0.3 g of an organic compound containing C, H and O on combustion yields 0.44 g of CO2 and 0.18 g of H20, with two O atoms per molecule.

Match the column →

1. One mole of compund contains 4 N_a atoms of Hydrogen
2. The emperical formula of the compound is same as its molecular formula.
3. Combustion of one mole of compound contains larger number of moles of CO₂ han that of H₂O.
4. CO₂ gas produced by the combustion of 0.25 mole of compound occupies a volume of 11.2 litre at NTP.

Answer is P and S. Please provide a complete solution.

To solve the problem regarding the combustion of the organic compound, we need to analyze the data provided and derive the necessary information step by step. Let's break it down systematically.

Step 1: Analyzing the Combustion Products

When the organic compound combusts, it produces carbon dioxide (CO2) and water (H2O). From the problem, we have:

• Mass of CO2 produced = 0.44 g
• Mass of H2O produced = 0.18 g

Step 2: Calculating Moles of CO2 and H2O

To find the number of moles of CO2 and H2O, we use the molar masses:

• Molar mass of CO2 = 44 g/mol
• Molar mass of H2O = 18 g/mol

Now, we can calculate the moles:

• Moles of CO2 = 0.44 g / 44 g/mol = 0.01 moles
• Moles of H2O = 0.18 g / 18 g/mol = 0.01 moles

Step 3: Determining the Moles of Carbon and Hydrogen

From the combustion products, we can deduce the number of moles of carbon and hydrogen:

• Each mole of CO2 contains 1 mole of carbon. Therefore, moles of carbon = 0.01 moles.
• Each mole of H2O contains 2 moles of hydrogen. Therefore, moles of hydrogen = 0.01 moles × 2 = 0.02 moles.

Step 4: Finding the Mass of Carbon and Hydrogen

Next, we calculate the mass of carbon and hydrogen:

• Mass of carbon = moles × molar mass = 0.01 moles × 12 g/mol = 0.12 g.
• Mass of hydrogen = moles × molar mass = 0.02 moles × 1 g/mol = 0.02 g.

Step 5: Calculating the Mass of Oxygen

Now, we can find the mass of oxygen in the compound. The total mass of the compound is given as 0.3 g:

• Mass of oxygen = Total mass - (Mass of carbon + Mass of hydrogen)
• Mass of oxygen = 0.3 g - (0.12 g + 0.02 g) = 0.16 g.

Step 6: Finding Moles of Oxygen

Using the molar mass of oxygen (16 g/mol), we can calculate the moles of oxygen:

• Moles of oxygen = 0.16 g / 16 g/mol = 0.01 moles.

Step 7: Determining the Empirical Formula

Now we have the moles of each element:

• Carbon: 0.01 moles
• Hydrogen: 0.02 moles
• Oxygen: 0.01 moles

To find the simplest ratio, we divide each by the smallest number of moles (0.01):

• Carbon: 0.01 / 0.01 = 1
• Hydrogen: 0.02 / 0.01 = 2
• Oxygen: 0.01 / 0.01 = 1

This gives us an empirical formula of CH2O.

Step 8: Molecular Formula and Additional Information

Given that the molecular formula has two oxygen atoms per molecule, we can deduce that the molecular formula is C2H4O2 (since the empirical formula is CH2O). This means the empirical formula is the same as the molecular formula in terms of the ratio of elements.

Step 9: Analyzing the Statements

Now, let's match the statements with our findings:

• One mole of the compound contains 4 atoms of hydrogen: This is true for C2H4O2.
• The empirical formula of the compound is the same as its molecular formula: This is false.
• Combustion of one mole of the compound contains a larger number of moles of CO2 than that of H2O: This is false.
• CO2 gas produced by the combustion of 0.25 mole of the compound occupies a volume of 11.2 liters at NTP: This is true.

Thus, the correct matches are P and S. This analysis confirms the relationships and properties of the compound based on the combustion data provided.