An organic compound on analysis gives the fol | vedclass

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

Question

(B) $1$. Calculate the percentage of oxygen: $O = 100 - (57.82 + 3.6) = 38.58 \%$. $2$. Determine the empirical formula: $C: \frac{57.82}{12} = 4.818

Vedclass · Accepted Answer

$C_8H_6O_4$

Vedclass · Answer

(B) $1$. Calculate the percentage of oxygen: $O = 100 - (57.82 + 3.6) = 38.58 \%$. $2$. Determine the empirical formula: $C: \frac{57.82}{12} = 4.818 \Rightarrow \frac{4.818}{2.41} = 2$ $H: \frac{3.6}{1} = 3.6 \Rightarrow \frac{3.6}{2.41} = 1.5 = \frac{3}{2}$ $O: \frac{38.58}{16} = 2.41 \Rightarrow \frac{2.41}{2.41} = 1$ Multiplying by $2$ to get whole numbers: $C = 4, H = 3, O = 2$. Empirical Formula ($E$.$F$.) = $C_4H_3O_2$. $3$. Calculate Empirical Formula Mass ($E$.$F$.$M$.): $(4 	imes 12) + (3 	imes 1) + (2 	imes 16) = 48 + 3 + 32 = 83$. $4$. Calculate Molecular Formula Mass ($M$.$F$.$M$.): $2 	imes 	ext{Vapour Density} = 2 	imes 83 = 166$. $5$. Calculate $n$: $n = \frac{M.F.M.}{E.F.M.} = \frac{166}{83} = 2$. $6$. Molecular Formula = $n 	imes E.F. = 2 	imes C_4H_3O_2 = C_8H_6O_4$.

An organic compound on analysis gives the following data: $C = 57.82 \%$,$H = 3.6 \%$ and the rest is oxygen. Its vapour density is $83$. Calculate its molecular formula.

Similar Questions

An organic compound on analysis gave $C = 48 \,g$, $H = 8 \,g$ and $N = 56 \,g$. The volume of $1.0 \,g$ of the compound was found to be $200 \,mL$ at $NTP$. What is the molecular formula of the compound?

On complete combustion,$0.492 \ g$ of an organic compound gave $0.792 \ g$ of $CO_2$. The $\%$ of carbon in the organic compound is $..................$ (Nearest integer)

Hemoglobin contains $0.34 \%$ of iron by mass. The number of $Fe$ atoms in $3.3 \ g$ of hemoglobin is. (Given : Atomic mass of $Fe$ is $56 \ u, N_A = 6.022 \times 10^{23} \ mol^{-1}$)

$A$ gas is found to contain $2.8 \ g$ of nitrogen and $3.2 \ g$ of oxygen. The simplest formula of the compound is:

An organic compound containing $C$ and $H$ has $92.3 \%$ of carbon. Its empirical formula is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module