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Showing posts with the label Stoichiometry

1 Mole = Kitne Particles? | Easy Mole Concept Explained for JEE Students

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  📘 1 Mole = Kitne Particles? Easy Explanation! "1 mole ka matlab sirf ek bada number nahi hai… ye concept JEE ke Stoichiometry ke sawalon ki chaabi hai! Agar aap mole ka matlab samajh gaye, toh chemical calculations bohot aasaan ho jaayengi." 🧪 What is 1 Mole? 1 mole = 6.022 × 10²³ particles Ye number ko Avogadro Number kehte hain. Particles ka matlab kya hota hai? 👉 Atoms, molecules, ions, electrons – jo bhi aap measure kar rahe ho, ussi ke hote hain particles. ⚖️ 1 Mole & Mass ka Connection: Har substance ka ek molar mass hota hai (grams mein). 👉 1 mole = molar mass in grams 💡 Example: Water (H₂O) ka molar mass = 18g/mol So, 18g H₂O = 1 mole = 6.022 × 10²³ molecules of water 🔁 Conversion Trick: Mole Bridge Ye ek shortcut technique hai: Mass ⇄ Moles ⇄ Number of Particles Jaise: Agar aapke paas mass ho → moles nikaalo → phir number of particles Ya particles ho → moles nikaalo → phir mass 📌 Shortcut Formula: No. of part...

Combustion Analysis | % of Oxygen in Organic Compound | JEE Chemistry Doubt | Doubtify JEE

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🔥 Combustion Analysis – Percentage of Oxygen in an Organic Compound | JEE Chemistry | Doubtify JEE

⚗️ Empirical Formula from Combustion Data – Chemistry JEE | Doubtify JEE

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🧪 Question: The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6:1. If one molecule of the above compound contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO₂ and H₂O, then what is the empirical formula of the compound? 📌 Why This Question is Important: This question combines two highly testable concepts in JEE Chemistry – combustion analysis and empirical formula derivation . You’re not only working with mass percent ratio , but also applying oxygen balancing logic in a combustion reaction. Such questions are frequently seen in organic chemistry chapters , particularly under Basic Principles of Organic Chemistry and Stoichiometry . Mastering this helps you crack more complex analysis-based questions in JEE Mains and even Advanced . 💡 Concept Used: To find the empirical formula , we first convert the mass percent ratio of C and H into moles , assuming 100 g of substance. Then we apply a stoichiometric appr...

🧲 Mole Concept in Combustion Reactions | JEE Chemistry | Doubtify JEE

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🧮 Question: 100 grams of propane is completely reacted with 1000 grams of oxygen. The mole fraction of carbon dioxide in the resulting mixture is X × 10⁻². The value of x is: 🖼️ Question Image: