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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306625 If two compounds have the same empirical formula but different molecular formula, they must have

1 different percentage composition.

2 different molecular weights.

3 same viscosity.

4 same vapour density.

Explanation:

Since, the molecular formula is $n$ times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306626 An organic compound has an empirical formula ${CH}_{2} O$ , its vapour density is 45 . The molecular formula of the compound is

1

{CH}_{2} O

2

C_{2} H_{5} O

3

C_{2} H_{2} O

4

C_{3} H_{6} O_{3}

Explanation:

Molecular weight $= 2 \times$ vapour density
$= 2 \times 45 = 90 g / mol$
Empirical formula weight
$= 30 (12 + 2 \times 1 + 16) g / mol$
$∴ n = \frac{m o l e c u l a r w e i g h t}{e m p i r i c a l f o r m u l a w e i g h t}$
$= \frac{90}{30} = 3$
$\begin{array}{l} ∴ M o l e c u l a r f o r m u l a o f t h e c o m p o u n d \\ = (C H_{2} O_{3}) = C_{3} H_{6} O_{3} \end{array}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306627 The empirical formula of an acid is $C H_{2} O_{2}$ , the probable molecular formula of acid may be:

1

C_{3} H_{6} O_{4}

2

C H_{2} O

3

C_{2} H_{4} O_{4}

4

C_{2} H_{4} O_{2}

Explanation:

The acid with empirical formula $C H_{2} O_{2}$ is formic acid, HCOOH.

KCET - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306628 An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1

C H_{2}

2

C H_{3}

3

C H_{4}

4

C H

Explanation:

Percentage of H = 100 – 78 = 22%
Element Mass Mole Simple ratio
C 78g $\frac{78}{12} = 6 .5$ $\frac{6 .5}{6 .5} = 1$
H 22g $\frac{22}{1} = 22$ $\frac{22}{6 .5} = 3 .38$
Empirical formula of organic compound $= C H_{3}$

NEET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306629 Find empirical formula of the compound, if $M = 68 %$ (atomic mass M = 34) and remaining $M = 32 %$ oxygen.

1

MO

2

M_{2} O

3

{MO}_{2}

4

M_{2} O_{3}

Explanation:

supporting img
Hence empirical formula of given compound is $M_{2} O_{2}$ or MO.

AIIMS - 2019

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306625 If two compounds have the same empirical formula but different molecular formula, they must have

1 different percentage composition.

2 different molecular weights.

3 same viscosity.

4 same vapour density.

Explanation:

Since, the molecular formula is $n$ times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306626 An organic compound has an empirical formula ${CH}_{2} O$ , its vapour density is 45 . The molecular formula of the compound is

1

{CH}_{2} O

2

C_{2} H_{5} O

3

C_{2} H_{2} O

4

C_{3} H_{6} O_{3}

Explanation:

Molecular weight $= 2 \times$ vapour density
$= 2 \times 45 = 90 g / mol$
Empirical formula weight
$= 30 (12 + 2 \times 1 + 16) g / mol$
$∴ n = \frac{m o l e c u l a r w e i g h t}{e m p i r i c a l f o r m u l a w e i g h t}$
$= \frac{90}{30} = 3$
$\begin{array}{l} ∴ M o l e c u l a r f o r m u l a o f t h e c o m p o u n d \\ = (C H_{2} O_{3}) = C_{3} H_{6} O_{3} \end{array}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306627 The empirical formula of an acid is $C H_{2} O_{2}$ , the probable molecular formula of acid may be:

1

C_{3} H_{6} O_{4}

2

C H_{2} O

3

C_{2} H_{4} O_{4}

4

C_{2} H_{4} O_{2}

Explanation:

The acid with empirical formula $C H_{2} O_{2}$ is formic acid, HCOOH.

KCET - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306628 An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1

C H_{2}

2

C H_{3}

3

C H_{4}

4

C H

Explanation:

Percentage of H = 100 – 78 = 22%
Element Mass Mole Simple ratio
C 78g $\frac{78}{12} = 6 .5$ $\frac{6 .5}{6 .5} = 1$
H 22g $\frac{22}{1} = 22$ $\frac{22}{6 .5} = 3 .38$
Empirical formula of organic compound $= C H_{3}$

NEET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306629 Find empirical formula of the compound, if $M = 68 %$ (atomic mass M = 34) and remaining $M = 32 %$ oxygen.

1

MO

2

M_{2} O

3

{MO}_{2}

4

M_{2} O_{3}

Explanation:

supporting img
Hence empirical formula of given compound is $M_{2} O_{2}$ or MO.

AIIMS - 2019

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306625 If two compounds have the same empirical formula but different molecular formula, they must have

1 different percentage composition.

2 different molecular weights.

3 same viscosity.

4 same vapour density.

Explanation:

Since, the molecular formula is $n$ times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306626 An organic compound has an empirical formula ${CH}_{2} O$ , its vapour density is 45 . The molecular formula of the compound is

1

{CH}_{2} O

2

C_{2} H_{5} O

3

C_{2} H_{2} O

4

C_{3} H_{6} O_{3}

Explanation:

Molecular weight $= 2 \times$ vapour density
$= 2 \times 45 = 90 g / mol$
Empirical formula weight
$= 30 (12 + 2 \times 1 + 16) g / mol$
$∴ n = \frac{m o l e c u l a r w e i g h t}{e m p i r i c a l f o r m u l a w e i g h t}$
$= \frac{90}{30} = 3$
$\begin{array}{l} ∴ M o l e c u l a r f o r m u l a o f t h e c o m p o u n d \\ = (C H_{2} O_{3}) = C_{3} H_{6} O_{3} \end{array}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306627 The empirical formula of an acid is $C H_{2} O_{2}$ , the probable molecular formula of acid may be:

1

C_{3} H_{6} O_{4}

2

C H_{2} O

3

C_{2} H_{4} O_{4}

4

C_{2} H_{4} O_{2}

Explanation:

The acid with empirical formula $C H_{2} O_{2}$ is formic acid, HCOOH.

KCET - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306628 An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1

C H_{2}

2

C H_{3}

3

C H_{4}

4

C H

Explanation:

Percentage of H = 100 – 78 = 22%
Element Mass Mole Simple ratio
C 78g $\frac{78}{12} = 6 .5$ $\frac{6 .5}{6 .5} = 1$
H 22g $\frac{22}{1} = 22$ $\frac{22}{6 .5} = 3 .38$
Empirical formula of organic compound $= C H_{3}$

NEET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306629 Find empirical formula of the compound, if $M = 68 %$ (atomic mass M = 34) and remaining $M = 32 %$ oxygen.

1

MO

2

M_{2} O

3

{MO}_{2}

4

M_{2} O_{3}

Explanation:

supporting img
Hence empirical formula of given compound is $M_{2} O_{2}$ or MO.

AIIMS - 2019

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306625 If two compounds have the same empirical formula but different molecular formula, they must have

1 different percentage composition.

2 different molecular weights.

3 same viscosity.

4 same vapour density.

Explanation:

Since, the molecular formula is $n$ times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306626 An organic compound has an empirical formula ${CH}_{2} O$ , its vapour density is 45 . The molecular formula of the compound is

1

{CH}_{2} O

2

C_{2} H_{5} O

3

C_{2} H_{2} O

4

C_{3} H_{6} O_{3}

Explanation:

Molecular weight $= 2 \times$ vapour density
$= 2 \times 45 = 90 g / mol$
Empirical formula weight
$= 30 (12 + 2 \times 1 + 16) g / mol$
$∴ n = \frac{m o l e c u l a r w e i g h t}{e m p i r i c a l f o r m u l a w e i g h t}$
$= \frac{90}{30} = 3$
$\begin{array}{l} ∴ M o l e c u l a r f o r m u l a o f t h e c o m p o u n d \\ = (C H_{2} O_{3}) = C_{3} H_{6} O_{3} \end{array}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306627 The empirical formula of an acid is $C H_{2} O_{2}$ , the probable molecular formula of acid may be:

1

C_{3} H_{6} O_{4}

2

C H_{2} O

3

C_{2} H_{4} O_{4}

4

C_{2} H_{4} O_{2}

Explanation:

The acid with empirical formula $C H_{2} O_{2}$ is formic acid, HCOOH.

KCET - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306628 An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1

C H_{2}

2

C H_{3}

3

C H_{4}

4

C H

Explanation:

Percentage of H = 100 – 78 = 22%
Element Mass Mole Simple ratio
C 78g $\frac{78}{12} = 6 .5$ $\frac{6 .5}{6 .5} = 1$
H 22g $\frac{22}{1} = 22$ $\frac{22}{6 .5} = 3 .38$
Empirical formula of organic compound $= C H_{3}$

NEET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306629 Find empirical formula of the compound, if $M = 68 %$ (atomic mass M = 34) and remaining $M = 32 %$ oxygen.

1

MO

2

M_{2} O

3

{MO}_{2}

4

M_{2} O_{3}

Explanation:

supporting img
Hence empirical formula of given compound is $M_{2} O_{2}$ or MO.

AIIMS - 2019

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306625 If two compounds have the same empirical formula but different molecular formula, they must have

1 different percentage composition.

2 different molecular weights.

3 same viscosity.

4 same vapour density.

Explanation:

Since, the molecular formula is $n$ times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306626 An organic compound has an empirical formula ${CH}_{2} O$ , its vapour density is 45 . The molecular formula of the compound is

1

{CH}_{2} O

2

C_{2} H_{5} O

3

C_{2} H_{2} O

4

C_{3} H_{6} O_{3}

Explanation:

Molecular weight $= 2 \times$ vapour density
$= 2 \times 45 = 90 g / mol$
Empirical formula weight
$= 30 (12 + 2 \times 1 + 16) g / mol$
$∴ n = \frac{m o l e c u l a r w e i g h t}{e m p i r i c a l f o r m u l a w e i g h t}$
$= \frac{90}{30} = 3$
$\begin{array}{l} ∴ M o l e c u l a r f o r m u l a o f t h e c o m p o u n d \\ = (C H_{2} O_{3}) = C_{3} H_{6} O_{3} \end{array}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306627 The empirical formula of an acid is $C H_{2} O_{2}$ , the probable molecular formula of acid may be:

1

C_{3} H_{6} O_{4}

2

C H_{2} O

3

C_{2} H_{4} O_{4}

4

C_{2} H_{4} O_{2}

Explanation:

The acid with empirical formula $C H_{2} O_{2}$ is formic acid, HCOOH.

KCET - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306628 An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]

1

C H_{2}

2

C H_{3}

3

C H_{4}

4

C H

Explanation:

Percentage of H = 100 – 78 = 22%
Element Mass Mole Simple ratio
C 78g $\frac{78}{12} = 6 .5$ $\frac{6 .5}{6 .5} = 1$
H 22g $\frac{22}{1} = 22$ $\frac{22}{6 .5} = 3 .38$
Empirical formula of organic compound $= C H_{3}$

NEET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306629 Find empirical formula of the compound, if $M = 68 %$ (atomic mass M = 34) and remaining $M = 32 %$ oxygen.

1

MO

2

M_{2} O

3

{MO}_{2}

4

M_{2} O_{3}

Explanation:

supporting img
Hence empirical formula of given compound is $M_{2} O_{2}$ or MO.

AIIMS - 2019

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