Question

При взаимодействии 1.74 грамм алкана с бромом образовалось 4.11 грамм монобромпроизводного.Определите молекулярную формулу алкана

Answer 1

Общая формула алканов

СnH2n+2

Запишем реакцию галогенирования – взаимодействие алкана с бромом

СnH2n+2 + Br2 ⇒CnH2n+1Br + HBr

n(CnH2n+2) = n(CnH2n+1Br) составим соотношение: 1,74 / (14n+2) = 4,11 / (14n+81)

57,54n + 8,22 = 24,36n + 140,94

33,18n = 132,72

n = 4            C4H10

Answer 2

Determining the Molecular Formula of the Alkane

To determine the molecular formula of the alkane, we need to use the given information about the mass of the alkane and the mass of the monobrominated product formed.

According to the given information: - The mass of the alkane is 1.74 grams. - The mass of the monobrominated product is 4.11 grams.

To find the molecular formula, we need to calculate the empirical formula first. The empirical formula represents the simplest whole-number ratio of atoms in a compound.

To calculate the empirical formula, we can use the molar masses of the elements involved. However, since we don't have the molar masses of the alkane and the monobrominated product, we can't directly calculate the empirical formula.

Without additional information or assumptions, it is not possible to determine the molecular formula of the alkane based solely on the given information.

If you have any additional information or assumptions, please provide them, and I'll be happy to assist you further.