A compound has the following percentage composition by mass-

Carbon-14.4%

Hydrogen-1.2%

Chlorine-85.5%

The relative molecular mass of this compound is 168.50 what ia is molecular formula?

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Solution

Element Percentage composition Atomic weight Relative No. of atoms Simplest ratio

Carbon 14.4 12 14.4/12 = 1.2 1.2/1.2 = 1

Hydrogen 1.2 1 1.2/1 = 1.2 1.2/1.2 = 1

Chlorine 84.5 35.5 84.5/35.5 = 2.38 2.38/1.2 = 2

Hence empirical formula is CHCl₂.
n = Molecular mass/Empirical formula mass
= 168/(12 + 1 + 70)
= 168/83
= 2.
Therefore, molecular formula = (CHCl₂)₂ = C₂H2Cl₄.

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Similar questions

Q. A compound contains carbon 14.4 %, hydrogen 1.2 % and chlorine 84.5 %. The relative molecular mass of this compound is 168, its molecular formula is:

[C = 12, H = 1, C l = 35.5]

(a) (i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)
(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?
(iii) By what type of reaction could this compound be obtained from ethyne?

(b) From the equation:
C + 2H₂SO₄ $\overset{}{\to}$ CO₂ + 2H₂O + 2SO₂
Calculate:
(i) the mass of carbon oxidised by 49 g of sulphuric acid (C = 12; relative molecular mass of sulphuric acid = 98).
(ii) the volume of sulphur dioxide measured at STP, liberated at the same time.
(Volume occupied by 1 mole of a gas at STP is 22.4 dm³).

A compound has the following percentage composition by mass: carbon 14.4%,hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. [H = 1;C=12;Cl = 35.5]

(a) from the equation"
$C + 2 H_{2} S O_{4} \to C O_{2} + 2 H_{2} O + 2 S O_{2}$
Calculate:
(i) hte mass of carbon oxidized by 49 g of sulphuric acid.
(ii) the volume of sulphur dioxide measured at STP, linerated at the same time.
(b) (i) A compound has the following percentage composition by mass; carbon 14.4 % hydrogen 1.2% and chlorine 84.5 % . Determine the empirical formula of this compound. Work correct to 1 decimal place. (H - 1; C = 12; Cl = 35.5)
(ii) The relativ molecular mass of this compound is 168, so what is its molecular formuka?

(a) Calculate the percentage of boron (B) in borax (Na₂B₄O₇·10H₂O). (H = 1, B = 11, O = 16, Na = 23).
(b) (i) The compound A has the following percentage composition by mass:
Carbon = 26.7%, oxygen = 71.1%, hydrogen = 2.2%. Determine the empirical formula of A (H = 1, C = 12, O = 16)
(ii) If the relative molecular mass of A is 90, what is the molecular formula of A?

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Element	Percentage composition	Atomic weight	Relative No. of atoms	Simplest ratio
Carbon	14.4	12	14.4/12 = 1.2	1.2/1.2 = 1
Hydrogen	1.2	1	1.2/1 = 1.2	1.2/1.2 = 1
Chlorine	84.5	35.5	84.5/35.5 = 2.38	2.38/1.2 = 2

A compound has the following percentage composition by mass- Carbon-14.4% Hydrogen-1.2% Chlorine-85.5% The relative molecular mass of this compound is 168.50 what ia is molecular formula?

A compound has the following percentage composition by mass-

Carbon-14.4%

Hydrogen-1.2%

Chlorine-85.5%

The relative molecular mass of this compound is 168.50 what ia is molecular formula?