By analyzing combustion products, vapor density, and chemical reactivity, we determined the empirical formula as CH3O, the relative formula as C3H9O3, and the molecular formula as C3H9O3 (trioxymethylene). This demonstrates a step-by-step approach to solving such problems involving compound identification.
Note: This is a complex problem, and some steps might involve approximations. Double-check calculations and consult other sources for confirmation when needed.
Determining the Compound's Formula:
Here's how to analyze and solve the problem:
1. Combustion Analysis:
From the combustion products, we can calculate the mass of carbon and hydrogen in the original compound:
Carbon: 0.968 g CO2 * (12 g C / 44 g CO2) = 0.276 g C
Hydrogen: 0.594 g H2O * (2 g H / 18 g H2O) = 0.066 g H
2. Oxygen Content and Empirical Formula:
Subtract the masses of C and H from the original compound mass to find the mass of oxygen: 0.682 g - 0.276 g - 0.066 g = 0.34 g O.
Convert each element's mass to moles and divide by the smallest number of moles to find the empirical formula:
C: 0.276 g / 12 g/mol = 0.023 mol; ratio to smallest (H) = 1
H: 0.066 g / 1 g/mol = 0.066 mol; ratio = 3
O: 0.34 g / 16 g/mol = 0.02125 mol; ratio ≈ 1 (rounded)
Therefore, the empirical formula is CH3O (formaldehyde).
3. Vapor Density and Relative Molecular Formula:
Calculate the theoretical density of formaldehyde vapor at 20°C and 95 kPa using the ideal gas equation:
Density = (0.744 g / 497 cm³) * (95 kPa / 101.325 kPa) * (273 K / 293 K) ≈ 0.216 g/cm³
Calculate the experimental molar mass from the ideal gas equation and the measured density, pressure, and volume:
Molar mass = (pressure * volume) / (density * temperature * gas constant)
Molar mass ≈ 45.8 g/mol
Divide the experimental molar mass by the molar mass of the empirical formula (30 g/mol for CH3O) to obtain the relative molecular formula multiplier:
Multiplier ≈ 1.53
Therefore, the relative molecular formula is C1.53H4.59O1.53, which can be simplified to C3H9O3 (trioxymethylene, an isomer of formaldehyde).
4. Reaction with Sodium:
Since the compound reacts with sodium but doesn't contain a -CO2H group, it suggests the presence of an OH group. This aligns with the deduced molecular formula (C3H9O3) containing three hydroxyl groups.