In a compound,C,H,and N atoms are present in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. Calculate the relative number of moles of each element by dividing the weight ratio by their respective atomic weights: $C = 9/12 = 0.75$,$H

Vedclass · Accepted Answer

$C_6H_8N_2$

Vedclass · Answer

(C) $1$. Calculate the relative number of moles of each element by dividing the weight ratio by their respective atomic weights: $C = 9/12 = 0.75$,$H = 1/1 = 1$,$N = 3.5/14 = 0.25$. $2$. Determine the simplest molar ratio by dividing by the smallest value $(0.25)$: $C = 0.75/0.25 = 3$,$H = 1/0.25 = 4$,$N = 0.25/0.25 = 1$. $3$. The empirical formula is $C_3H_4N$. $4$. Calculate the empirical formula weight: $(3 	imes 12) + (4 	imes 1) + (1 	imes 14) = 36 + 4 + 14 = 54$. $5$. Determine the value of $n = 	ext{Molecular weight} / 	ext{Empirical formula weight} = 108 / 54 = 2$. $6$. The molecular formula is $n 	imes (C_3H_4N) = 2 	imes (C_3H_4N) = C_6H_8N_2$.

In a compound,$C$,$H$,and $N$ atoms are present in $9 : 1 : 3.5$ by weight. The molecular weight of the compound is $108$. What is the molecular formula of the compound?

Similar Questions

The percentage of nitrogen by mass in ammonium sulphate is closest to $.....$ $\%$ (atomic masses of $H = 1, N = 14, O = 16, S = 32$ ).

Percentage of gold in $14$ carat gold is (in $\%$)

$120 \ g$ of an organic compound that contains only carbon and hydrogen gives $330 \ g$ of $CO_{2}$ and $270 \ g$ of water on complete combustion. The percentage of carbon and hydrogen,respectively,are ...... .

An organic compound containing $C$,$H$,and $N$ gave the following analysis: $C = 40\%$,$H = 13.33\%$,and $N = 46.67\%$. Its empirical formula would be:

When equal weights of the two fertilizers,urea and ammonium sulphate are taken,urea contains

Vedclass Test Series

Exam Paper Generator

Online Exam Module