Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the molecular formula of a compound, whose empirical formula is \(\mathrm{CH}_{2} \mathrm{O}\) and molecular weight is 90 ? (a) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{3}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{3} \mathrm{O}_{4}\)

Step-by-Step Solution

Answer

The molecular formula is (c) \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \).

Step 1: Determine Empirical Formula Weight

Calculate the weight of the empirical formula \( \text{CH}_2\text{O} \). The atomic weights are approximately \( \text{C} = 12 \), \( \text{H} = 1 \), and \( \text{O} = 16 \). Therefore, the empirical formula weight is: \[ 12 + 2(1) + 16 = 30. \]

Step 2: Divide Molecular Weight by Empirical Formula Weight

Divide the given molecular weight by the empirical formula weight to find how many empirical units are in the molecular formula: \[ \frac{90}{30} = 3. \]

Step 3: Multiply Empirical Formula by Factor

Multiply each subscript in the empirical formula \( \text{CH}_2\text{O} \) by 3 (the factor found in Step 2) to get the molecular formula: \[ \text{C}_{3}\text{H}_{6}\text{O}_{3}. \]

Step 4: Identify Matching Molecular Formula

Compare \( \text{C}_{3}\text{H}_{6}\text{O}_{3} \) with the given options and find it matches option (c). The molecular formula is \( \text{C}_{3}\text{H}_{6}\text{O}_{3} \).

Key Concepts

Molecular Formula

The molecular formula of a compound illustrates the actual number of each type of atom present in a molecule of that compound. It provides a complete picture of the composition of the molecule. Unlike the empirical formula, which shows the simplest whole-number ratio of atoms in a compound, the molecular formula reveals the exact number of atoms. To determine the molecular formula, you often start with the empirical formula and then calculate a multiplying factor. This factor, when applied to the empirical formula, gives the molecular formula. For a compound with an empirical formula of \( \text{CH}_2\text{O} \) and a known molecular weight of 90, you divide the molecular weight by the empirical formula weight. This step helps you discover how many empirical formula units fit into the molecular weight, leading you to the full molecular formula.

Molecular Weight

Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. For instance, to find the molecular weight of a compound like \( \text{CH}_2\text{O} \), you must add together the weights of all the atoms: the carbon (C), the hydrogen atoms (H), and the oxygen (O). Using approximate atomic weights, we get \( 12 + 2(1) + 16 = 30 \) for the empirical formula \( \text{CH}_2\text{O} \). If the molecular weight of the compound is given as 90, this indicates the molecular formula is a multiple of the empirical formula. By dividing the given molecular weight (90) by the empirical formula weight (30), you deduce that there are three times as many atoms of each type, which adjusts the empirical formula to the full molecular formula.

Organic Chemistry

Organic chemistry is a branch of chemistry that deals with the structure, properties, and reactions of organic compounds, which contain carbon in covalent bonding. In organic chemistry, compounds like \( \text{CH}_2\text{O} \) serve as building blocks for more complex molecules. Organic chemists often work with empirical and molecular formulas to understand and create complex molecules found in living organisms or synthetic substances. The process of determining the molecular formula from the empirical formula is a fundamental technique in organic synthesis and analysis, allowing chemists to construct molecules with very specific properties and functions. This knowledge is essential as it helps chemists design drugs, develop new materials, and explore the vast potential of carbons in forming diverse chemical structures.

Chemical Calculations

Chemical calculations are essential in chemistry for understanding how molecules and reactions work. These calculations often involve determining the amounts and relationships of different atoms in a molecule or a reaction, much like in the process of finding a molecular formula from an empirical one. To perform such calculations, you start by knowing the molecular and empirical formula weights. Then, divide the molecular weight by the empirical formula weight to find a factor. Multiply the empirical formula by this factor to find the molecular formula, revealing the exact composition of the compounds. These calculations form the backbone of many activities in chemistry, supporting accurate data analysis and helping predict how substances might behave under different conditions.