Question

сожгли кислородосодержащее вещество массой 6,9г и получили углекислый газ массой 6,6 и воду массой 2,7г. относительная плотность паров этого вещества по воздуху равна 1,6. установите молекулярную формулу данного вещества.

Answer 1

М(вещ.) = 29г/моль * 1,6 = 46г моль
n(CO2) = 6,6g / 44= 0,15 mol
n(C) = n(CO2) = 0,15 mol
n(H2O) = 2,7g / 18 = 0,15 mol
n(H) = n(H2O) *2 = 0,3 mol
m(C) = 0,15 mol *12= 1,8g
m(H) = 0,3mol *1 = 0,3g
m(O) = 6,9 - 1,8 - 0,3 = 4,8 g
n(O) = 4,8g / 16 = 0,3 mol
C:H:O = 0,15:0,3:0,3 => CH2O2
По молярным массам совпадает, выходит что ответ  CH2O2

Answer 2

Calculation of the Molecular Formula

To determine the molecular formula of the substance, we need to analyze the given information.

According to the problem statement, a oxygen-containing substance with a mass of 6.9g was burned, resulting in the production of carbon dioxide with a mass of 6.6g and water with a mass of 2.7g. The relative density of the vapor of this substance with respect to air is given as 1.6.

To find the molecular formula, we need to determine the number of moles of each element present in the given masses and then find the simplest whole number ratio between them.

Let's start by calculating the number of moles of carbon dioxide, water, and oxygen in the given masses.

1. Moles of carbon dioxide (CO2): - Molar mass of carbon dioxide (CO2) = 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen) = 44.01 g/mol - Moles of carbon dioxide = mass of carbon dioxide / molar mass of carbon dioxide = 6.6g / 44.01 g/mol = 0.15 mol

2. Moles of water (H2O): - Molar mass of water (H2O) = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol - Moles of water = mass of water / molar mass of water = 2.7g / 18.02 g/mol = 0.15 mol

3. Moles of oxygen (O2): - Moles of oxygen = moles of carbon dioxide - moles of water = 0.15 mol - 0.15 mol = 0 mol

Since the moles of oxygen are zero, it means that all the oxygen in the original substance was converted to water during the combustion process.

Now, let's calculate the moles of carbon and hydrogen in the water to determine the empirical formula.

4. Moles of carbon in water: - Moles of carbon = 0 mol

5. Moles of hydrogen in water: - Molar mass of hydrogen (H) = 1.01 g/mol - Moles of hydrogen = mass of water / molar mass of hydrogen = 2.7g / 1.01 g/mol = 2.67 mol

The empirical formula of the substance is therefore CH2O, which represents the simplest whole number ratio of carbon, hydrogen, and oxygen atoms in the substance.

To determine the molecular formula, we need to find the ratio between the empirical formula mass and the molar mass of the substance.

6. Empirical formula mass of CH2O: - Molar mass of carbon (C) = 12.01 g/mol - Molar mass of hydrogen (H) = 1.01 g/mol - Molar mass of oxygen (O) = 16.00 g/mol - Empirical formula mass = (12.01 g/mol * 1) + (1.01 g/mol * 2) + (16.00 g/mol * 1) = 30.03 g/mol

7. Molar mass of the substance: - Molar mass of the substance = mass of the substance / number of moles of the substance - Molar mass of the substance = 6.9g / 0.15 mol = 46.00 g/mol

8. Ratio between the empirical formula mass and the molar mass of the substance: - Ratio = molar mass of the substance / empirical formula mass = 46.00 g/mol / 30.03 g/mol = 1.53

The ratio between the empirical formula mass and the molar mass of the substance is approximately 1.53.

Since the relative density of the vapor of this substance with respect to air is given as 1.6, which is close to the ratio we calculated, we can conclude that the molecular formula of the substance is C1.5H3O1.5.

Please note that the molecular formula is an approximation based on the given information and the assumption that the substance contains only carbon, hydrogen, and oxygen atoms. The actual molecular formula may differ depending on the specific compound.

Conclusion

Based on the given information, the molecular formula of the substance is approximately C1.5H3O1.5.