Question

A compound with the empirical formula \(\mathrm{CH}_{4} \mathrm{O}\) was found in a subsequent experiment to have a molar mass of approximately \(192 \mathrm{g}\). What is the molecular formula of the compound?

Short answer

The molecular formula of the compound is \(\mathrm{C}_{6} \mathrm{H}_{24} \mathrm{O}_{6}\).

Step-by-step solution

Calculate the molar mass of the empirical formula

To calculate the molar mass of CH4O, we need to find the atomic masses of the elements and add them up, considering the number of atoms of each element in the empirical formula. Carbon (C): 1 atom × 12.01 g/mol = 12.01 g/mol Hydrogen (H): 4 atoms × 1.01 g/mol = 4.04 g/mol Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol So, the molar mass of the empirical formula (CH4O) = 12.01 g/mol + 4.04 g/mol + 16.00 g/mol = 32.05 g/mol.

Calculate the whole number ratio

Now we need to find the whole number ratio between the molecular formula molar mass (192 g) and the empirical formula molar mass (32.05 g). Whole number ratio = (Molecular formula molar mass) / (Empirical formula molar mass) Whole number ratio = 192 g / 32.05 g ≈ 6

Find the molecular formula

Now that we have found the whole number ratio (6), we can determine the molecular formula by multiplying the empirical formula (CH4O) by this ratio: C(1 × 6)H(4 × 6)O(1 × 6) = C6H24O6 So, the molecular formula of the compound is \(\mathrm{C}_{6} \mathrm{H}_{24} \mathrm{O}_{6}\).

Key concepts

Empirical Formula

The empirical formula of a compound gives the simplest whole number ratio of atoms present. It's derived from elemental analysis and provides the minimum basic unit, representing how different atoms are combined in a compound. To find the empirical formula, consider the relative amounts of each element present.
For instance, in the compound with the empirical formula \(\mathrm{CH}_{4}\mathrm{O}\), there is one atom of Carbon (C), four atoms of Hydrogen (H), and one atom of Oxygen (O). This simple formula doesn't necessarily reflect the actual numbers of atoms in a molecule, but rather the smallest ratio in which these atoms combine in the compound.
To use this empirical formula to find a molecular formula, one needs more information, such as the total molar mass, which gives clues about how many of these empirical units are actually combined in the real molecular structure.

Molar Mass Calculation

Molar mass is a physical property that allows for the calculation of the mass of one mole of a given compound. To calculate the molar mass of the empirical formula \(\mathrm{CH}_{4}\mathrm{O}\), we sum up the atomic masses in proportion to their quantity in the formula.

• Carbon (C): One atom, mass = 12.01 g/mol
• Hydrogen (H): Four atoms, mass = 4.04 g/mol (1.01 g/mol per H atom)
• Oxygen (O): One atom, mass = 16.00 g/mol

Adding these, the molar mass of \(\mathrm{CH}_{4}\mathrm{O}\) is 32.05 g/mol. This calculation is a fundamental step in determining how many empirical units fit into the molecular formula, especially when compared against the compound's known molar mass of 192 g/mol.

Chemical Compound Analysis

Chemical compound analysis often involves determining not just the composition of elements but also the true molecular structure. With both the empirical formula molar mass and the compound's total molar mass, one can determine a molecular formula.
To find the exact molecular formula, we first determine the ratio between the actual molar mass and the empirical formula mass. For example, given a molar mass of 192 g/mol for an unknown compound and an empirical formula molar mass of 32.05 g/mol, computing the ratio gives:
\[\text{Whole number ratio} = \frac{192}{32.05} \approx 6\]This tells us the molecular formula includes 6 times the number of atoms for each element in the empirical formula. Thus, the molecular formula derived from \(\mathrm{CH}_{4}\mathrm{O}\) becomes \(\mathrm{C}_{6}\mathrm{H}_{24}\mathrm{O}_{6}\). Understanding this process is pivotal in analyzing and predicting chemical structure based on empirical data.