chemistry-class-11-unit-12-chapter-07-organic-chemistry-basic-principles-and-techniques-l-07-7-vdsd0bfuid4

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Duman method </primary>
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- $C_xH_yN_z\xrightarrow [Cu ~ metal]{CuO} xCO_2 + \frac {y}{2}+\frac {z}{2}N_2$

- Volume of $N_2$ liberates from the organic compound m grams of an Organic compund is subjected to Duman method $\frac{x100}{xm}$ % of N

- Vol of $N_2$ gas collected = $V_1$ ml at $T_1P_1$

- $\frac{P_1V_1}{T_1} = \frac {P_2V_2}{T_2}$

- $V_2=\frac {P_1V_1\times 273}{T_1\times P_2(760 ml of Hg)}$

- 22400 ml of $N_2$ at STP = 28g of $N_2$ (1 mole weight of $N_2$)

- wt of $N_2$ Corresponding to $V_2$ ml of $N_2$ =$\frac {28\times V_2\times 100}{22400\times m}$% of N

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<footer style = "font-size: 18px"> $\rightarrow$ Basic principles and techniques $\rightarrow$ <span style = "color:blue"> Duman method </span> $\rightarrow$ Duman method $\rightarrow$ Kjeldhal's method estimation of N </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Duman method </primary>
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- From 0.3g of organic Substance 50ml of$ N_2$ is evolved at 300 k and 715 mm of Hgwater vapour pressure at 300k = 15 mm of Hg

- $P_1$ = 715-15 = 700 mm of $Hg$

- $V_2=\frac {P_1V_1\times T_2}{T_1\times P_2}$

- $V_2=\frac {700\times 50\times 273}{300K\times 710}$

- $V_2$= 41.9 ml of $N_2$ at STP

- Wt of $N_2 = \frac {28\times 41.9}{22400}$

- % of $N_2= \frac {28\times 41.9\times 100}{22400\times 0.3}$

- % of $N_2$=17.46 % of N

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<footer style = "font-size: 18px">Basic principles and techniques $\rightarrow$ Duman method $\rightarrow$ <span style = "color:blue"> Duman method </span> $\rightarrow$ Kjeldhal's method estimation of N $\rightarrow$ Kjeldhal's method estimation of N </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Kjeldhal's method estimation of N </primary>
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- 0.257g of organic compd was treated with $H_2SO_4/CuSO_4$ and liberated$ NH_3$ on treatmentwith NaOH absorbed with 50 ml of$\frac {M}{10}$of HCl

- The excess acid acid  require 23.2 ml of $\frac {M}{10}$ of NaOH for neutralization

- Calculate % of N in the organic compound

- Vol of$\frac {M}{10}$ Na OH = 23.2 ml = 23.2ml of $\frac{M}{10}$ of HCUnused HCl = 23.2 ml of HCl $\frac {M}{10}$

- HCl used of $NH_3$ absorption = 50-23.2

- HCl used of $NH_3$ absorption = 26.8 ml of $\frac{M}{10}$ HCl

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<footer style = "font-size: 18px">Duman method $\rightarrow$ Kjeldhal's method estimation of N $\rightarrow$ <span style = "color:blue"> Kjeldhal's method estimation of N </span> $\rightarrow$ Kjeldhal's method estimation of N $\rightarrow$ Kjeldhal's Method estimation of N </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Kjeldhal's method estimation of N </primary>
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- 6.8 ml of $\frac {M}{10}$ HCL = 26.8ml of $\frac{M}{10} NH_3$

- Solution

- 1000 ml of 1M $NH_3$ = 14 g of N

- $NH_3$ 14 grams

- 1000ml of $\frac {M}{10}NH_3$= 1.4g of N

- wt of N =$\frac{1.4\times 26.8}{1000}$

- % of N =$\frac{1.4\times 26.8\times 100}{1000\times 0.257}$ = 14.6 % of N

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<footer style = "font-size: 18px">Kjeldhal's method estimation of N $\rightarrow$ Kjeldhal's method estimation of N $\rightarrow$ <span style = "color:blue"> Kjeldhal's method estimation of N </span> $\rightarrow$ Kjeldhal's Method estimation of N $\rightarrow$ Kjeldhal's Method estimation of N </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Kjeldhal's Method estimation of N </primary>
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- 0.35 g of an organic Substance was kjeldhalised and the $NH_3$ obtained was passed into 100 ml of $\frac {M}{10}H_2SO_4$

- The excess and required 154 ml of $\frac {M}{10}$ NaOH

- calculate the % of N in the organic compound

- 154 ml of NaOH of $\frac{M}{10}$ concentrated = $\frac{154}{2}ml$ of $\frac {M}{10}H_2SO_4$ = 77ml

- $H_2SO_4$ used for $NH_3$ absorption = 100-17=23ml of $\frac{M}{10}H_2SO_4$

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<footer style = "font-size: 18px">Kjeldhal's method estimation of N $\rightarrow$ Kjeldhal's method estimation of N $\rightarrow$ <span style = "color:blue"> Kjeldhal's Method estimation of N </span> $\rightarrow$ Kjeldhal's Method estimation of N $\rightarrow$ Carius method of estimation of halogen </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Kjeldhal's Method estimation of N </primary>
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- 2 NaOH+$H_2SO_4\rightarrow  Na_2SO_4 +2H_2O$

- $2NH_4OH + H_2SO_4\rightarrow (NH_4)_2SO_4+2H_2O$

- 23ml of $\frac{M}{10}H_2SO_4 =$ 46ml of $NH_3$ of $\frac{M}{10}$ conc

- 1000ml of $ NH_3$ (1M)= 14g of N

- 1000ml of $\frac{M}{10}NH_3$ = 1.4 g of N

- Wt of N =$\frac{1.4 \times 4.6}{1000}$

- % N =$\frac{1.4\times 46\times 100}{1000\times 0.35}$ = 18.4 % of N

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<footer style = "font-size: 18px">Kjeldhal's method estimation of N $\rightarrow$ Kjeldhal's Method estimation of N $\rightarrow$ <span style = "color:blue"> Kjeldhal's Method estimation of N </span> $\rightarrow$ Carius method of estimation of halogen $\rightarrow$ Carius method of estimation of halogen </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Carius method of estimation of halogen </primary>
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- A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube

- Dry weight of Agx formed is estimated

- Agcl(143.5) $\rightarrow$ 35.5 g of $Cl$

- $x gm$ of AGCl $\rightarrow \frac{35.5\times x \times 100}{143.5 \times m}$ % of $Cl$

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<footer style = "font-size: 18px">Kjeldhal's Method estimation of N $\rightarrow$ Kjeldhal's Method estimation of N $\rightarrow$ <span style = "color:blue"> Carius method of estimation of halogen </span> $\rightarrow$ Carius method of estimation of halogen $\rightarrow$ Carius Method of estimation of halogen </footer>

- 0.15 g of an organic compound gave

- 0.12 g of AgBr using carius method

- hat is the % of Br in the org compound

- Ag Br

- 108 + 80 = 188 g/mole

- 188 g/mole of Ag Br = 80g of Br

- % of Br in the compound =$\frac{80 \times 0.12 \times 100}{188 \times 0.15}$ = 34.04 %

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<footer style = "font-size: 18px">Kjeldhal's Method estimation of N $\rightarrow$ Carius method of estimation of halogen $\rightarrow$ <span style = "color:blue"> Carius method of estimation of halogen </span> $\rightarrow$ Carius Method of estimation of halogen $\rightarrow$ Carius method of estimation of halogen </footer>

### <Success style="font-size:25px"> Organic Chemistry Basic Principles And Techniques L-07 </Success>
### <primary style="font-size:24px"> Carius Method of estimation of halogen </primary>
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- **Estimation of sulphur**

- A known mass of an organic compound is heated in a Carius tube with sodium peroxideor fuming nitric acid Sulphur present in thecompound is oxidised to sulphuric acid

- It is precipitated as barium sulphate by addingexcess of barium chloride solution in water

- The precipitate is filtered, washed, dried andweighed

- $BaSO_4$

- 137 + 32 + $C_4= 233g BaSO_4$ (precipitated)

- 233g of $BaSO_4$ = 32g of S

- $x$ $gr$ of $BaSO_4$ =$\frac{32 \times x}{233}$

- % of S= $\frac{32 \times x \times 100}{233 \times m}$ = % of S

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<footer style = "font-size: 18px">Carius method of estimation of halogen $\rightarrow$ Carius method of estimation of halogen $\rightarrow$ <span style = "color:blue"> Carius Method of estimation of halogen </span> $\rightarrow$ Carius method of estimation of halogen $\rightarrow$ Thank you </footer>

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- **Estimation of sulphur**

- 0.157 of an organic compd gave 0.4813 g of $BaSO_4$ What isthe % of S in the organic Compdound

- 233g of $BaSo_4$ = 32g of s

- 0.4813 of $BaSo_4$ =$\frac{32\times 0.4813}{233}$

- 0.157 g of organic substance gave =$\frac{32\times0.4813}{233}$ of S

- 100 g of organic substance gave =$\frac{32\times 0.4813\times 100}{233\times 0.157}$ = 42.10 % of S

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<footer style = "font-size: 18px">Carius method of estimation of halogen $\rightarrow$ Carius Method of estimation of halogen $\rightarrow$ <span style = "color:blue"> Carius method of estimation of halogen </span> $\rightarrow$ Thank you $\rightarrow$  </footer>