Question

При сжигании 2,6г вещества образовалось 8,8г оксида углерода(IV) и 1,8г воды. Плотность паров этого вещества по водороду равна 39. Выведите молекулярную формулу этого соединения.

Answer 1

M(CxHy) = D(H2)xM(H2) = 39x2 = 78 г/моль
CxHy + xyO2 = xCO2 + y/2H2O
n(CO2) = m(CO2)/M(CO2) = 8,8/44 = 0,2 моль = x
n(H2O) = m(H2O)/M(H2O) = 1,8/18 = 0,1 моль = y/2, y = 0,1x2 = 0,2 моль 
n(CxHy) = m/M = 2,6/78 = 0,033 моль
x:y = n(CO2)/n(CxHy) : n(H2O)/n(CxHy) = 0,2/0,033 : 0,2/0,033 = 6:6
Следовательно, формула УВ: C6H6, M(C6H6) = 12x6 + 1x6 = 72 + 6 = 78 г/моль, решение верно.

Answer 2

Given Information:

- Mass of substance burned: 2.6g - Mass of carbon dioxide (CO2) formed: 8.8g - Mass of water (H2O) formed: 1.8g - Density of the vapor of the substance with respect to hydrogen: 39

Calculating the Moles of Carbon Dioxide and Water:

To determine the molecular formula of the compound, we need to calculate the moles of carbon dioxide and water formed during the combustion.

The molar mass of carbon dioxide (CO2) is 44 g/mol, and the molar mass of water (H2O) is 18 g/mol.

The moles of carbon dioxide can be calculated using the formula:

moles of CO2 = mass of CO2 / molar mass of CO2

Substituting the given values:

moles of CO2 = 8.8g / 44 g/mol = 0.2 mol

Similarly, the moles of water can be calculated using the formula:

moles of H2O = mass of H2O / molar mass of H2O

Substituting the given values:

moles of H2O = 1.8g / 18 g/mol = 0.1 mol

Calculating the Moles of Carbon and Hydrogen:

Since carbon dioxide (CO2) is formed by the combustion of carbon (C) and water (H2O) is formed by the combustion of hydrogen (H), we can calculate the moles of carbon and hydrogen present in the compound.

The molar mass of carbon (C) is 12 g/mol, and the molar mass of hydrogen (H) is 1 g/mol.

The moles of carbon can be calculated using the formula:

moles of C = moles of CO2

Substituting the calculated value:

moles of C = 0.2 mol

The moles of hydrogen can be calculated using the formula:

moles of H = 2 * moles of H2O

Substituting the calculated value:

moles of H = 2 * 0.1 mol = 0.2 mol

Determining the Empirical Formula:

The empirical formula represents the simplest whole-number ratio of atoms in a compound. To determine the empirical formula, we need to find the ratio of moles of carbon and hydrogen.

The ratio of moles of carbon to hydrogen can be calculated as:

C:H = moles of C : moles of H

Substituting the calculated values:

C:H = 0.2 mol : 0.2 mol = 1:1

Therefore, the empirical formula of the compound is CH.

Determining the Molecular Formula:

To determine the molecular formula, we need to know the molar mass of the compound. However, the molar mass is not provided in the given information.

Since the density of the vapor of the substance with respect to hydrogen is given as 39, we can use this information to estimate the molar mass of the compound.

The molar mass of hydrogen (H) is 1 g/mol.

The molar mass of the compound can be calculated using the formula:

molar mass of compound = density of vapor * molar mass of hydrogen

Substituting the given values:

molar mass of compound = 39 * 1 g/mol = 39 g/mol

Now, we can compare the molar mass of the compound to the empirical formula's molar mass (12 g/mol for carbon and 1 g/mol for hydrogen) to determine the molecular formula.

The molar mass of the empirical formula (CH) is 12 g/mol + 1 g/mol = 13 g/mol.

Since the molar mass of the compound (39 g/mol) is greater than the molar mass of the empirical formula (13 g/mol), the molecular formula must be a multiple of the empirical formula.

To find the multiple, we divide the molar mass of the compound by the molar mass of the empirical formula:

multiple = molar mass of compound / molar mass of empirical formula

Substituting the calculated values:

multiple = 39 g/mol / 13 g/mol = 3

Therefore, the molecular formula of the compound is C3H3.

Conclusion:

The molecular formula of the compound formed by the combustion of 2.6g of the substance, which produced 8.8g of carbon dioxide (CO2) and 1.8g of water (H2O), is C3H3.