W(C) =53.33%
w(H) =15.56%
w(N)=31.11%
D(Ne)=2.25
Найти:CxHyNz

Предыдущий
вопрос Следующий
вопрос

To find the empirical formula of the compound, we first need to convert the weight percentages into moles.

Given:
W(C) = 53.33%
w(H) = 15.56%
w(N) = 31.11%

Let's assume we have 100g of the compound:

Mass of C in 100g = 53.33gMass of H in 100g = 15.56gMass of N in 100g = 31.11g

Now, let's convert the masses of the elements into moles using their atomic masses:

Moles of C = 53.33g / 12.01g/mol ≈ 4.44 molesMoles of H = 15.56g / 1.01g/mol ≈ 15.41 molesMoles of N = 31.11g / 14.01g/mol ≈ 2.22 moles

Next, we need to find the simplest ratio of the moles of the elements by dividing each value by the smallest number of moles (2.22 moles in this case):

C ≈ 4.44 / 2.22 ≈ 2H ≈ 15.41 / 2.22 ≈ 7N = 2.22 / 2.22 = 1

Therefore, the empirical formula of the compound is C2H7N.