Percent yield, theoretical yield, actual yield

1.3.4 · Chemistry › Chemical Reactions & Stoichiometry

YE TEEN QUANTITIES KYA HAIN?

THEORETICAL YIELD KO FIRST PRINCIPLES SE KAISE DERIVE KAREIN

Yaad karne ke liye koi magic formula nahi hai — theoretical yield bas atoms ki careful bookkeeping hai. Is chain ko follow karo:

$mass reactant \div M mol reactant \times \frac{coeff. product}{coeff. reactant} mol product \times M mass product$

Limiting reagent woh reactant hai jo pehle khatam hota hai, isliye yeh cap karta hai ki kitni product banti hai. Ise dhundhne ke liye: calculate karo ki har reactant akela kitni product bana sakta hai; sabse chota jawab hi theoretical yield set karta hai.

Figure — Percent yield, theoretical yield, actual yield

Worked Example 1 — simple single-reactant case

8.0 g methane ko completely jalao: $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ . Oxygen excess mein hai. Tum 19.0 g $CO_{2}$ collect karte ho. Percent yield nikaalo.

Step 1 — $CH_{4}$ ke moles: $n = \frac{8.0}{16.0} = 0.50$ mol. Yeh step kyun? Grams ko recipe se compare nahi kar sakte; moles se kar sakte hain.

Step 2 — $CO_{2}$ ke saath mole ratio: ratio $1 : 1$ hai, toh theoretically $0.50$ mol $CO_{2}$ . Yeh step kyun? Ek carbon andar, ek carbon bahar — atoms ka conservation.

Step 3 — theoretical mass: $0.50 \times 44.0 = 22.0$ g. Yeh step kyun? Balance grams padhta hai, isliye lab result se compare karne ke liye hum wapas grams mein convert karte hain.

Step 4 — percent yield: $\frac{19.0}{22.0} \times 100 = 86.4%$ . Yeh step kyun? Actual ÷ theoretical, same units (g), ×100.

Worked Example 2 — limiting reagent theoretical yield decide karta hai

$N_{2} + 3 H_{2} \to 2 NH_{3}$ . 28 g $N_{2}$ aur 9.0 g $H_{2}$ react karo. Actual $NH_{3}$ obtained = 25 g. Percent yield nikaalo.

Step 1 — moles: $n (N_{2}) = \frac{28}{28} = 1.0$ mol; $n (H_{2}) = \frac{9.0}{2.0} = 4.5$ mol.

Step 2 — har ek kitni product bana sakta hai:

$N_{2}$ se: $1.0 \times \frac{2}{1} = 2.0$ mol $NH_{3}$ .
$H_{2}$ se: $4.5 \times \frac{2}{3} = 3.0$ mol $NH_{3}$ .

Yeh step kyun? Jo kam product deta hai wahi asli bottleneck hai — tum apni sabse scarce ingredient se zyada product nahi bana sakte.

Step 3 — limiting reagent = $N_{2}$ (2.0 < 3.0). Theoretical = 2.0 mol → $2.0 \times 17.0 = 34.0$ g.

Step 4 — percent yield: $\frac{25}{34.0} \times 100 = 73.5%$ .

Worked Example 3 — percent yield se backwards kaam karna

Ek reaction ki known 90.0% yield hai aur tumhe 18.0 g product chahiye ( $M = 60$ g/mol). Theoretical yield kitne moles ki honi chahiye?

Step 1: theoretical mass $= \frac{actual}{yield fraction} = \frac{18.0}{0.900} = 20.0$ g. Yeh step kyun? Percent-yield formula ko rearrange karo: theoretical = actual ÷ (%/100).

Step 2: $n = \frac{20.0}{60} = 0.333$ mol kaagaz par chahiye. Yeh step kyun? Ab tum apne reactants ko scale up karte ho taaki 10% loss ke bawajood is target ko hit kar sako.

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bacche ko samjhao

Soch ki tum paper airplanes bana rahe ho. Kaagaz ki apni stack se tum could 20 planes bana sakte the — yahi "theoretical" number hai, sapna. Lekin kuch kaagaz phaat jaata hai, aur tumhara mann bhar jaata hai, toh tum sirf 15 finish karte ho — yahi tumne "actually" banaye. Percent yield hai 15 out of 20 = 75%, tumhara score ki real world ne sapne se kitna match kiya. Aur agar kaagaz glue se pehle khatam ho gaya, toh kaagaz woh "limiting" cheez hai jo tumhara sapna ka number decide karti hai.

Active Recall Flashcards

#flashcards/chemistry

Percent yield formula?

(actual yield ÷ theoretical yield) × 100%, dono same unit mein.

Theoretical yield kis reactant se calculate ki jaati hai?

Limiting reagent se.

Limiting reagent ko kaise identify karte hain?

Compute karo ki har reactant kitni product bana sakta hai; jo sabse kam product deta hai woh limit karta hai.

Dono yields same unit mein kyun honi chahiye?

Taaki units cancel ho jaayein aur ratio ek pure fraction ban jaaye; coefficients moles relate karte hain grams nahi.

Percent yield >100% kya indicate karta hai?

Measured product mein impurity ya moisture; real yield 100% exceed nahi kar sakti.

Mole ratio use karne se pehle grams ko moles mein kyun convert karte hain?

Coefficients particles (moles) count karte hain, masses nahi; conservation of atoms moles mein kaam karta hai.

Actual yield kaise obtain ki jaati hai?

Lab mein balance par experimentally measure ki jaati hai.

Agar yield 80% hai aur tumhe 40 g product chahiye, toh kitna theoretical mass chahiye?

40 ÷ 0.80 = 50 g.

Kya excess reagent theoretical yield ko affect karta hai?

Nahi — sirf limiting reagent usse set karta hai; excess leftover reh jaata hai.

Connections

Limiting Reagent — theoretical yield ki ceiling set karta hai.
Mole Concept — mole har stoichiometric conversion ki currency hai.
Balancing Chemical Equations — coefficients mole ratio dete hain.
Molar Mass — grams aur moles ke beech dono directions mein convert karta hai.
Conservation of Mass — woh reason ki atoms in = atoms out.
Empirical & Molecular Formula — mass–mole reasoning dobara use hoti hai.