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) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (d) \(\mathrm{CH}_{2}\)

Short answer

The molecular formula is \(\mathrm{C}_{3} \mathrm{H}_{6}\), which is option (a).

Step-by-step solution

Calculate Molar Mass of Empirical Formula

First, determine the molar mass of the empirical formula \(\mathrm{CH}_{2}\). The atomic mass of carbon (C) is approximately 12 g/mol, and hydrogen (H) is approximately 1 g/mol. Thus, \(\mathrm{CH}_{2}\) has a molar mass of \(12 + 2(1) = 14 \text{ g/mol}\).

Determine the Factor of Multiplication

To find how many times the empirical formula fits into the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula. That is \(\frac{42}{14} = 3\). This means the empirical formula needs to be multiplied by 3.

Derive the Molecular Formula

Multiply the subscripts in the empirical formula \(\mathrm{C}_{1} \mathrm{H}_{2}\) by the factor found in Step 2. Thus, the molecular formula is \(\mathrm{C}_{3} \mathrm{H}_{6}\).

Verify with Given Options

Finally, compare the derived molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6}\) with the options provided to ensure it matches Option (a).

Key concepts

Empirical Formula Calculation

The empirical formula of a compound provides the simplest integer ratio of the elements present in the compound. To determine it, the actual number of atoms of each element is not considered, but rather their ratio. This ratio is obtained after finding out the percentage or mass of each element in a compound, then converting these to moles.
The empirical formula is always the simplest form and a stepping stone to finding the molecular formula. In our example, the compound has the empirical formula \(\text{CH}_2\), implying there is one carbon atom for every two hydrogen atoms. Even if the compound's molecular formula may be larger, this ratio remains consistent.

• Identify each element's amount in mass or percentage.
• Convert these amounts to moles using their atomic masses.
• Divide all mole values by the smallest mole fraction to find a ratio.
• Convert this ratio to its simplest whole-number form for the empirical formula.

This gives a base to compare the larger molecular formula and see if there is any multiple relation.

Molar Mass Calculation

Calculating the molar mass of a compound is key to understanding the relationship between its empirical and molecular formulas. Each atom of an element has a specific mass, often found on the periodic table. For example, the atomic mass of Carbon (C) is 12 g/mol, while Hydrogen (H) is 1 g/mol.
To calculate the molar mass of the empirical formula, sum the masses of each individual atom according to their ratios in the formula.

• For \(\text{CH}_2\), the calculation is \(12 + 2(1) = 14 \text{ g/mol}\).
• Ensure all atoms present are accounted for in the sum.

Once the molar mass of the empirical formula is known, it can be compared to the molecular formula's molar mass as given in the problem, which is 42 g/mol in our example. This comparison helps in determining how many times the empirical formula's mass fits into the molecular formula's mass.

Stoichiometry

Stoichiometry is the math of chemistry. It involves calculations based on the quantities of reactants and products in a chemical reaction. Here, it helps determine the multiplication factor needed to relate the empirical formula to the molecular formula.
It is all about using ratios. For our example, we find the factor by dividing the compound's total molar mass by the empirical formula's molar mass.

• Calculate \(\frac{42}{14} = 3\).
• The result, 3, indicates the empirical formula must be multiplied by this factor to obtain the molecular formula.

Ultimately, stoichiometry helps in quantitatively understanding chemical equations and allows predictions about the amounts of substances consumed and generated. To apply this understanding, multiply the subscripts in the empirical formula by this factor, yielding the molecular formula \(\text{C}_3\text{H}_6\). By knowing this, it is easier to relate chemical quantities and formulas efficiently.