g/mol. nHydrogen (H) has a molar mass of about 1.008 g/mol.

The molar mass of CH3 is calculated as:

Molar Mass of CH3 12.01 g/mol 3 × 1.008 g/mol 15.034 g/mol

Step 2: Determine the Ratio of the Molar Mass of the Compound to the Molar Mass of the Empirical Formula

Molar Mass of the Compound 30.0 g/mol

Molar Mass of CH3 15.034 g/mol

Calculate the ratio:

Ratio Molar Mass of Compound / Molar Mass of Empirical Formula 30.0 g/mol / 15.034 g/mol ≈ 1.996

This ratio is approximately 2.

Step 3: Use the Ratio to Determine the Molecular Formula

Since the ratio is approximately 2, multiply the subscripts in the empirical formula by 2:

Molecular Formula C1×2H3×2 C2H6

Therefore, the molecular formula of the compound is CH6, which is actually ethane (C2H6).

Additional Examples and Calculations

Let's look at a few additional examples and calculations to reinforce the concept.

Example 1

Molar mass 30 g/mol

Molecular mass 30 u

Empirical formula CH3 mass 12 3 × 1 15 u

Molecular mass/Empirical mass 30 u / 15 u 2

So, molecular formula Empirical formula × 2

CH3 × 2 C2H6

Example 2

Ratio 30.0 g / empirical formula mass of CH3

30.0 g / 15 g 2

The molecular formula is: C2H6

Example 3

Calculate the Molar Mass of CH3:

Molecular formula empirical formula × 2

empirical formula mass × 2

Divide 30g by the above answer. You should get an integer.

Multiply the empirical formula CH3 by that integer. Be sure to multiply each atom by the integer.

Empirical formula CH3

Empirical formula mass 12 3 × 1 15 u

Molar mass 30 g/mol

Molecular mass 30 u

Molecular mass/Empirical formula mass 30 / 15 2

Molecular formula empirical formula × 2

empirical formula mass × 2

15 u × 2 C2H6

Conclusion

By following these steps, you can accurately determine the molecular formula of a compound from its molar mass and empirical formula. Understanding this process is crucial for chemists and chemical engineers in fields such as materials science, pharmaceuticals, and environmental chemistry. Remember that the molecular formula provides a complete account of the types and numbers of atoms in each molecule, while the empirical formula gives the simplest whole-number ratio of these atoms.