An inorganic compound contains Carbon $12.6 %$ , Hydrogen $2.13 %$ and Bromine $85.11 %$ . Determine the empirical formula of the compound. $[C = 12, H = 1, B r = 80]$

C H_{2} B r

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{3} B r_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{2} B r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H B r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $C H_{2} B r$
Let the compound is of $100 g$
$∴$ Carbon= $12.6$ % = $12.6 g$ = $\frac{12.6}{12}$ = $1.05$ moles
Hydrogen= $2.13$ % = $2.13 g$ = $2.13$ moles
Bromine= $85.11$ % = $\frac{85.11}{80}$ = $1.06$ moles
$∴$ $n_{C} {: n}_{H} {: n}_{B r}$ = $1.05 : 2.13 : 1.06$
= $1 : 2 : 1$
$∴$ Empirical formula= $C H_{2} B r$

Suggest Corrections

Similar questions

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula.
[Atomic mass: C = 12, H = 1, Br = 80]

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13% and Br = 85.11%. Find the molecular formula.

[Atomic mass: C = 12, H = 1, Br = 80]

An organic compound has the following percentage composition: C = 12.76%, H = 2.13%, Br = 85.11%. The vapour density of the compound is 94. Find out its molecular formula [C = 12, H = 1, Br = 80].

A compound X consists of 4.8% carbon and 95.2% bromine by mass.

Determine the empirical formula of this compound working correct to one decimal place.

[C= 12; Br = 80]

A compound X consists fo 4.8 % carbon and 95.2 % bromine by mass:
(i) Determine the empirical formula of this compound working correctly to one decimal place (C = 12; Br = 80).
(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?

Join BYJU'S Learning Program

An inorganic compound contains Carbon 12.6%, Hydrogen 2.13% and Bromine 85.11%. Determine the empirical formula of the compound. [C=12,H=1,Br=80]

The correct option is A CH2BrLet the compound is of 100g∴ Carbon= 12.6% =12.6g =12.612 =1.05 molesHydrogen= 2.13% =2.13g =2.13 molesBromine= 85.11% =85.1180 =1.06 moles∴ nC:nH:nBr= 1.05:2.13:1.06 =1:2:1∴ Empirical formula= CH2Br

An inorganic compound contains Carbon $12.6 %$ , Hydrogen $2.13 %$ and Bromine $85.11 %$ . Determine the empirical formula of the compound. $[C = 12, H = 1, B r = 80]$