Question

При горении органического бескислородного вещества массой 23,54 г было получено 67,76 г углекислого газа, 17,82 мл воды и азот. Известно, что оно может вступить в реакцию с 3 моль брома в водном растворе. Определите молекулярную и структурную формулу вещества, составьте уравнение его реакции с избытком бромной воды.​

Answer 1

Ответ:

n(CO2)=67.76/44=1,54 моль  n(H2O)=17.82/18=0,99 моль

m(C)=1.54*12=18,48 грамм m(H)=0.99*2=1,98 грамм m(N)=23.54-18.48-1.98=3,08

Доли компонентов в соединении

С - 18,48/(23,54*12)=0,06542

H - 1.98/23.54=0,0841

N - 3.08/(23.54*14)=0,00934

Отношение компонентов

С/H/N - 0,06542 /0,0841 /0,00934 - 7/9/1

С7H9N

С 3 моль брома в водном растворе может вступить соединение с тремя двойными связями 1-амин циклгептатриен 2,4,6 (Циклогептатриен-2,4,6-иламин) в результат получится 2,3,4,5,6,7-гексабромциклогептанамин

C7H9N + 3Br2 = C7H9NBr6

Answer 2

Given Information:

- Mass of organic substance burned: 23.54 g - Mass of carbon dioxide produced: 67.76 g - Volume of water produced: 17.82 mL - Nitrogen is also produced - The substance can react with 3 moles of bromine in an aqueous solution

Finding the Molecular and Structural Formula:

To determine the molecular and structural formula of the substance, we need to analyze the products of the reaction.

1. Carbon Dioxide (CO2): - The mass of carbon dioxide produced is 67.76 g. - The molar mass of carbon dioxide is 44.01 g/mol. - Using the equation: mass = molar mass × moles, we can calculate the moles of carbon dioxide produced: moles = mass / molar mass. - Substituting the values, we get: moles = 67.76 g / 44.01 g/mol = 1.54 mol.

2. Water (H2O): - The volume of water produced is 17.82 mL. - The density of water is 1 g/mL. - Using the equation: mass = volume × density, we can calculate the mass of water produced: mass = 17.82 mL × 1 g/mL = 17.82 g. - The molar mass of water is 18.02 g/mol. - Using the equation: moles = mass / molar mass, we can calculate the moles of water produced: moles = 17.82 g / 18.02 g/mol = 0.99 mol.

3. Nitrogen (N2): - Nitrogen is also produced in the reaction, but we don't have any specific information about its mass or volume. Therefore, we cannot determine the moles of nitrogen produced.

4. Bromine (Br2): - The substance can react with 3 moles of bromine in an aqueous solution. This information suggests that the substance contains a functional group that can react with bromine.

Based on the given information, we can conclude that the substance contains carbon, hydrogen, and possibly nitrogen. The molecular formula of the substance cannot be determined without additional information.

Reaction with Excess Bromine Water:

To write the reaction equation with excess bromine water, we need to know the functional group present in the substance. Unfortunately, the given information does not provide any details about the functional group.

Without knowing the specific functional group, it is not possible to write the balanced equation for the reaction with excess bromine water.

In summary, the molecular and structural formula of the substance cannot be determined without additional information. The reaction equation with excess bromine water cannot be written without knowing the functional group present in the substance.

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