Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.2\) gram of an organic compound containing carbon, hydrogen and oxygen, on combustion, yielded \(0.147\) gram carbon dioxide and \(0.12\) gram water. What will be the content of oxygen in the substance? (a) \(73.29 \%\) (b) \(68.45 \%\) (c) \(85.23 \%\) (d) \(89.50 \%\)

Step-by-Step Solution

Answer

The content of oxygen in the substance is 73.29%, option (a).

Step 1: Calculate Moles of Carbon in CO2

To find the moles of carbon in the CO2 produced, we use the molar mass of CO2 (44 g/mol). The mass of CO2 given is 0.147 g. The moles of CO2 (and thus carbon) are calculated by dividing the mass by the molar mass:\[ n_{CO2} = \frac{0.147 \, \text{g}}{44 \, \text{g/mol}} = 0.00334 \, \text{mol} \]Every mole of CO2 contains one mole of carbon, so:\[ n_{C} = 0.00334 \, \text{mol} \]

Step 2: Calculate Mass of Carbon

Next, we calculate the mass of carbon from its moles using its atomic mass (12 g/mol):\[ m_{C} = n_{C} \times 12 \, \text{g/mol} = 0.00334 \, \text{mol} \times 12 \, \text{g/mol} = 0.04008 \, \text{g} \]

Step 3: Calculate Moles of Hydrogen in H2O

For water, we need to calculate the moles of H2O produced, using its molar mass (18 g/mol). The mass of H2O given is 0.12 g:\[ n_{H2O} = \frac{0.12 \, \text{g}}{18 \, \text{g/mol}} = 0.00667 \, \text{mol} \]Each mole of H2O contains two moles of hydrogen, so:\[ n_{H} = 2 \times 0.00667 \, \text{mol} = 0.01334 \, \text{mol} \]

Step 4: Calculate Mass of Hydrogen

Now, calculate the mass of hydrogen using its atomic mass (1 g/mol):\[ m_{H} = n_{H} \times 1 \, \text{g/mol} = 0.01334 \, \text{mol} \times 1 \, \text{g/mol} = 0.01334 \, \text{g} \]

Step 5: Determine Mass of Oxygen in the Compound

To find the mass of oxygen, we subtract the mass of carbon and hydrogen from the total mass of the compound:\[ m_{O} = 0.2 \, \text{g} - (m_{C} + m_{H}) = 0.2 \, \text{g} - (0.04008 \, \text{g} + 0.01334 \, \text{g}) = 0.14658 \, \text{g} \]

Step 6: Calculate Percentage of Oxygen

Finally, the percentage composition of oxygen in the compound can be calculated as:\[ \%O = \left(\frac{m_{O}}{0.2 \, \text{g}}\right) \times 100 = \left(\frac{0.14658 \, \text{g}}{0.2 \, \text{g}}\right) \times 100 = 73.29\% \]

Key Concepts

Moles Calculation

Understanding the concept of moles is fundamental in chemistry. A mole is essentially a unit of measurement used to express amounts of a chemical substance. The term "mole" is used to count atoms, molecules, ions, or other particles in a given sample when dealing with a chemical reaction. When calculating the number of moles, we use the formula:\[ n = \frac{m}{M} \]where:

• \( n \) is the number of moles
• \( m \) is the mass of the substance (in grams)
• \( M \) is the molar mass of the substance (in grams per mole)

Let's see this in action with our example. We calculated the moles of carbon dioxide (\( n_{CO2} \)) by dividing the mass of CO2 produced (0.147 g) by the molar mass of CO2 (44 g/mol):\[ n_{CO2} = \frac{0.147 \, \text{g}}{44 \, \text{g/mol}} = 0.00334 \, \text{mol} \] Since each molecule of CO2 contains one atom of carbon, the moles of carbon are the same as the moles of CO2: 0.00334 mol.

Percentage Composition

Percentage composition gives an insight into the relative amount of each element within a compound, expressed as a percentage of the compound's total mass. It describes how much of the compound's mass comes from a particular element.The formula for percentage composition is:\[ \text{Percentage} = \left( \frac{\text{mass of element}}{\text{total mass of compound}} \right) \times 100 \]In the given exercise, after finding the masses of carbon and hydrogen, we can determine the mass of oxygen by subtraction from the total mass of the compound. The total mass was given as 0.2 g. The calculated mass of oxygen was found to be 0.14658 g.To find the percentage composition of oxygen, we plug these values into our formula:\[ \%O = \left( \frac{0.14658 \, \text{g}}{0.2 \, \text{g}} \right) \times 100 = 73.29\% \]This means that oxygen makes up 73.29% of the compound's mass.

Mass of Elements

Knowing the mass of individual elements within a compound is crucial for understanding its chemical structure and behavior. In combustion analysis, the goal is to find the mass of each element contained in a compound.From our example, we first calculated the mass of carbon by using its moles and its atomic mass (12 g/mol):\[ m_{C} = 0.00334 \, \text{mol} \times 12 \, \text{g/mol} = 0.04008 \, \text{g} \]Next, we calculated the mass of hydrogen. Since water, H2O, contains two hydrogen atoms per molecule, we first found the moles of hydrogen by: \[ n_{H} = 2 \, \times 0.00667 \, \text{mol} = 0.01334 \, \text{mol} \] Using the atomic mass of hydrogen (1 g/mol), we found:\[ m_{H} = 0.01334 \, \text{mol} \times 1 \, \text{g/mol} = 0.01334 \, \text{g} \]Finally, to find oxygen’s mass, subtracting the summed mass of carbon and hydrogen from the total (0.2 g) gives the mass of oxygen as 0.14658 g. These individual masses help identify the order and quantity in which atoms are present in the compound.