масові частки Карбону та гідрогену, які входять до складу невідомого вуглеводню, відповідно
дорівнюють 85,7% і 14,3%. Відносна густина цієї речовини за повітрям 1,45. Визначити молекулярну формулу речовини. 30 баловОтветы на вопрос
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To determine the molecular formula of the unknown hydrocarbon, we can use the given information about the mass percentages of carbon and hydrogen, as well as the relative density of the substance.
Given Information:
- Mass percentage of carbon: 85.7% - Mass percentage of hydrogen: 14.3% - Relative density of the substance: 1.45Steps to Determine the Molecular Formula:
1. Convert the mass percentages to moles: - Assume we have 100 grams of the substance. - Mass of carbon = 85.7 grams - Mass of hydrogen = 14.3 gramsTo convert the masses to moles, we need to divide them by the molar masses of carbon and hydrogen, respectively.
- Molar mass of carbon (C) = 12.01 g/mol - Molar mass of hydrogen (H) = 1.01 g/mol
Number of moles of carbon = 85.7 g / 12.01 g/mol Number of moles of hydrogen = 14.3 g / 1.01 g/mol
2. Determine the empirical formula: The empirical formula represents the simplest whole number ratio of atoms in a compound. To find it, divide the number of moles of each element by the smallest number of moles obtained.
Let's assume the empirical formula is CxHy.
- Number of moles of carbon = x - Number of moles of hydrogen = y
Divide the number of moles of each element by the smallest number of moles obtained to get the simplest ratio.
- x = Number of moles of carbon / Number of moles of hydrogen - y = Number of moles of hydrogen / Number of moles of hydrogen
3. Calculate the empirical formula mass: The empirical formula mass is the sum of the atomic masses of all the atoms in the empirical formula.
- Empirical formula mass = (x * molar mass of carbon) + (y * molar mass of hydrogen)
4. Determine the molecular formula: The molecular formula represents the actual number of atoms of each element in a molecule. To find it, we need to know the molar mass of the compound.
- Molar mass of the compound = Relative density * Empirical formula mass
Divide the molar mass of the compound by the empirical formula mass to find the ratio.
- Molecular formula ratio = Molar mass of the compound / Empirical formula mass
Multiply the subscripts in the empirical formula by the molecular formula ratio to obtain the molecular formula.
Calculation:
Let's calculate the molecular formula using the given information.1. Convert the mass percentages to moles: - Mass of carbon = 85.7 grams - Mass of hydrogen = 14.3 grams
Number of moles of carbon = 85.7 g / 12.01 g/mol = 7.13 mol Number of moles of hydrogen = 14.3 g / 1.01 g/mol = 14.16 mol
2. Determine the empirical formula: Divide the number of moles of each element by the smallest number of moles obtained.
- x = Number of moles of carbon / Number of moles of hydrogen = 7.13 mol / 7.13 mol = 1 - y = Number of moles of hydrogen / Number of moles of hydrogen = 14.16 mol / 7.13 mol = 2
The empirical formula is CH2.
3. Calculate the empirical formula mass: Empirical formula mass = (1 * molar mass of carbon) + (2 * molar mass of hydrogen) Empirical formula mass = (1 * 12.01 g/mol) + (2 * 1.01 g/mol) = 14.03 g/mol
4. Determine the molecular formula: Molar mass of the compound = Relative density * Empirical formula mass Molar mass of the compound = 1.45 * 14.03 g/mol = 20.34 g/mol
Molecular formula ratio = Molar mass of the compound / Empirical formula mass Molecular formula ratio = 20.34 g/mol / 14.03 g/mol = 1.45
Multiply the subscripts in the empirical formula by the molecular formula ratio to obtain the molecular formula.
- Molecular formula = C(1 * 1.45)H(2 * 1.45) = C1.45H2.9
Therefore, the molecular formula of the unknown hydrocarbon is C1.45H2.9.
Please note that the molecular formula is typically expressed with whole numbers. In this case, we can multiply all the subscripts by 2 to obtain the simplest whole number ratio:
- Molecular formula = C(1.45 * 2)H(2.9 * 2) = C2.9H5.8
So, another way to represent the molecular formula is C2.9H5.8.
I hope this helps! Let me know if you have any further questions.
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