Question

Массовая доля углерода и водорода в веществе =92,31% и 7,69% Плотность паров по водороду =39 . Формула -?

Answer 1

Дано:

w (C) = 92,31%

w (H) = 7,69%

D(H2) = 39

Найти: CxHy

Решение:

Mr (в-ва) = 39∙ 2 = 78

n (э) = w(э) ∙ Mr(в-ва) \Ar(э) ∙ 100%

n (C) = 92,31% ∙ 78 \12 ∙ 100% = 6

n (H) = 7,69% ∙ 78\1 ∙ 100% = 6

n (C) : n (H) = 6:6 = CxHy =  С6H6

ответ: С6H6 – бензол

Answer 2

Chemical Composition and Vapor Density Calculation

The given substance has a mass fraction of 92.31% carbon and 7.69% hydrogen. The vapor density with respect to hydrogen is 39.

To find the formula of the substance, we can use the given mass fractions of carbon and hydrogen to calculate the empirical formula.

Using the given mass fractions, we can calculate the number of moles of carbon and hydrogen in 100g of the substance. Then, we can find the ratio of moles of carbon to moles of hydrogen and use it to determine the empirical formula.

The calculation can be done as follows:

1. Calculate the moles of carbon: - \( \text{Moles of Carbon} = \frac{\text{Mass % of Carbon}}{\text{Molar mass of Carbon}} \)

2. Calculate the moles of hydrogen: - \( \text{Moles of Hydrogen} = \frac{\text{Mass % of Hydrogen}}{\text{Molar mass of Hydrogen}} \)

3. Determine the ratio of moles of carbon to moles of hydrogen.

Let's perform the calculations:

- Moles of Carbon: - \( \text{Moles of Carbon} = \frac{92.31}{12} = 7.692 \)

- Moles of Hydrogen: - \( \text{Moles of Hydrogen} = \frac{7.69}{1} = 7.69 \)

- Ratio of moles of carbon to moles of hydrogen: - \( \frac{7.692}{7.69} \approx 1:1 \)

Based on the calculated ratio, the empirical formula of the substance is CH.

Conclusion

Therefore, the empirical formula of the substance is CH.