Question

При спалюванні органічної сполуки отримали карбон (ІV) оксид кількістю речовини 0,1 моль та воду кількістю речовини 0,1 моль. Пари цієї речовини масою 0,05 г займають об’єм 16 мл за нормальних умов. Визначити молекулярну формулу сполуки, вказати, до яких гомологічних рядів можна віднести дану речовину.

Answer 1

Решение задачи см. во вложении      

Answer 2

Molecular Formula of the Compound and its Homologous Series

The given information states that when an organic compound is burned, it produces 0.1 moles of carbon (IV) oxide and 0.1 moles of water. The vapor of this substance, with a mass of 0.05 g, occupies a volume of 16 ml under normal conditions.

To determine the molecular formula of the compound, we can start by calculating the molar mass of the compound and then use the given information to find the molecular formula.

First, let's calculate the molar mass of the compound using the given data: - Moles of carbon (IV) oxide produced = 0.1 moles - Moles of water produced = 0.1 moles - Mass of vapor = 0.05 g - Volume of vapor = 16 ml

Using this information, we can calculate the molar mass of the compound.

Calculation of Molar Mass: - Molar mass of carbon (IV) oxide (CO2) = 44.01 g/mol - Molar mass of water (H2O) = 18.02 g/mol

The total moles of the products are 0.1 moles of CO2 and 0.1 moles of H2O, which gives a total of 0.2 moles of products.

The total mass of the products can be calculated as follows: - Mass of CO2 = 0.1 moles * 44.01 g/mol = 4.401 g - Mass of H2O = 0.1 moles * 18.02 g/mol = 1.802 g - Total mass of products = 4.401 g + 1.802 g = 6.203 g

Now, we can calculate the molar mass of the compound using the ideal gas law: - PV = nRT - n = m/M, where n is the number of moles, m is the mass, and M is the molar mass

Given: - Mass of vapor (m) = 0.05 g - Volume of vapor (V) = 16 ml = 0.016 L (converted to liters for the ideal gas law) - R = 0.0821 L.atm/mol.K (gas constant) - T = 273 K (room temperature)

Using the ideal gas law, we can calculate the molar mass of the compound.

Calculation of Molar Mass using Ideal Gas Law: - n = PV/RT - n = (0.05 g) / (0.0821 L.atm/mol.K * 273 K) - n ≈ 0.002 mol