QBManager

GENERAL ORGANIC CHEMISTRY

25275 $60 g$ of a compound on analysis gave $C = 24 g,$ $H = 4 g$ and $O = 32 g$ . Its Empirical formula is

1

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

2

C_{2} H_{2} O

3

C H_{2} O_{2}

4

C H_{2} O

Explanation:

(d)
Element No. of Moles Simple Ratio

$C = 24$

$24 / 12 = 2$

$1$

$H = 4$

$4 / 1 = 4$

$2$

$O = 32$

$32 / 16 = 2$

$1$

Therefore $C H_{2} O$ .

GENERAL ORGANIC CHEMISTRY

25276 An organic compound contains $C = 38.8 %,$ $H = 16 %$ and $N = 45.2 %$ . Empirical formula of the compound is

1

C H_{3} N H_{2}

2

C H_{3} C N

3

C_{2} H_{5} C N

4

C H_{2} (N H)_{2}

Explanation:

(a)
Element No. of Moles Simple Ratio

$C = 38.8$

$38.8 / 12 = 3.2$

$1$

$H = 16$

$16 / 1 = 16$

$5$

$N = 45.2$

$45.2 / 14 = 3.2$

$1$

Therefore, Empirical formula = $C H_{5} N$ or $C H_{3} N H_{2}$ .

GENERAL ORGANIC CHEMISTRY

25277 In Kjeldahl's method for the estimation of nitrogen, the formula used is

1

% N = \frac{1.4 V W}{N}

2

% N = \frac{1.4 N W}{V}

3

% N = \frac{V N W}{1.8}

4

% N = \frac{1.4 V N}{W}

Explanation:

(d) $%$ of $N = \frac{1.4 \times V \times N}{W}$
where $V$ = Volume of acid used
$N$ = Normality of acid, $W$ = Weight of substance

GENERAL ORGANIC CHEMISTRY

25278 An organic compound on analysis gave the following results : $C = 54.5 %, O = 36.4 %, H = 9.1 %$ . The Empirical formula of the compound is

1

C H_{3} O

2

C_{2} H_{4} O

3

C_{3} H_{4} O

4

C_{4} H_{8} O

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 54.5$

$54.5 / 12 = 4.54$

$2$

$H = 9.1$

$9.1 / 1 = 9.1$

$4$

$O = 36.4$

$36.4 / 16 = 2.27$

$1$

Hence, $C_{2} H_{4} O$ .

GENERAL ORGANIC CHEMISTRY

25275 $60 g$ of a compound on analysis gave $C = 24 g,$ $H = 4 g$ and $O = 32 g$ . Its Empirical formula is

1

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

2

C_{2} H_{2} O

3

C H_{2} O_{2}

4

C H_{2} O

Explanation:

(d)
Element No. of Moles Simple Ratio

$C = 24$

$24 / 12 = 2$

$1$

$H = 4$

$4 / 1 = 4$

$2$

$O = 32$

$32 / 16 = 2$

$1$

Therefore $C H_{2} O$ .

GENERAL ORGANIC CHEMISTRY

25276 An organic compound contains $C = 38.8 %,$ $H = 16 %$ and $N = 45.2 %$ . Empirical formula of the compound is

1

C H_{3} N H_{2}

2

C H_{3} C N

3

C_{2} H_{5} C N

4

C H_{2} (N H)_{2}

Explanation:

(a)
Element No. of Moles Simple Ratio

$C = 38.8$

$38.8 / 12 = 3.2$

$1$

$H = 16$

$16 / 1 = 16$

$5$

$N = 45.2$

$45.2 / 14 = 3.2$

$1$

Therefore, Empirical formula = $C H_{5} N$ or $C H_{3} N H_{2}$ .

GENERAL ORGANIC CHEMISTRY

25277 In Kjeldahl's method for the estimation of nitrogen, the formula used is

1

% N = \frac{1.4 V W}{N}

2

% N = \frac{1.4 N W}{V}

3

% N = \frac{V N W}{1.8}

4

% N = \frac{1.4 V N}{W}

Explanation:

(d) $%$ of $N = \frac{1.4 \times V \times N}{W}$
where $V$ = Volume of acid used
$N$ = Normality of acid, $W$ = Weight of substance

GENERAL ORGANIC CHEMISTRY

25278 An organic compound on analysis gave the following results : $C = 54.5 %, O = 36.4 %, H = 9.1 %$ . The Empirical formula of the compound is

1

C H_{3} O

2

C_{2} H_{4} O

3

C_{3} H_{4} O

4

C_{4} H_{8} O

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 54.5$

$54.5 / 12 = 4.54$

$2$

$H = 9.1$

$9.1 / 1 = 9.1$

$4$

$O = 36.4$

$36.4 / 16 = 2.27$

$1$

Hence, $C_{2} H_{4} O$ .

GENERAL ORGANIC CHEMISTRY

25275 $60 g$ of a compound on analysis gave $C = 24 g,$ $H = 4 g$ and $O = 32 g$ . Its Empirical formula is

1

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

2

C_{2} H_{2} O

3

C H_{2} O_{2}

4

C H_{2} O

Explanation:

(d)
Element No. of Moles Simple Ratio

$C = 24$

$24 / 12 = 2$

$1$

$H = 4$

$4 / 1 = 4$

$2$

$O = 32$

$32 / 16 = 2$

$1$

Therefore $C H_{2} O$ .

GENERAL ORGANIC CHEMISTRY

25276 An organic compound contains $C = 38.8 %,$ $H = 16 %$ and $N = 45.2 %$ . Empirical formula of the compound is

1

C H_{3} N H_{2}

2

C H_{3} C N

3

C_{2} H_{5} C N

4

C H_{2} (N H)_{2}

Explanation:

(a)
Element No. of Moles Simple Ratio

$C = 38.8$

$38.8 / 12 = 3.2$

$1$

$H = 16$

$16 / 1 = 16$

$5$

$N = 45.2$

$45.2 / 14 = 3.2$

$1$

Therefore, Empirical formula = $C H_{5} N$ or $C H_{3} N H_{2}$ .

GENERAL ORGANIC CHEMISTRY

25277 In Kjeldahl's method for the estimation of nitrogen, the formula used is

1

% N = \frac{1.4 V W}{N}

2

% N = \frac{1.4 N W}{V}

3

% N = \frac{V N W}{1.8}

4

% N = \frac{1.4 V N}{W}

Explanation:

(d) $%$ of $N = \frac{1.4 \times V \times N}{W}$
where $V$ = Volume of acid used
$N$ = Normality of acid, $W$ = Weight of substance

GENERAL ORGANIC CHEMISTRY

25278 An organic compound on analysis gave the following results : $C = 54.5 %, O = 36.4 %, H = 9.1 %$ . The Empirical formula of the compound is

1

C H_{3} O

2

C_{2} H_{4} O

3

C_{3} H_{4} O

4

C_{4} H_{8} O

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 54.5$

$54.5 / 12 = 4.54$

$2$

$H = 9.1$

$9.1 / 1 = 9.1$

$4$

$O = 36.4$

$36.4 / 16 = 2.27$

$1$

Hence, $C_{2} H_{4} O$ .

GENERAL ORGANIC CHEMISTRY

25275 $60 g$ of a compound on analysis gave $C = 24 g,$ $H = 4 g$ and $O = 32 g$ . Its Empirical formula is

1

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

2

C_{2} H_{2} O

3

C H_{2} O_{2}

4

C H_{2} O

Explanation:

(d)
Element No. of Moles Simple Ratio

$C = 24$

$24 / 12 = 2$

$1$

$H = 4$

$4 / 1 = 4$

$2$

$O = 32$

$32 / 16 = 2$

$1$

Therefore $C H_{2} O$ .

GENERAL ORGANIC CHEMISTRY

25276 An organic compound contains $C = 38.8 %,$ $H = 16 %$ and $N = 45.2 %$ . Empirical formula of the compound is

1

C H_{3} N H_{2}

2

C H_{3} C N

3

C_{2} H_{5} C N

4

C H_{2} (N H)_{2}

Explanation:

(a)
Element No. of Moles Simple Ratio

$C = 38.8$

$38.8 / 12 = 3.2$

$1$

$H = 16$

$16 / 1 = 16$

$5$

$N = 45.2$

$45.2 / 14 = 3.2$

$1$

Therefore, Empirical formula = $C H_{5} N$ or $C H_{3} N H_{2}$ .

GENERAL ORGANIC CHEMISTRY

25277 In Kjeldahl's method for the estimation of nitrogen, the formula used is

1

% N = \frac{1.4 V W}{N}

2

% N = \frac{1.4 N W}{V}

3

% N = \frac{V N W}{1.8}

4

% N = \frac{1.4 V N}{W}

Explanation:

(d) $%$ of $N = \frac{1.4 \times V \times N}{W}$
where $V$ = Volume of acid used
$N$ = Normality of acid, $W$ = Weight of substance

GENERAL ORGANIC CHEMISTRY

25278 An organic compound on analysis gave the following results : $C = 54.5 %, O = 36.4 %, H = 9.1 %$ . The Empirical formula of the compound is

1

C H_{3} O

2

C_{2} H_{4} O

3

C_{3} H_{4} O

4

C_{4} H_{8} O

Explanation:

(b)
Element No. of Moles Simple Ratio

$C = 54.5$

$54.5 / 12 = 4.54$

$2$

$H = 9.1$

$9.1 / 1 = 9.1$

$4$

$O = 36.4$

$36.4 / 16 = 2.27$

$1$

Hence, $C_{2} H_{4} O$ .

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