Question

Задача:При спалюванні 11,4 г вуглецю утворилося 35,2 г оксиду вуглецю (IV) та 16,2 кг пари води.Відносна молекулярна маса речовини дорівнює114.Яка молекулярна формула речовини?

Answer 1

я не всё поняла, можно по- русски, тогда помогу

Answer 2

Problem Analysis

We are given the following information: - Mass of carbon burned = 11.4 g - Mass of carbon dioxide formed = 35.2 g - Mass of water vapor formed = 16.2 kg - Relative molecular mass of the substance = 114

We need to determine the molecular formula of the substance.

Solution

To solve this problem, we need to calculate the number of moles of carbon, carbon dioxide, and water vapor formed. Then, we can use these mole ratios to determine the empirical formula of the substance. Finally, we can calculate the molecular formula using the relative molecular mass.

1. Calculate the number of moles of carbon: - The molar mass of carbon (C) is 12.01 g/mol. - Number of moles of carbon = mass of carbon / molar mass of carbon - Number of moles of carbon = 11.4 g / 12.01 g/mol

2. Calculate the number of moles of carbon dioxide: - The molar mass of carbon dioxide (CO2) is 44.01 g/mol. - Number of moles of carbon dioxide = mass of carbon dioxide / molar mass of carbon dioxide - Number of moles of carbon dioxide = 35.2 g / 44.01 g/mol

3. Calculate the number of moles of water vapor: - The molar mass of water (H2O) is 18.02 g/mol. - Number of moles of water vapor = mass of water vapor / molar mass of water - Number of moles of water vapor = 16.2 kg / 18.02 g/mol

4. Determine the empirical formula: - The mole ratio between carbon and carbon dioxide is 1:1. - The mole ratio between carbon and water vapor is 1:2. - The mole ratio between carbon dioxide and water vapor is 35.2 g / 44.01 g/mol : 16.2 kg / 18.02 g/mol - Simplifying the ratios, we find that the empirical formula is C:H2:O.

5. Calculate the empirical formula mass: - The empirical formula mass can be calculated by adding the molar masses of the elements in the empirical formula. - Empirical formula mass = (1 * molar mass of carbon) + (2 * molar mass of hydrogen) + (1 * molar mass of oxygen)

6. Calculate the number of empirical formula units in the relative molecular mass: - Number of empirical formula units = relative molecular mass / empirical formula mass

7. Determine the molecular formula: - The molecular formula is a whole number multiple of the empirical formula. - Multiply the subscripts in the empirical formula by the number of empirical formula units calculated in step 6.

Let's perform the calculations:

1. Number of moles of carbon = 11.4 g / 12.01 g/mol = 0.949 moles 2. Number of moles of carbon dioxide = 35.2 g / 44.01 g/mol = 0.799 moles 3. Number of moles of water vapor = 16.2 kg / 18.02 g/mol = 899.0 moles

The empirical formula is C:H2:O.

4. Empirical formula mass = (1 * 12.01 g/mol) + (2 * 1.01 g/mol) + (1 * 16.00 g/mol) = 30.03 g/mol

5. Number of empirical formula units = 114 / 30.03 = 3.797

6. The molecular formula is obtained by multiplying the subscripts in the empirical formula by the number of empirical formula units: - Empirical formula: C:H2:O - Molecular formula: C3:H6:O3

Therefore, the molecular formula of the substance is C3H6O3.

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