Home
Class 11
CHEMISTRY
A compound on analysis gave the followin...

A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09% 0 = 36.36. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.

Text Solution

Verified by Experts

Calculation of empirical formula

Empirical formula is `C_2H_4O`.
Calculation of Moleuclar formula
Empirical formula mass `= 12 xx 2 + 1 xx 4 + 16 xx 1 = 44 `
Molecular mass `= 2 xx ` Vapour density
` =2 xx 44 =88`
`n=(" Molecular mass ")/("Empirical Formula mass ")=(88)/(44)=2`
Molecular formula = Empirical formula ` xx ` n
`=C_2H_4O xx 2 `
`= C_4 H_(8)O_2`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL CALCULATION

    NCERT TAMIL|Exercise PROBLEM|4 Videos

  • CHEMICAL CALCULATION

    NCERT TAMIL|Exercise EXAMPLE|3 Videos

  • CHEMICAL BONDING

    NCERT TAMIL|Exercise EVALUATION |30 Videos

  • CHEMICAL EQUILIBRIUM - I

    NCERT TAMIL|Exercise QUESTION|22 Videos

Similar Questions

Explore conceptually related problems

A compound on analysis gave the following percentage composition: C = 24.47%, H= 4.07 %, Cl = 71.65%. Find out its empirical formula.

An organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44

A compound on analysis gave the following percentage composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

A laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen. Find out the Empirical Formula of the compound.

A substance on analysis, gave the following percentage composition, Na = 43.4%, C = 11.3%, 0 = 43.3% calculate its empirical formula [Na = 23, C = 12, O = 16].

An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?