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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains $C = 74.0 %,$ $H = 8.65 %$ and $N = 17.3 %$ . Its Empirical formula is

1

C_{5} H_{8} N

2

C_{10} H_{12} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 74$

$74 / 12 = 6.1$

$6.1 / 1.2 = 5.08$ or $5$

$H = 8.65$

$8.65 / 1 = 8.65$

$8.6 / 1.2 = 7.16$ or $7$

$N = 17.3$

$17.3 / 14 = 1.2$

$1.2 / 1.2 =$ $1$ or $1$

Therefore Empirical formula $= C_{5} H_{7} N$ .

GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method

2 Diffusion method

3 Vapour pressure method

4 Victor Meyer's method

GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight

\times

basicity

2

\frac{E q u i v a l e n t w e i g h t}{B a s i c i t y}

3

\frac{B a s i c i t y}{E q u i v a l e n t w e i g h t}

4 Equivalent weight

\times

valency

Explanation:

(a) Mol. mass of an acid $=$ Equivalent wt. $\times$ basicity.

GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition

2 Different molecular weight

3 Same viscosity

4 Same vapour density

GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is $C_{2} H_{5} O$ and its molecular weight is $90$ . Molecular formula of the compound is

1

C_{2} H_{5} O

2

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

3

C_{4} H_{10} O_{2}

4

C_{5} H_{14} O

Explanation:

(c)Empirical formula mass = $C_{2} H_{5} O$ = $24 + 5 + 16 = 45.$
$n = \frac{M o l . m a s s}{E m p . m a s s} = \frac{90}{45} = 2$
Mol. formula = $(C_{2} H_{5} O)_{2} = C_{4} H_{10} O_{2}$ .

GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains $C = 74.0 %,$ $H = 8.65 %$ and $N = 17.3 %$ . Its Empirical formula is

1

C_{5} H_{8} N

2

C_{10} H_{12} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 74$

$74 / 12 = 6.1$

$6.1 / 1.2 = 5.08$ or $5$

$H = 8.65$

$8.65 / 1 = 8.65$

$8.6 / 1.2 = 7.16$ or $7$

$N = 17.3$

$17.3 / 14 = 1.2$

$1.2 / 1.2 =$ $1$ or $1$

Therefore Empirical formula $= C_{5} H_{7} N$ .

GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method

2 Diffusion method

3 Vapour pressure method

4 Victor Meyer's method

GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight

\times

basicity

2

\frac{E q u i v a l e n t w e i g h t}{B a s i c i t y}

3

\frac{B a s i c i t y}{E q u i v a l e n t w e i g h t}

4 Equivalent weight

\times

valency

Explanation:

(a) Mol. mass of an acid $=$ Equivalent wt. $\times$ basicity.

GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition

2 Different molecular weight

3 Same viscosity

4 Same vapour density

GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is $C_{2} H_{5} O$ and its molecular weight is $90$ . Molecular formula of the compound is

1

C_{2} H_{5} O

2

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

3

C_{4} H_{10} O_{2}

4

C_{5} H_{14} O

Explanation:

(c)Empirical formula mass = $C_{2} H_{5} O$ = $24 + 5 + 16 = 45.$
$n = \frac{M o l . m a s s}{E m p . m a s s} = \frac{90}{45} = 2$
Mol. formula = $(C_{2} H_{5} O)_{2} = C_{4} H_{10} O_{2}$ .

GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains $C = 74.0 %,$ $H = 8.65 %$ and $N = 17.3 %$ . Its Empirical formula is

1

C_{5} H_{8} N

2

C_{10} H_{12} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 74$

$74 / 12 = 6.1$

$6.1 / 1.2 = 5.08$ or $5$

$H = 8.65$

$8.65 / 1 = 8.65$

$8.6 / 1.2 = 7.16$ or $7$

$N = 17.3$

$17.3 / 14 = 1.2$

$1.2 / 1.2 =$ $1$ or $1$

Therefore Empirical formula $= C_{5} H_{7} N$ .

GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method

2 Diffusion method

3 Vapour pressure method

4 Victor Meyer's method

GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight

\times

basicity

2

\frac{E q u i v a l e n t w e i g h t}{B a s i c i t y}

3

\frac{B a s i c i t y}{E q u i v a l e n t w e i g h t}

4 Equivalent weight

\times

valency

Explanation:

(a) Mol. mass of an acid $=$ Equivalent wt. $\times$ basicity.

GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition

2 Different molecular weight

3 Same viscosity

4 Same vapour density

GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is $C_{2} H_{5} O$ and its molecular weight is $90$ . Molecular formula of the compound is

1

C_{2} H_{5} O

2

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

3

C_{4} H_{10} O_{2}

4

C_{5} H_{14} O

Explanation:

(c)Empirical formula mass = $C_{2} H_{5} O$ = $24 + 5 + 16 = 45.$
$n = \frac{M o l . m a s s}{E m p . m a s s} = \frac{90}{45} = 2$
Mol. formula = $(C_{2} H_{5} O)_{2} = C_{4} H_{10} O_{2}$ .

GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains $C = 74.0 %,$ $H = 8.65 %$ and $N = 17.3 %$ . Its Empirical formula is

1

C_{5} H_{8} N

2

C_{10} H_{12} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 74$

$74 / 12 = 6.1$

$6.1 / 1.2 = 5.08$ or $5$

$H = 8.65$

$8.65 / 1 = 8.65$

$8.6 / 1.2 = 7.16$ or $7$

$N = 17.3$

$17.3 / 14 = 1.2$

$1.2 / 1.2 =$ $1$ or $1$

Therefore Empirical formula $= C_{5} H_{7} N$ .

GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method

2 Diffusion method

3 Vapour pressure method

4 Victor Meyer's method

GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight

\times

basicity

2

\frac{E q u i v a l e n t w e i g h t}{B a s i c i t y}

3

\frac{B a s i c i t y}{E q u i v a l e n t w e i g h t}

4 Equivalent weight

\times

valency

Explanation:

(a) Mol. mass of an acid $=$ Equivalent wt. $\times$ basicity.

GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition

2 Different molecular weight

3 Same viscosity

4 Same vapour density

GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is $C_{2} H_{5} O$ and its molecular weight is $90$ . Molecular formula of the compound is

1

C_{2} H_{5} O

2

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

3

C_{4} H_{10} O_{2}

4

C_{5} H_{14} O

Explanation:

(c)Empirical formula mass = $C_{2} H_{5} O$ = $24 + 5 + 16 = 45.$
$n = \frac{M o l . m a s s}{E m p . m a s s} = \frac{90}{45} = 2$
Mol. formula = $(C_{2} H_{5} O)_{2} = C_{4} H_{10} O_{2}$ .

GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains $C = 74.0 %,$ $H = 8.65 %$ and $N = 17.3 %$ . Its Empirical formula is

1

C_{5} H_{8} N

2

C_{10} H_{12} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 74$

$74 / 12 = 6.1$

$6.1 / 1.2 = 5.08$ or $5$

$H = 8.65$

$8.65 / 1 = 8.65$

$8.6 / 1.2 = 7.16$ or $7$

$N = 17.3$

$17.3 / 14 = 1.2$

$1.2 / 1.2 =$ $1$ or $1$

Therefore Empirical formula $= C_{5} H_{7} N$ .

GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method

2 Diffusion method

3 Vapour pressure method

4 Victor Meyer's method

GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight

\times

basicity

2

\frac{E q u i v a l e n t w e i g h t}{B a s i c i t y}

3

\frac{B a s i c i t y}{E q u i v a l e n t w e i g h t}

4 Equivalent weight

\times

valency

Explanation:

(a) Mol. mass of an acid $=$ Equivalent wt. $\times$ basicity.

GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition

2 Different molecular weight

3 Same viscosity

4 Same vapour density

GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is $C_{2} H_{5} O$ and its molecular weight is $90$ . Molecular formula of the compound is

1

C_{2} H_{5} O

2

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

3

C_{4} H_{10} O_{2}

4

C_{5} H_{14} O

Explanation:

(c)Empirical formula mass = $C_{2} H_{5} O$ = $24 + 5 + 16 = 45.$
$n = \frac{M o l . m a s s}{E m p . m a s s} = \frac{90}{45} = 2$
Mol. formula = $(C_{2} H_{5} O)_{2} = C_{4} H_{10} O_{2}$ .