Butyric acid contains only C , H and O . A 4 . 24 mg sample of butyric acid is completely burnt. It gives 8 . 45 mg of CO 2 and 3 . 46 mg of H 2 O . The molecular mass of butyric acid was determined by experiment to be 88 amu . What is molecular formula?

Given, 4 . 24 mg butyric acid gives 8 . 45 mg CO 2 and 3 . 46 mg H 2 O . Molecular mass of CO 2 = 12 + 2 × 16 = 12 + 32 = 44 g mol - 1 . 44 g of CO 2 contains 12 g C . ∴ 44 mg of CO 2 contains 12 mg C . Thus, 1 mg of CO 2 contains 12 44 × 1 = 0 . 273 mg C . 8 . 45 mg of CO 2 contains 0 . 273 × 8 . 45 = 2 . 306 mg C . It is present in 4 . 24 mg compound. Therefore, the percentage of carbon in the given compound will be as follows, % of C = 2 . 306 4 . 24 × 100 = 54 . 4 % . Molecular mass of H 2 O = 2 × 1 + 16 = 18 g mol - 1 18 g of H 2 O contains 2 g H . ∴ 18 mg of H 2 O contains 2 mg H . Thus, 3 . 46 mg of H 2 O contains 2 18 × 3 . 46 = 0 . 384 mg H . Therefore, the percentage of hydrogen in the given compound will be as follows, % of H = 0 . 384 4 . 24 × 100 = 9 . 1 % . The percentage of oxygen, % O = 100 - % C + % H % O = 100 - 54 . 4 + 9 . 1 % O = 100 - 63 . 5 = 36 . 5 C = 54 . 4 12 = 4 . 53 ⇒ 4 . 53 2 . 28 = 1 . 99 ⇒ 2 . H = 9 . 1 1 = 9 . 1 ⇒ 9 . 1 2 . 28 = 3 . 99 ⇒ 4 . O = 36 . 5 16 = 2 . 28 ⇒ 2 . 28 2 . 28 = 1 ⇒ 1 . C : H : O = 2 : 4 : 1 . Empirical formula = C 2 H 4 O 1 Empirical formula mass C 2 H 4 O 1 = 2 × 12 + 4 × 1 + 16 = 24 + 4 + 16 = 44 u . Molecular mass of butyric acid = 88 u . n = Molecular mass Empirical formula mass n = 88 44 = 2 . Therefore, the molecular formula, M . F = n × E . F M . F = 2 × C 2 H 4 O M . F = C 4 H 8 O 2 .

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IMPORTANT

Earn 100

Calculate the empirical formula of gold chloride which contains $35.1 %$ of chlorine. At. Mass of $Au = 197$ .

Important Questions on Some Basic Concepts of Chemistry

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 6

Calculate the empirical formula of a mineral having the following compositions.

$CaO = 48.0 %; P_{2} O_{5} = 41.3 %; {CaCl}_{2} = 10.7 %;$

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 1

A crystalline salt when heated becomes anhydrous and loses $51.2 %$ of its weight. The anhydrous salt on analysis gave the percentage composition as:

$Mg = 20.0 %; S = 26.66 %; O = 53.33 %;$ Calculate the molecular formula of the anhydrous salt and the crystalline salt. Molecular mass of the anhydrous salt is $120$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 2

A compound containing carbon, hydrogen and oxygen gave the following analytical data: $C = 40.0 %$ and $H = 6.67 %$ .

Calculate the molecular formula of the compound if its molecular mass is $180$ .

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11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 3

On analysis, a substance was found to have the following percentage compositions:

K = 31.84; Cl = 28.98; O = 39.18;

Calculate its molecular formula if its molecular mass is

122.5 g {mol}^{- 1}

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 4

An organic liquid having carbon, hydrogen, nitrogen and oxygen was found to contain

C = 41.37 %; H = 5.75 %; N = 16.09 %;

and the rest oxygen. Calculate the molecular formula of the liquid if its vapour density (

V . D

) is

43.3

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 5

A chemical compound is found to have the following composition:

C = 19.57 %; Fe = 15.2 %; N = 22.83 %; K = 42.39 %;

Calculate the empirical formula of the compound. What will be its molecular formula if the molecular mass of the compound is

368

? Name the compound.

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 6

Butyric acid contains only

C, H

and

O

. A

4.24 mg

sample of butyric acid is completely burnt. It gives

8.45 mg

{CO}_{2}

and

3.46 mg

H_{2} O

. The molecular mass of butyric acid was determined by experiment to be

88 amu

. What is molecular formula?

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 1 - Some Basic Concepts of Chemistry>PROBLEMS FOR PRACTICE>Q 1

Balance the following equations by Hit and Trial Method:

{SO}_{2} + H_{2} S \to S + H_{2} O

Calculate the empirical formula of gold chloride which contains 35.1 % of chlorine. At. Mass of Au=197.

Important Questions on Some Basic Concepts of Chemistry

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Calculate the empirical formula of gold chloride which contains $35.1 %$ of chlorine. At. Mass of $Au = 197$ .