Question

The empirical formula of a compound is \(\mathrm{CH}_{2} \mathrm{O}\). If \(0.0832\) mole of the compound contains \(1.0 \mathrm{~g}\) of hydrogen, then the molecular formula of the compound is: (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (b) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)

Short answer

The molecular formula is \( \text{C}_6\text{H}_{12}\text{O}_6 \).

Step-by-step solution

Determine Molar Mass of Empirical Formula

First, calculate the molar mass of the empirical formula \( \mathrm{CH}_2\mathrm{O} \). The atomic masses are: Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol, and Oxygen (O) = 16 g/mol. Thus, the molar mass of \( \mathrm{CH}_2\mathrm{O} \) is \( 12 + 2 \times 1 + 16 = 30 \) g/mol.

Calculate Total Moles of Hydrogen

Since 0.0832 moles of the compound contain 1.0 g of hydrogen, we first determine how many moles of hydrogen atoms this is. The molar mass of hydrogen is 1 g/mol, so 1.0 g of hydrogen corresponds to \( 1.0 \text{ g} \div 1 \text{ g/mol} = 1.0 \text{ mole of hydrogen atoms} \).

Relate Empirical Formula to Moles of Hydrogen

The empirical formula \( \mathrm{CH}_2\mathrm{O} \) indicates that each mole of the compound contains 2 moles of hydrogen. Therefore, 0.0832 moles of \( \mathrm{CH}_2\mathrm{O} \) would contain \( 0.0832 \times 2 = 0.1664 \) moles of hydrogen, which must be scaled up to match the required 1.0 mole of hydrogen.

Determine the Scaling Factor for Moles

To find how many times the empirical formula fits into the molecular formula, calculate the scaling factor. With 0.1664 moles accounting for all the hydrogen in 0.0832 moles of compound, to achieve 1 mole of hydrogen, you should scale by \( 1.0 \div 0.1664 \approx 6 \).

Find the Molecular Formula

Multiply each subscript in the empirical formula \( \mathrm{CH}_2\mathrm{O} \) by the scaling factor found in the previous step, \( 6 \):- Carbon: \( 1 \times 6 = 6 \)- Hydrogen: \( 2 \times 6 = 12 \)- Oxygen: \( 1 \times 6 = 6 \)Thus, the molecular formula is \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \).

Key concepts

Empirical Formula

The starting point in determining the molecular formula is understanding the empirical formula. An empirical formula provides the simplest whole-number ratio of atoms in a compound. For example,

• The empirical formula \( \mathrm{CH}_2\mathrm{O} \) indicates that for every carbon atom, there are two hydrogen atoms and one oxygen atom in the molecule.

It doesn't tell you the actual number of atoms in the molecule, just the ratio. This is crucial for understanding compound composition and serves as the foundation for calculating the molecular formula.
The empirical formula represents the compound's basic chemical makeup, and it is derived from experimental data through elemental analysis. When the molar mass is known, it helps refine this to the actual molecular formula by acting as a base from which we grow our calculations.

Molar Mass Calculation

To convert an empirical formula into a molecular formula, knowing the molar mass of the compound is vital. Molar mass is the mass of one mole of a substance, and it's determined by summing the atomic masses of all atoms in the molecule using periodic table values.For the empirical formula \( \mathrm{CH}_2\mathrm{O} \):

• Carbon (\(\text{C}\)) has an atomic mass of 12 g/mol.
• Hydrogen (\(\text{H}\)) has an atomic mass of 1 g/mol, and since there are two hydrogen atoms, their total contribution is \(2 \times 1 = 2\) g/mol.
• Oxygen (\(\text{O}\)) has an atomic mass of 16 g/mol.

Adding these together gives the empirical molar mass: \(12 + 2 + 16 = 30 \text{ g/mol}\)This calculated molar mass helps determine how many times the empirical formula's structure must be repeated to reach the molecular formula's mass.

Scaling Factor

The scaling factor is a crucial step in determining the number of times the empirical formula needs to be multiplied to achieve the molecular formula. It is like a multiplier that adjusts the quantity found in the empirical formula to reflect the molecular formula's actual size.Start by finding out how many moles of a particular element are present in the compound. Here, we're focusing on hydrogen since the information given is based on 1.0 g of hydrogen in 0.0832 mole of the compound, which means 1 mole of hydrogen atoms overall. Next, utilizing the empirical formula \( \mathrm{CH}_2\mathrm{O} \):

• The formula suggests that each mole of \( \mathrm{CH}_2\mathrm{O} \) contains 2 moles of hydrogen.
• So, 0.0832 moles of the compound would have \(0.0832 \times 2 = 0.1664\) moles of hydrogen.

To fit 1 mole of hydrogen into this scenario, we calculate:\[\text{Scaling Factor} = \frac{1.0}{0.1664} \approx 6\]The scaling factor tells us that each element in the empirical formula should be multiplied by 6 to give the correct mole ratio for the molecular formula.

Chemical Composition Analysis

Through careful analysis of chemical composition, one can derive the molecular formula from the empirical formula. This process involves relating empirical data to molecular-scale realities, essentially decoding the actual structure of the compound from the simplest form.When determining the molecular formula, we apply the scaling factor to the empirical formula:

• Carbon: \(1 \times 6 = 6\)
• Hydrogen: \(2 \times 6 = 12\)
• Oxygen: \(1 \times 6 = 6\)

This gives us \[\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\]Hence, by analyzing the chemical composition and respecting the ratios given by the empirical formula, you infer the actual quantity of each atom in the molecular formula. This deep understanding of the compound's makeup informs chemists about potential reactions and properties.