Question

При сгорании 0,31 г органического вещества выделялось 0,224 л СО2, 0,45 г Н2О и 0,112 N2 .Плотность вещества равна 1,384 г/л. Найдите формулу.Помогите пожалуйста,срочно надо)

Answer 1

Пусть органическое соединение имеет состав CxHyOzNu.
M(CxHyOzNu)=DN2(CxHyOzNu)*M(N2)=1,607*28г/моль=45г/моль
1)Количество элементов,входящих в состав вещества
а)n(CO2)=V/VM=0,448л/(22,4л/моль)=0,02моль
С-->CO2
1 моль 1 моль
n©=n(CO2)=0,02 моль
m©=0,02моль*12г/моль=0,24г
б)n(N2)=0,112л/(22,4л/моль)=5ммоль
2N-->N2
2 моль 1 моль
n(N)=2n(N2)=0,01моль
m(N)=0,01моль*14г/моль=0,14г
в)n(H2O)=0,45г/(18г/моль)=0,025моль
2H-->H2
2 моль 1 моль
n(H)=2n(H2)=0,07моль
m(H)=0,07моль*1г/моль=0,07г
г)Т.к. m(N)+m(H)+m©=m(CxHyOzNu),то формулу органического вещества можно представить в виде CxHyNu
2)Определение формулы вещества
x:y:u=n©:n(H):n(N)
x:y:u=0,02:0,07:0,01;
x:y:u=2:7:1
Простейшая формула вещества-C2H7N
Mпрост.=Mист.
Ответ:C2H7N

Answer 2

Finding the Formula of the Organic Substance

To find the formula of the organic substance, we need to analyze the products of its combustion and use stoichiometry to determine the ratio of elements present.

Given: - Mass of organic substance burned = 0.31 g - Volume of CO2 produced = 0.224 L - Mass of H2O produced = 0.45 g - Mass of N2 produced = 0.112 g - Density of the substance = 1.384 g/L

Let's start by calculating the number of moles of each product:

1. Moles of CO2: - The molar mass of CO2 is 44.01 g/mol. - Using the ideal gas law, we can convert the volume of CO2 to moles at standard temperature and pressure (STP). - The molar volume of a gas at STP is 22.4 L/mol. - Therefore, the number of moles of CO2 produced is: 0.224 L / 22.4 L/mol = 0.01 mol.

2. Moles of H2O: - The molar mass of H2O is 18.02 g/mol. - The number of moles of H2O produced is: 0.45 g / 18.02 g/mol = 0.025 mol.

3. Moles of N2: - The molar mass of N2 is 28.02 g/mol. - The number of moles of N2 produced is: 0.112 g / 28.02 g/mol = 0.004 mol.

Next, we need to determine the number of moles of carbon (C), hydrogen (H), and nitrogen (N) in the organic substance.

1. Moles of C: - From the balanced equation for the combustion of organic compounds, we know that each mole of CO2 produced corresponds to one mole of carbon in the organic substance. - Therefore, the number of moles of carbon is also 0.01 mol.

2. Moles of H: - From the balanced equation for the combustion of organic compounds, we know that each mole of H2O produced corresponds to two moles of hydrogen in the organic substance. - Therefore, the number of moles of hydrogen is 2 * 0.025 mol = 0.05 mol.

3. Moles of N: - From the balanced equation for the combustion of organic compounds, we know that each mole of N2 produced corresponds to two moles of nitrogen in the organic substance. - Therefore, the number of moles of nitrogen is 2 * 0.004 mol = 0.008 mol.

Now, let's calculate the molar mass of the organic substance:

- Molar mass = (mass of carbon) + (mass of hydrogen) + (mass of nitrogen) - Molar mass = (moles of carbon * molar mass of carbon) + (moles of hydrogen * molar mass of hydrogen) + (moles of nitrogen * molar mass of nitrogen) - Molar mass = (0.01 mol * 12.01 g/mol) + (0.05 mol * 1.01 g/mol) + (0.008 mol * 14.01 g/mol) - Molar mass = 0.1201 g + 0.0505 g + 0.1121 g - Molar mass = 0.2827 g

Finally, let's calculate the empirical formula of the organic substance:

- Empirical formula = (moles of carbon / total moles) : (moles of hydrogen / total moles) : (moles of nitrogen / total moles) - Empirical formula = (0.01 mol / 0.068 mol) : (0.05 mol / 0.068 mol) : (0.008 mol / 0.068 mol) - Empirical formula = 0.147 : 0.735 : 0.118

To simplify the empirical formula, we can divide all the subscripts by the smallest subscript (0.118):

- Empirical formula = 0.147 / 0.118 : 0.735 / 0.118 : 0.118 / 0.118 - Empirical formula = 1.245 : 6.220 : 1

Rounding these ratios to the nearest whole number, we get the empirical formula of the organic substance: C1H6N1.

Please note that the empirical formula represents the simplest whole-number ratio of atoms in a compound. It does not provide information about the actual molecular formula, which may be a multiple of the empirical formula.

I hope this helps! Let me know if you have any further questions.