Question

Даю максимальное количество баллов за помощь в решении задачи. 10 класс. При сгорании алкина массой 32.8г в кислороде образуется углекислый газ, занимающий при нормальных условиях объем 53.76л. Определите молекулярную формулу алкина. Приведите три различных изомера для данного соединения. Прошу как можно подробнее, в решении химических задач я, к сожалению, полный ноль.

Answer 1

Problem Analysis

To determine the molecular formula of the alkyne and provide three different isomers, we need to use the given information about the mass of the alkyne and the volume of carbon dioxide produced during its combustion.

Solution

1. Calculate the number of moles of carbon dioxide produced: - The volume of carbon dioxide is given as 53.76 L at normal conditions (STP). - At STP, 1 mole of any gas occupies 22.4 L. - Therefore, the number of moles of carbon dioxide produced can be calculated using the following equation: ``` moles of CO2 = volume of CO2 / molar volume at STP ``` - Substitute the given values: ``` moles of CO2 = 53.76 L / 22.4 L/mol ```

2. Calculate the number of moles of the alkyne: - The balanced chemical equation for the combustion of an alkyne is: ``` 2C2H2 + 5O2 -> 4CO2 + 2H2O ``` - From the equation, we can see that 2 moles of alkyne produce 4 moles of carbon dioxide. - Therefore, the number of moles of the alkyne can be calculated using the following equation: ``` moles of alkyne = (moles of CO2 produced) / 4 ``` - Substitute the calculated value of moles of CO2 produced from step 1.

3. Calculate the molar mass of the alkyne: - The molar mass of the alkyne can be calculated using the following equation: ``` molar mass = mass of alkyne / moles of alkyne ``` - Substitute the given mass of the alkyne and the calculated moles of the alkyne.

4. Determine the empirical formula of the alkyne: - The empirical formula represents the simplest whole-number ratio of atoms in a compound. - To determine the empirical formula, divide the molar mass of the alkyne by the molar mass of the empirical formula. - The empirical formula mass of carbon is 12 g/mol, and the empirical formula mass of hydrogen is 1 g/mol. - Divide the molar mass of the alkyne by the empirical formula mass of carbon and hydrogen to find the ratio of carbon to hydrogen atoms.

5. Determine the molecular formula of the alkyne: - The molecular formula represents the actual number of atoms of each element in a compound. - To determine the molecular formula, we need to know the empirical formula and the molar mass of the alkyne. - Divide the molar mass of the alkyne by the molar mass of the empirical formula to find the ratio of the molecular formula to the empirical formula. - Multiply the subscripts in the empirical formula by this ratio to obtain the molecular formula.

Calculation

Let's perform the calculations step by step using the given values:

1. Calculate the number of moles of carbon dioxide produced: ``` moles of CO2 = 53.76 L / 22.4 L/mol = 2.4 mol ```

2. Calculate the number of moles of the alkyne: ``` moles of alkyne = (2.4 mol) / 4 = 0.6 mol ```

3. Calculate the molar mass of the alkyne: ``` molar mass = 32.8 g / 0.6 mol = 54.67 g/mol ```

4. Determine the empirical formula of the alkyne: - To determine the empirical formula, we need to know the ratio of carbon to hydrogen atoms. - The molar mass of carbon is 12 g/mol, and the molar mass of hydrogen is 1 g/mol. - Let's assume the empirical formula has x carbon atoms and y hydrogen atoms. - The molar mass of the empirical formula can be calculated as follows: ``` empirical formula mass = (x * molar mass of carbon) + (y * molar mass of hydrogen) ``` - Substitute the molar masses of carbon and hydrogen: ``` 12x + 1y = 54.67 ``` - Since the empirical formula represents the simplest whole-number ratio, we can divide the equation by the molar mass of hydrogen (1 g/mol) to simplify it: ``` 12x + y = 54.67 ``` - We need to find the values of x and y that satisfy this equation. Let's solve it.

- One possible solution is x = 3 and y = 2, which gives an empirical formula of C3H2. - Therefore, the empirical formula of the alkyne is C3H2.

5. Determine the molecular formula of the alkyne: - To determine the molecular formula, we need to know the molar mass of the alkyne. - The molar mass of the alkyne is 54.67 g/mol, as calculated in step 3. - Let's assume the molecular formula has a carbon-to-hydrogen ratio of a:b. - The molar mass of the molecular formula can be calculated as follows: ``` molecular formula mass = (a * molar mass of carbon) + (b * molar mass of hydrogen) ``` - Substitute the molar masses of carbon and hydrogen: ``` 12a + 1b = 54.67 ``` - We need to find the values of a and b that satisfy this equation. Let's solve it.

- One possible solution is a = 6 and b = 4, which gives a molecular formula of C6H4. - Therefore, one possible molecular formula for the alkyne is C6H4.

Isomers

An isomer is a compound that has the same molecular formula but different structural arrangements of atoms. Here are three different isomers for the alkyne with the molecular formula C6H4:

1. 1,3-Hexadiyne: This isomer has a triple bond between carbon atoms 1 and 3, with single bonds between the other carbon atoms. The structural formula is H-C≡C-C≡C-C≡C-H.

2. 1,4-Hexadiyne: This isomer has a triple bond between carbon atoms 1 and 4, with single bonds between the other carbon atoms. The structural formula is H-C≡C-C-C≡C-C≡C-H.

3. 1,5-Hexadiyne: This isomer has a triple bond between carbon atoms 1 and 5, with single bonds between the other carbon atoms. The structural formula is H-C≡C-C-C-C≡C-C≡C-H.

These are three different isomers for the alkyne with the molecular formula C6H4.

I hope this explanation helps you understand the solution to the problem. If you have any further questions, please feel free to ask!