Empirical formula vs molecular formula — determination from % composition

1.1.14 · Chemistry › Matter, Measurement & the Mole

% COMPOSITION SE EMPIRICAL FORMULA KAISE DERIVE KAREIN

Logic: percent → grams → moles → ratio → whole numbers.

Step-by-step recipe (har step KYU hai):

100 g assume karo → percent value grams ban jati hai. (Ratios ko concrete masses mein badalta hai.)
Har mass ko us element ki atomic mass se divide karo → moles milenge. (Kyunki equal moles = equal atom counts; grams akele jhooth bolte hain, kyunki ek C atom ≠ mass of an H atom.)
Har mole value ko sabse chhoti value se divide karo → ratio scale ho jata hai jaise smallest = 1. (Ratio ko aasaan numbers mein reveal karta hai.)
Whole numbers mein convert karo (multiply karo agar $.5, .33, .25$ … mile). (Atoms fractional nahi ho sakte.)

Figure — Empirical formula vs molecular formula — determination from % composition

Worked Example 1 — Glucose

Diya gaya: $40.0% C, 6.7% H, 53.3% O$ , molar mass $= 180 g/mol$ .

Element	grams (100 g assume karo)	÷ atomic mass	moles	÷ smallest (0.417)
C	40.0	÷12.01	3.33	1
H	6.7	÷1.008	6.65	2
O	53.3	÷16.00	3.33	1

0.417 se divide kyun? Nahi — yahan sabse chhoti mole value 3.33 hai (C aur O). Sabse chhoti se divide kyun? Taaki least-abundant element 1 ban jaye, aur baaki ratios ek nazar mein padhne laayak ho jaayein.

Empirical formula $= C H_{2} O$ , empirical mass $= 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol$ .

$n = \frac{180}{30.03} \approx 6 \Rightarrow Molecular formula = C_{6} H_{12} O_{6}$

5.99 ko 6 pe round kyun? Kyunki $n$ pure empirical units count karta hai; thodi si deviation atomic masses/data ki rounding se hai.

Worked Example 2 — Fraction jo 1 nahi hai

Diya gaya: $43.6% P, 56.4% O$ , molar mass $= 283.9 g/mol$ .

Element	grams	÷ atomic mass	moles	÷ smallest (1.408)
P	43.6	÷30.97	1.408	1.00
O	56.4	÷16.00	3.525	2.50

2.50 ko 3 pe round kyun nahi? Kyunki $2.50$ ek clean half hai — yeh signal hai ki true ratio $2 : 5$ hai, $2 : 6$ nahi. Dono ko 2 se multiply karo: $P_{2} O_{5}$ .

Empirical mass $= 2 (30.97) + 5 (16.00) = 141.94$ . $n = 283.9/141.94 \approx 2 \Rightarrow P_{4} O_{10}$ .

Yeh step kyun? Measured molar mass (283.9) empirical mass ka double hai, toh molecule do $P_{2} O_{5}$ units se bana hai.

Common Mistakes (Steel-man + Fix)

Active Recall

Recall Answer padhne se pehle predict karo: kaunsa extra data ek empirical formula ko molecular formula mein badalta hai?

The molar mass (ya molecular mass), kyunki $n = M_{molar} / M_{empirical}$ aur sirf $n$ hi dono ko alag karta hai.

Recall Forecast: do compounds ka empirical formula

$C H_{2}$ hai. Ek ka $M = 42$ , doosre ka $M = 84$ . Molecular formulas kya hain? $M_{emp} = 14$ . $n = 42/14 = 3 \Rightarrow C_{3} H_{6}$ (propene). $n = 84/14 = 6 \Rightarrow C_{6} H_{12}$ . Same simplest ratio, alag molecules — isliye empirical akela sufficient nahi hai.

Recall Exactly 100 g kyun assume karte hain, 50 g ya 1 g kyun nahi?

Kyunki 100 g ke saath har percent seedha grams ban jaata hai, arithmetic bachta hai. Koi bhi mass same ratio deta hai, lekin 100 g sabse aasaan correct choice hai.

Feynman (ek 12-saal ke bacche ko samjhao)

Recall Simple mein samjhao

Socho ek recipe kehti hai "2 part atta, 1 part cheeni." Woh ratio (empirical) tumhe nahi batata ki cake chhota hai ya bada. Agar main tumhe yeh bhi bataaun ki poore cake ka weight kuch khaas amount hai, toh tum actual cups atta aur cheeni nikal sakte ho (molecular). Percent composition recipe ratio hai; molar mass cake ka size hai.

Flashcards

Empirical formula definition

Ek compound mein har element ke atoms ka sabse simple whole-number ratio.

Molecular formula definition

Ek molecule mein har element ke atoms ki actual number.

Dono ke beech relationship

Molecular = n × Empirical, jahan n = molar mass ÷ empirical formula mass.

% composition problems mein 100 g kyun assume karte hain

Har percentage value seedha grams ban jaati hai, arithmetic simple ho jaata hai; ratio sample size se affect nahi hota.

Grams se pehla conversion step

Har mass ko element ki atomic mass se divide karo taaki moles milein (mass ratio ≠ atom ratio).

Moles ko sabse chhoti value se divide kyun karte hain

Ratio ko scale karne ke liye taaki least-abundant element 1 ban jaye, ratio padhna aasaan ho.

.5 pe end hone wale ratio ke saath kya karein

Sab ko 2 se multiply karo half clear karne ke liye (yeh asli fraction hai, rounding error nahi).

Molecular formula ke liye extra data kya chahiye

Measured molar (ya molecular) mass.

Glucose ka empirical formula

CH2O.

n for glucose (M=180, emp mass=30.03)

n ≈ 6, jo C6H12O6 deta hai.

Connections

The Mole and Avogadro's Number — moles mass se atom count tak ka bridge hain.
Percent Composition by Mass — is poore procedure ka input data.
Atomic and Molecular Mass — moles aur empirical mass dono compute karne ke liye zaroori.
Combustion Analysis — % composition paane ka ek experimental tarika.
Law of Definite Proportions — woh physical law jo fixed ratios (empirical formulas) ko meaningful banata hai.
Stoichiometry — molecular formulas seedha balanced-equation calculations mein jaate hain.