116 g of a substance upon dissociation react | vedclass

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

Question

(C) Calculate the moles of each element present in $116 \ g$ of the substance:Moles of $H = \frac{7.5 \ g}{1 \ g/mol} = 7.5 \ mol$Moles of $O = \frac{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Calculate the moles of each element present in $116 \ g$ of the substance:Moles of $H = \frac{7.5 \ g}{1 \ g/mol} = 7.5 \ mol$Moles of $O = \frac{60 \ g}{16 \ g/mol} = 3.75 \ mol$Moles of $C = \frac{48.5 \ g}{12 \ g/mol} \approx 4.04 \ mol$Dividing by the smallest value $(3.75)$:$H: \frac{7.5}{3.75} = 2$$O: \frac{3.75}{3.75} = 1$$C: \frac{4.04}{3.75} \approx 1.08 \approx 1$The empirical formula is $CH_2O$.Now,check the given options for the empirical formula $CH_2O$:$(A)$ $CH_3COOH$ is $C_2H_4O_2$,which simplifies to $CH_2O$.$(B)$ $HCHO$ is $CH_2O$,which simplifies to $CH_2O$.$(C)$ $CH_3OOCH_3$ is $C_2H_6O_2$,which does not simplify to $CH_2O$.$(D)$ $CH_3CHO$ is $C_2H_4O$,which does not simplify to $CH_2O$.Thus,$2$ formulae ($A$ and $B$) agree with the empirical formula $CH_2O$.

Similar Questions

$A$ gas is found to have a formula $[CO]_x$. If its vapour density is $70$,the value of $x$ is

The formula which represents the simplest ratio of atoms of different elements present in a molecule of a substance is called:

$C_4H_6O_2$ does not represent :

$4 \ g$ of a hydrocarbon on complete combustion gave $12.571 \ g$ of $CO_2$ and $5.143 \ g$ of water. What is the empirical formula of the hydrocarbon?

$A$ hydrocarbon contains $10.5 \, g$ of carbon and $1 \, g$ of hydrogen. If the volume of $2.4 \, g$ of this hydrocarbon is $1 \, L$ at $127 \, ^\circ C$ and $1 \, atm$ pressure,then the hydrocarbon is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module