Question

A compound with the empirical formula \(\mathrm{CH}_{2}\) was found to have a molar mass of approximately \(84 \mathrm{g}\) What is the molecular formula of the compound?

Short answer

The molecular formula of the given compound with the empirical formula \(\mathrm{CH}_{2}\) and a molar mass of approximately \(84 \mathrm{g}\) is \(\mathrm{C}_{6}\mathrm{H}_{12}\).

Step-by-step solution

Determine the empirical formula molar mass

First, we need to find the molar mass of the empirical formula by adding the molar masses of carbon (C) and hydrogen (H) atoms in the \(\mathrm{CH}_{2}\) empirical formula: Empirical formula molar mass = (1 × molar mass of C) + (2 × molar mass of H) Using the molar mass values: C=12.01 g/mol and H=1.01 g/mol, we get Empirical formula molar mass = (1 × 12.01) + (2 × 1.01) = 14.03 g/mol

Calculate the "n" factor

Now we need to find the "n" factor by dividing the molar mass of the molecular formula by the molar mass of the empirical formula: n = (Molecular formula molar mass) / (Empirical formula molar mass) Given that the molecular formula molar mass is approximately 84 g/mol, we get n = (84 g/mol) / (14.03 g/mol) ≈ 6 Since n is approximately 6, the molecular formula has six times the relative amounts of atoms as the empirical formula.

Determine the molecular formula

Finally, multiply the subscripts in the empirical formula by the "n" factor to obtain the molecular formula: Molecular formula = \(\mathrm{C}(\mathrm{1}\times n) \mathrm{H}(\mathrm{2} \times n)\) Molecular formula = \(\mathrm{C}(\mathrm{1}\times 6) \mathrm{H}(\mathrm{2} \times 6)\) Molecular formula = \(\mathrm{C}_{6}\mathrm{H}_{12}\) The molecular formula of the given compound is \(\mathrm{C}_{6}\mathrm{H}_{12}\).

Key concepts

Understanding Empirical Formulas

An empirical formula represents the simplest whole-number ratio of atoms in a chemical compound. It gives insight into the basic composition of the compound without specifying the exact number of each atom present. For instance, the compound \( \mathrm{CH}_2 \) is an empirical formula, indicating a ratio of one carbon atom to two hydrogen atoms. However, this does not imply the presence of only three atoms in the entire compound.

• It provides a foundational understanding of the compound's composition.
• Useful for determining the molecular formula when the molar mass is known.
• Not indicative of the compound's molecular structure or geometry.

When solving problems involving empirical formulas, the primary goal is often to determine the molecular formula. This requires knowing the empirical formula itself and additional information such as the compound's molar mass.

Steps in Molar Mass Calculation

Calculating the molar mass is essential for converting the empirical formula to the molecular formula. Molar mass is the mass of one mole of a compound, typically expressed in grams per mole (g/mol). For empirical formula calculations, the molar masses of individual elements are used. As shown in the solution:

• Carbon (C) has a molar mass of 12.01 g/mol.
• Hydrogen (H) has a molar mass of 1.01 g/mol.

To find the molar mass of the empirical formula \( \mathrm{CH}_2 \), you add the molar mass of one carbon atom to that of two hydrogen atoms, resulting in:
\( 12.01 + (2 \times 1.01) = 14.03 \).
This empirical formula molar mass helps us calculate the 'n' factor, necessary for scaling the empirical formula to derive the molecular formula.

Exploring Chemical Compounds and Their Formulas

Chemical compounds can have both empirical and molecular formulas, each serving a different purpose. While the empirical formula is useful for representing the simplest atom ratio, the molecular formula provides the exact number of each atom in a compound. In the given exercise, once we determined the empirical formula's molar mass, it was possible to find the molecular formula by:

• Calculating the 'n' factor, which is the ratio of the molar mass of the compound to the empirical formula molar mass.
• Multiplying the atom ratio by this 'n' factor, scaling the empirical formula to a larger unit.

In this instance, the initial empirical formula \( \mathrm{CH}_2 \) was scaled by the factor 6 to obtain the molecular formula \( \mathrm{C}_6\mathrm{H}_{12} \). Hence, knowing these forms allows chemists to convey detailed information about the composition and structure of compounds.