Question

The empirical formula of a compound is \(\mathrm{CH}_{2}\). One mole of the compound has a mass of \(42 \mathrm{~g}\), its molecular formula is: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

Short answer

None of the provided options match the calculated molecular formula, \(\mathrm{C}_3\mathrm{H}_6\).

Step-by-step solution

Determine the Molar Mass of the Empirical Formula

First, calculate the molar mass of the empirical formula, \(\mathrm{CH}_2\). The atomic mass of carbon (C) is \(12\, \mathrm{g/mol}\) and that of hydrogen (H) is \(1\, \mathrm{g/mol}\). Thus, the molar mass of \(\mathrm{CH}_2\) is: \[ 12\, \mathrm{g/mol} + 2 \times 1\, \mathrm{g/mol} = 14\, \mathrm{g/mol} \]

Find the Multiplication Factor

To find the actual molecular formula, determine how many empirical unit masses are in the molar mass of the compound. Divide the molar mass of the compound (\(42\, \mathrm{g/mol}\)) by the molar mass of the empirical formula (\(14\, \mathrm{g/mol}\)): \[ \text{Factor} = \frac{42\, \mathrm{g/mol}}{14\, \mathrm{g/mol}} = 3 \]

Calculate the Molecular Formula

Multiply the subscripts in \(\mathrm{CH}_2\) by the factor obtained. Since the factor is 3, the molecular formula is: \[ \mathrm{C}_{1 imes 3} \mathrm{H}_{2 imes 3} = \mathrm{C}_3\mathrm{H}_6 \]

Validate Molecular Formula Against Options

Check which option matches the calculated molecular formula. The calculated molecular formula \(\mathrm{C}_3\mathrm{H}_6\) corresponds to none of the provided choices. It appears there might be a mistake in the choices provided as no option matches. However, a close substitute might be re-evaluated for error in either the option or calculation.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is not necessarily the same as the molecular formula, but it provides a basic framework of how many atoms from each element combine to form a molecule. To understand an empirical formula, consider it as a reduced recipe for a compound, showcasing only the proportions between elements, rather than the exact number of atoms.

For example, if a compound has an empirical formula of \( \mathrm{CH}_2 \), it means there is one carbon atom for every two hydrogen atoms. This ratio does not indicate the actual number of atoms present in the molecule, but rather the proportionate relationship between them. To obtain an empirical formula, you usually take the following steps:

• Determine the percent composition of each element in the compound.
• Convert the percentages to grams (assuming a 100 g sample works well).
• Convert the grams to moles by dividing by each element’s atomic mass.
• Find the smallest whole number ratio by dividing all moles by the smallest number of moles calculated.

Understanding the empirical formula is crucial as it sets the stage for deducing the molecular formula, especially in chemical problem-solving contexts.

Molar Mass Calculation

Molar mass is essential for linking the microscopic world of atoms to the macroscropic world we interact with. To calculate molar mass, simply sum the atomic masses of all the atoms in the compound's formula found on the periodic table. Each element's atomic mass contributes to the total molar mass, represented in grams per mole (g/mol). This enables us to convert between the number of molecules present and the mass of the substance.

For instance, with \( \mathrm{CH}_2 \), calculate the molar mass as follows:

• Carbon (C) has an atomic mass of 12 g/mol.
• Hydrogen (H), having 2 atoms, contributes \(2 \times 1\) g/mol.

The total molar mass of \( \mathrm{CH}_2 \) becomes \(12 + 2 \times 1 = 14\) g/mol.

Conducting molar mass calculations correctly is fundamental before progressing to solve for more complex molecular formulas, and it is especially crucial for balancing chemical equations, interpreting stoichiometric coefficients, and making conversions in chemical reactions.

Chemical Problem Solving

Chemical problem-solving involves systematic steps to solve questions involving compositions, structures, and properties of substances. A classic approach in deriving a molecular formula from an empirical formula requires logical and thorough analysis, usually performed as follows:

First, determine the empirical formula as a precursor based on the stoichiometric ratios within the compound. Next, calculate the molar mass of the empirical formula. Subsequently, find the multiplication factor (or integer multiplier) by dividing the compound's molar mass by the empirical formula's molar mass. The integer obtained helps convert the empirical formula to the molecular formula.

For example, if the empirical formula \( \mathrm{CH}_2 \) has a molar mass of 14 g/mol and the compound's mass is 42 g/mol, divide to get the factor:\[ \text{Factor} = \frac{42}{14} = 3 \]Use this factor to scale up the empirical formula, multiplying the subscripts. Thus, the molecular formula becomes \( \mathrm{C}_3 \mathrm{H}_6 \), by multiplying each subscript in \( \mathrm{CH}_2 \) by 3.

• Verify against available options to ensure correctness.
• Re-evaluate any discrepancies, such as incorrect options, through backtracking calculations or reviewing assumptions.

This logical reasoning and sequential methodology form the backbone of effective chemical problem solving, leading to accurate chemical equations and understanding complex reactions.