English

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]

Advertisements
Advertisements

Question

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]

Numerical
Advertisements

Solution

Element Percentage Molecules Simple ratio Ration Simple whole ratio
Carbon 82.76 `82.76/12` = 6.9 `6.89/6.89` = 1 `6.9/6.9` = 1 2
Hydrogen 17.24 `17.24/1` = 17.24 `17.24/6.89` = 2.5 `17.24/6.9` = 2.5 5

Empirical formula = C2H5

Empirical formula mass = (12 × 2) + (1 × 5)

= 24 + 5 

= 29

Vapour density = 29    ...(Given)

Molecular weight = 2 × Vapour density

Molecular weight = 2 × 29

= 58 gm

Molecular formula mass = n × Empirical formula mass

⇒ n = `"Molecular formula mass"/"Empirical formula mass"`

= `58/29`

= 2

Molecular formula = n × Empirical formula

= 2 × C2H5

= C4H10

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Mole Concept and Stoichiometry - EXERCISE [Page 109]

APPEARS IN

Lakhmir Singh Chemistry [English] Class 10 ICSE
Chapter 5 Mole Concept and Stoichiometry
EXERCISE | Q 21. | Page 109

RELATED QUESTIONS

Identify the term or substance based on the descriptions given below:

The property by virtue of which the compound has the same molecular formula but different structural formulae.


Consider the following reaction and based on the reaction answer the questions that follow:

Calculate:

1) the quantity in moles of (NH4)2Cr2O7 if 63gm of(NH4)2Cr2O7 is heated.

2) the quantity in moles of nitrogen formed.

3) the volume in liters or dm3 of N2 evolved at S.T.P.

4) the mass in grams of Cr2O3 formed at the same time

(Atomic masses: H=1, Cr= 52, N=14]


Explain the terms empirical formula and molecular formula.


Give example of compound whose:
Empirical formula is different from the molecular formula.


A compound of lead has following percentage composition, Pb = 90.66%, O = 9.34%. Calculate empirical formula of a compound. [Pb = 207, O = 16]


A compound contains 87.5% by mass of nitrogen and 12.5% by mass of hydrogen. Determine the empirical formula of this compound.


The compound A has the following percentage composition by mass: C =26.7%, O = 71.1%, H = 2.2%.
Determine the empirical formula of A.(Answer to one decimal place)  (H=1,C=12,O=16)


Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen.[N=14,O=16,H=1]


A compound 'X' consists of 4.8% of C and 95.2% of Br by mass.
Determine the empirical formula of this compound.
[C = 12, Br = 80]


Pratik heated 11.2 grams of element ‘M’ (atomic weight 56) with 4.8 grams of element 'N' (atomic weight 16) to form a compound. Find the empirical formula of the compound obtained by Pratik.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×