Problem 48 - A Complete Resource Book in Chemistry for JEE Main

Question

The percentage of sulphur in an organic compound whose amount of \(0.32 \mathrm{~g}\) produces \(0.233 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\) (Atomic weight of \(\mathrm{Ba}=137, \mathrm{~S}=32\) ) is (a) \(1.0\) (b) \(10.0\) (c) \(25.3\) (d) \(32.1\)

Step-by-Step Solution

Answer

The percentage of sulfur in the compound is 10.0% (option b).

Step 1: Find the molar mass of \(\mathrm{BaSO}_{4}\)

First, calculate the molar mass of barium sulfate (\(\mathrm{BaSO}_{4}\)). Add the atomic masses: \(137\) for Ba, \(32\) for S, and \(4\times16=64\) for the O atoms. This gives: \(137 + 32 + 64 = 233\) g/mol.

Step 2: Calculate moles of \(\mathrm{BaSO}_{4}\) produced

Use the mass and molar mass of \(\mathrm{BaSO}_{4}\) to find the number of moles produced: \[ \text{moles of } \mathrm{BaSO}_4 = \frac{0.233 \text{ g}}{233 \text{ g/mol}} = 0.001 \text{ mol}. \]

Step 3: Determine mass of sulfur in \(\mathrm{BaSO}_{4}\)

Find the mass of sulfur in the produced \(\mathrm{BaSO}_{4}\) by multiplying the moles by the atomic mass of sulfur: \[ \text{mass of S} = 0.001 \text{ mol} \times 32 \text{ g/mol} = 0.032 \text{ g}. \]

Step 4: Calculate the percentage of sulfur in the compound

Calculate the percentage of sulfur in the original compound by dividing the mass of sulfur by the original mass of the sample and multiplying by 100: \[ \text{Percentage of S} = \left(\frac{0.032 \text{ g}}{0.32 \text{ g}}\right) \times 100 = 10.0\%. \]

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a compound is essential for understanding the quantities of each element within it. Molar mass is the mass of one mole of a given substance, typically expressed in g/mol. To find it, you sum the atomic masses of each element in the compound according to how many of each are present.

For example, with barium sulfate (\(\mathrm{BaSO}_{4}\)), you add:

• 137 g/mol for barium (Ba),
• 32 g/mol for sulfur (S),
• 64 g/mol from the four oxygen atoms (4 x 16 g/mol).

The overall molar mass of barium sulfate is therefore 233 g/mol. This foundational step allows us to see how much of each atom there is relative to a mole of the entire compound. This calculation is used in determining concentrations and is a starting point for various chemical calculations.

Understanding molar mass is crucial when moving to more complex concepts like stoichiometry and percentage composition.

Chemical Stoichiometry

Stoichiometry connects the dots between reactants and products in a chemical reaction by using the concept of the mole. It allows us to calculate the amounts involved in chemical reactions based on balanced equations. For instance, when barium sulfate (\(\mathrm{BaSO}_{4}\)) forms, stoichiometry helps in calculating how many moles of sulfur are present in a given sample.

We use the relationship between mass and moles to find the amount of sulfur. First, calculate moles of \(\mathrm{BaSO}_{4}\) by dividing its mass (0.233 g) by its molar mass (233 g/mol). This yields 0.001 moles. Since the molecular formula shows that every mole of \(\mathrm{BaSO}_{4}\) contains one mole of sulfur, the stoichiometric coefficients tell us directly how much sulfur is in these 0.001 moles.

This exploration through stoichiometry allows us to quantify any substance within a chemical reaction, making it an invaluable tool in chemistry for both qualitative and quantitative analysis.

Percentage Composition

Percentage composition is a way of expressing how much of each element is present within a compound. It is calculated by comparing the mass of the element to the mass of the compound, then multiplying by 100 to get the percentage.

In the example with the organic compound containing sulfur, we found the mass of sulfur present in \(\mathrm{BaSO}_{4}\) (0.032 g of S). We then compare it to the total mass of the original sample (0.32 g). The percentage is calculated by:

• Taking the ratio of sulfur's mass to the sample's mass: \( \frac{0.032 \text{ g}}{0.32 \text{ g}} \)
• Multiplying the result by 100: \( \left(\frac{0.032 \text{ g}}{0.32 \text{ g}}\right) \times 100 \)

Which gives us a percentage composition of 10% sulfur in the original sample.

This method provides insights into the makeup of compounds. It's widely used in analytical chemistry to understand the constituent elements of unknown samples.