On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g $H_2O$ and 0.307 g $CO_2$. The percentages of hydrogen and oxygen in the given organic compound respectively are:
Options:
A) 7.55, 43.85
B) 6.72, 53.41
C) 6.72, 39.87
D) 53.41, 39.6
Options:
A) 7.55, 43.85
B) 6.72, 53.41
C) 6.72, 39.87
D) 53.41, 39.6
✓
Solution Steps
1
Given Data Collect Karein
Organic compound ka mass ($w$) = $0.210$ g. $CO_2$ ka mass = $0.307$ g aur $H_2O$ ka mass = $0.127$ g.
2
Hydrogen ki Percentage Calculate Karein
Formula use karein: $\%H = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$
$$\%H = \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = \frac{0.127}{9 \times 0.210} \times 100 \approx 6.719\% \approx 6.72\%$$
3
Carbon ki Percentage Calculate Karein
Formula: $\%C = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$
$$\%C = \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = \frac{3 \times 0.307}{11 \times 0.210} \times 100 \approx 39.87\%$$
4
Oxygen ki Percentage ka Calculation
Compound mein sirf C, H aur O hain. Isliye Oxygen ki percentage difference se niklegi:
$\%O = 100 – (\%C + \%H)$
5
Values Substitute Karein
Values put karne par:
$\%O = 100 – (39.87 + 6.72)$
6
Final Subtraction
$\%O = 100 – 46.59 = 53.41\%$
7
Option Match Karein
Humne nikala $\%H = 6.72$ aur $\%O = 53.41$. Ye Option B se match karta hai.
Result: 6.72%, 53.41%
📚
Theory
1. Liebig’s Combustion Method
Liebig’s combustion method organic compounds ke element estimation ka ek fundamental tarika hai. Ismein compound ko excess oxygen aur copper(II) oxide ($CuO$) ki presence mein heat kiya jata hai. Is process mein saara Carbon $CO_2$ mein aur saara Hydrogen $H_2O$ mein convert ho jata hai. Nikli hui $CO_2$ ko potassium hydroxide ($KOH$) solution absorb karta hai, jabki $H_2O$ ko anhydrous calcium chloride ($CaCl_2$) absorb karta hai. In absorbers ke mass mein aaya change hi products ka mass batata hai.
2. Stoichiometry of CO2 and H2O
Chemical reactions mein mass hamesha conserved rehta hai. Jab hum $\%C$ nikalte hain, toh factor $12/44$ ka use karte hain kyunki $44$g $CO_2$ mein exact $12$g carbon hota hai. Similary, $18$g $H_2O$ mein $2$g Hydrogen atoms hote hain, isliye $\%H$ ke liye $2/18$ ka factor use hota hai. Ye stoichiometry humein accurately batati hai ki original organic molecule mein kitna element present tha, chahe wo molecule kitna bhi complex kyun na ho.
3. Oxygen Estimation by Difference
Combustion analysis mein oxygen ko directly measure karna mushkil hota hai kyunki combustion ke liye bahar se oxygen gas supply ki jati hai. Isliye standard practice ye hai ki Carbon, Hydrogen, Nitrogen, Halogens aur Sulfur jaise baaki sabhi elements ki percentage nikali jaye. In sabka sum $100$ se subtract karne par bachi hui value Oxygen ki hoti hai. Agar sum $100$ ke bahut kareeb hai, toh iska matlab compound mein oxygen nahi hai.
4. Significance in Empirical Formula
Percentage composition se hi hum kisi bhi compound ka Empirical Formula (saraltam sutra) nikalte hain. In percentages ko unke atomic mass se divide karke humein atoms ka relative molar ratio mil jata hai. Organic chemistry ke research mein jab koi naya compound synthesize hota hai, toh uski identity confirm karne ke liye combustion analysis pehla step hota hai. Isse ye verify ho jata hai ki compound ki purity kitni hai.
❓
FAQs
1
Oxygen percentage 100 se minus kyu karte hain?
Kyunki combustion mein bahar ki air se oxygen use hoti hai, isliye products se direct mass nahi mil sakta.
2
Agar nitrogen hota toh kya procedure change hota?
Haan, nitrogen ke liye Dumas ya Kjeldahl method use karte hain.
3
H2O absorb karne ke liye CaCl2 kyu use karte hain?
CaCl2 ek hygroscopic substance hai jo moisture ko effectively absorb kar leta hai.
4
CO2 absorb karne ke liye KOH hi kyu?
KOH acid-base reaction karke CO2 ko Potassium Carbonate mein badal deta hai.
5
Kya ye method sabhi organic compounds par kaam karta hai?
Haan, jinmein C aur H ho. High temperature par ye complete oxidise ho jate hain.
6
Atomic masses kya use karne chahiye?
C=12.01, H=1.008, O=16. Lekin JEE ke liye 12, 1 aur 16 use hote hain.
7
Calculation mein factor 2/18 ki jagah 1/18 le sakte hain?
Nahi, kyunki water ($H_2O$) ke ek molecule mein do hydrogen atoms hote hain.
8
Agar calculation ka sum 100 se kam aaye toh?
Iska matlab ya toh koi error hai ya compound mein koi aur element (jaise Cl, S) present hai.
9
Ye method kisne invent kiya tha?
Justus von Liebig ne 1831 mein ise modernize kiya tha.
10
Is problem mein Carbon ki percentage puchi hoti toh?
Humne step 3 mein nikala hai, Carbon ki percentage 39.87% hai.
🔗
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