Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is (a) \(\mathrm{CH}_{4} \mathrm{O}\) (b) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula of the compound is (d) \(\mathrm{CH}_2 \mathrm{O}\).

Step 1: Determine Moles of Each Element

Calculate the number of moles for each element given their masses and atomic weights. - Carbon: Atomic weight = 12, Moles = \( \frac{24}{12} = 2 \) moles.- Hydrogen: Atomic weight = 1, Moles = \( \frac{4}{1} = 4 \) moles.- Oxygen: Atomic weight = 16, Moles = \( \frac{32}{16} = 2 \) moles.

Step 2: Find the Simplest Mole Ratio

Identify the simplest whole-number ratio among the moles of the elements.- Divide each element's moles by the smallest number of moles calculated: - Carbon: \( \frac{2}{2} = 1 \) - Hydrogen: \( \frac{4}{2} = 2 \) - Oxygen: \( \frac{2}{2} = 1 \)This gives us the ratio of C:H:O as 1:2:1.

Step 3: Determine Empirical Formula

Translate the mole ratio into an empirical formula.- Using the mole ratio C:H:O = 1:2:1, the empirical formula is \( \mathrm{CH}_2 \mathrm{O} \).

Key Concepts

Mole Ratio

The mole ratio is a crucial concept in chemistry that helps us understand the proportions of elements in a compound based on their moles. Simply put, it allows us to compare the amounts of different substances in a reaction or compound by looking at their mole numbers. In this exercise, we have calculated the moles for each element: carbon, hydrogen, and oxygen.

To get the simplest whole-number mole ratio, we divide each element's moles by the smallest number found among the elements' moles. Once you've calculated the moles of each, the process of finding their simplest ratio is straightforward. For example, if carbon has 2 moles, hydrogen has 4 moles, and oxygen also has 2 moles, we divide each by 2, giving a simple ratio of 1:2:1.

This ratio reflects the proportion of each element in the empirical formula of the compound, forming the basis of further chemical analysis.

Molecular Formula Determination

The determination of the molecular formula is a step further from identifying the empirical formula. While the empirical formula gives the simplest whole-number ratio of atoms in a compound, the molecular formula reveals the exact number of each type of atom present in a molecule.

After establishing the mole ratio and empirical formula, determining the molecular formula typically involves knowing the molar mass of the compound. If the molar mass is given or determined through experimental methods like mass spectrometry, you can divide the compound's molar mass by the empirical formula mass to find a multiplication factor. This factor is then used to scale the empirical formula, giving the molecular formula.

• Empirical formula example: (C₁H₂O₁)
• If molar mass = 60 g/mol, empirical formula mass = 30 g/mol, then 60 / 30 = 2
• Molecular formula becomes C₂H₄O₂ by multiplying each subscript in the empirical formula by 2.

Understanding this distinction ensures accuracy in defining chemical compounds and interpreting their molecular structures.

Chemical Composition Analysis

Chemical composition analysis involves determining the amount and proportion of each element present in a compound. This is essential for identifying substances and understanding their chemical properties.

In this specific exercise, you began with given masses of carbon, hydrogen, and oxygen resulting from the breakdown of the compound. By converting these masses to moles, you can analyze the chemical composition on a more accessible and comparable level, since moles account for the atomic weights of elements.

By understanding chemical composition, you can deduce structural aspects of the compound and predict how it might react with other substances. Regular chemical composition analysis further helps in ensuring the purity and quality of chemical products. In research and industry, this process is vital for new compound synthesis, quality control, and verifying product standards.

• Initial step: Mass conversion to moles.
• Second step: Mole analysis to find ratios for empirical formulas.
• Allows inference of compound purity and reaction possibilities.

Through systematic chemical composition analysis, chemists can demystify the intricate nature of compounds.