Cheat Sheet: Some Basic Concepts of Chemistry | Chemistry Class 11 - NEET PDF Download

 Page 1


1 Basic Concepts
1.1 Key Definitions
• Matter : An ything with mass and v olume; exists as solid, liquid, or gas.
• Elemen t : Pure substance with one t yp e of atom (e.g., H, O).
• Comp ound : Substance with t w o or more elemen ts in a fixed ratio (e.g., H
2
O).
• Mixture : Com bination of substances; homogeneous (uniform) or heterogeneous
(non-uniform).
• Mole : Amoun t of substance con taining 6.022×10
23
particles (A v ogadro’s n um b er,
N
A
).
• Molar Mass : Mass of one mole of a substance (g/mol); for elemen ts, it equals
atomic m ass; for comp ounds, sum of atomic masses.
1.2 La ws of Chemical Com bination
• La w of Conserv ation of Mass : Mass of reactan ts equals mass of pro ducts.
• La w of Definite Prop ortions : A comp ound has a fixed elemen t ratio b y mass.
• La w of Multiple Prop ortions : Elemen ts com bine in simple whole-n um b er ratios.
• Ga y-Lussac’s La w : Gases com bine in simple v olume ratios at constan t T and P .
• A v ogadro’s La w : Equal v olumes of gases at same T and P con tain equal moles.
2 Mole Concept and Stoic hiometry
2.1 Key F orm ulas
• Num b er of m oles: n =
mass (g)
molar mass (g/mol)
• Num b er of pa rticles: N =n×N
A
, whereN
A
= 6.022×10
23
• Molar v olume (gases, STP): 22.4 L/mol at 0°C, 1 atm
• Empirical form ula: Simplest whole-n um b er ratio of atoms
• Molecular form ula: n× empirical form ula, wheren =
molecular mass
empirical form ula mass
• Stoic hiometry: Use balanced equations to find reactan t/pro duct ratios
2.2 Concen tration T erms
• Molarit y (M) :M =
moles of solute
v olume of solution (L)
• Molalit y (m) : m =
moles of solute
mass of solv en t (kg)
• Mole fraction (X ) : X
A
=
n
A
n
A
+n
B
• Mass p er cen t : % =
mass of solute
mass of solution
×100
1
Some	Basic	Concepts	of	Chemistry	Cheat	Sheet
(EduRev)
Page 2


1 Basic Concepts
1.1 Key Definitions
• Matter : An ything with mass and v olume; exists as solid, liquid, or gas.
• Elemen t : Pure substance with one t yp e of atom (e.g., H, O).
• Comp ound : Substance with t w o or more elemen ts in a fixed ratio (e.g., H
2
O).
• Mixture : Com bination of substances; homogeneous (uniform) or heterogeneous
(non-uniform).
• Mole : Amoun t of substance con taining 6.022×10
23
particles (A v ogadro’s n um b er,
N
A
).
• Molar Mass : Mass of one mole of a substance (g/mol); for elemen ts, it equals
atomic m ass; for comp ounds, sum of atomic masses.
1.2 La ws of Chemical Com bination
• La w of Conserv ation of Mass : Mass of reactan ts equals mass of pro ducts.
• La w of Definite Prop ortions : A comp ound has a fixed elemen t ratio b y mass.
• La w of Multiple Prop ortions : Elemen ts com bine in simple whole-n um b er ratios.
• Ga y-Lussac’s La w : Gases com bine in simple v olume ratios at constan t T and P .
• A v ogadro’s La w : Equal v olumes of gases at same T and P con tain equal moles.
2 Mole Concept and Stoic hiometry
2.1 Key F orm ulas
• Num b er of m oles: n =
mass (g)
molar mass (g/mol)
• Num b er of pa rticles: N =n×N
A
, whereN
A
= 6.022×10
23
• Molar v olume (gases, STP): 22.4 L/mol at 0°C, 1 atm
• Empirical form ula: Simplest whole-n um b er ratio of atoms
• Molecular form ula: n× empirical form ula, wheren =
molecular mass
empirical form ula mass
• Stoic hiometry: Use balanced equations to find reactan t/pro duct ratios
2.2 Concen tration T erms
• Molarit y (M) :M =
moles of solute
v olume of solution (L)
• Molalit y (m) : m =
moles of solute
mass of solv en t (kg)
• Mole fraction (X ) : X
A
=
n
A
n
A
+n
B
• Mass p er cen t : % =
mass of solute
mass of solution
×100
1
Some	Basic	Concepts	of	Chemistry	Cheat	Sheet
(EduRev)
• V olume p ercen t : % =
v olume of solute
v olume of solution
×100
3 A tomic and Molecular Masses
• A tomic Mass : Mass of an atom in atomic mass units (am u); 1 am u =
1
12
mass of
C-12.
• A v erage A tomic M ass : A vg. mass =
?
( fractional abundance× isotopic mass)
• Molecular Mas s : Sum of atomic masses of all atoms in a molecule.
4 Stoic hiometry Quic k Reference
T yp e F orm ula/Relation
Mass-Mass Use mole ratio from balanced equation;
m =n× molar mass
Mass-V olume (gas) Use molar v olume at STP (22.4 L/mol)
V olume-V olume (gas) Use v olume ratios from balanced equation
(A v ogadro’s la w)
Limiting Reagen t Reactan t that pro duces least pro duct moles
P ercen t Yield % yield =
actual yi eld
theoretical yield
×100
5 Solv ed Examples
5.1 Example 1: Mole Calculation
Problem : Calculate the n um b er of moles in 32 g of O
2
(molar mass = 32 g/mol).
Solution : n =
mass
molar mass
=
32
32
= 1 mol.
Answ er : 1 mol.
5.2 Example 2: Empirical F orm ula
Problem : A comp ound has 40% C, 6.67% H, 53.33% O. Find its empirical form ula (C
= 12, H = 1, O = 16).
Solution :
• Moles: C =
40
12
= 3.33 , H =
6.67
1
= 6.67 , O =
53.33
16
= 3.33 .
• Ratio: Divide b y smallest (3.33): C = 1, H = 2, O = 1.
• Empirical form ula: CH
2
O.
Answ er : CH
2
O.
5.3 Example 3: Limiting Reagen t
Problem : 4 g H
2
reacts with 32 g O
2
to form w ater. Find the limiting reagen t and mass
of w ater (H
2
= 2 g/mol, O
2
= 32 g/mol, H
2
O = 18 g/mol).
Solution :
• Balanced e quation: 2 H
2
+ O
2
? 2 H
2
O.
• Moles: H
2
=
4
2
= 2 , O
2
=
32
32
= 1 .
• Mole rat io: 2 H
2
: 1 O
2
. H
2
needs 1 mol O
2
, a v ailable = 1 mol (su?icien t).
2
Page 3


1 Basic Concepts
1.1 Key Definitions
• Matter : An ything with mass and v olume; exists as solid, liquid, or gas.
• Elemen t : Pure substance with one t yp e of atom (e.g., H, O).
• Comp ound : Substance with t w o or more elemen ts in a fixed ratio (e.g., H
2
O).
• Mixture : Com bination of substances; homogeneous (uniform) or heterogeneous
(non-uniform).
• Mole : Amoun t of substance con taining 6.022×10
23
particles (A v ogadro’s n um b er,
N
A
).
• Molar Mass : Mass of one mole of a substance (g/mol); for elemen ts, it equals
atomic m ass; for comp ounds, sum of atomic masses.
1.2 La ws of Chemical Com bination
• La w of Conserv ation of Mass : Mass of reactan ts equals mass of pro ducts.
• La w of Definite Prop ortions : A comp ound has a fixed elemen t ratio b y mass.
• La w of Multiple Prop ortions : Elemen ts com bine in simple whole-n um b er ratios.
• Ga y-Lussac’s La w : Gases com bine in simple v olume ratios at constan t T and P .
• A v ogadro’s La w : Equal v olumes of gases at same T and P con tain equal moles.
2 Mole Concept and Stoic hiometry
2.1 Key F orm ulas
• Num b er of m oles: n =
mass (g)
molar mass (g/mol)
• Num b er of pa rticles: N =n×N
A
, whereN
A
= 6.022×10
23
• Molar v olume (gases, STP): 22.4 L/mol at 0°C, 1 atm
• Empirical form ula: Simplest whole-n um b er ratio of atoms
• Molecular form ula: n× empirical form ula, wheren =
molecular mass
empirical form ula mass
• Stoic hiometry: Use balanced equations to find reactan t/pro duct ratios
2.2 Concen tration T erms
• Molarit y (M) :M =
moles of solute
v olume of solution (L)
• Molalit y (m) : m =
moles of solute
mass of solv en t (kg)
• Mole fraction (X ) : X
A
=
n
A
n
A
+n
B
• Mass p er cen t : % =
mass of solute
mass of solution
×100
1
Some	Basic	Concepts	of	Chemistry	Cheat	Sheet
(EduRev)
• V olume p ercen t : % =
v olume of solute
v olume of solution
×100
3 A tomic and Molecular Masses
• A tomic Mass : Mass of an atom in atomic mass units (am u); 1 am u =
1
12
mass of
C-12.
• A v erage A tomic M ass : A vg. mass =
?
( fractional abundance× isotopic mass)
• Molecular Mas s : Sum of atomic masses of all atoms in a molecule.
4 Stoic hiometry Quic k Reference
T yp e F orm ula/Relation
Mass-Mass Use mole ratio from balanced equation;
m =n× molar mass
Mass-V olume (gas) Use molar v olume at STP (22.4 L/mol)
V olume-V olume (gas) Use v olume ratios from balanced equation
(A v ogadro’s la w)
Limiting Reagen t Reactan t that pro duces least pro duct moles
P ercen t Yield % yield =
actual yi eld
theoretical yield
×100
5 Solv ed Examples
5.1 Example 1: Mole Calculation
Problem : Calculate the n um b er of moles in 32 g of O
2
(molar mass = 32 g/mol).
Solution : n =
mass
molar mass
=
32
32
= 1 mol.
Answ er : 1 mol.
5.2 Example 2: Empirical F orm ula
Problem : A comp ound has 40% C, 6.67% H, 53.33% O. Find its empirical form ula (C
= 12, H = 1, O = 16).
Solution :
• Moles: C =
40
12
= 3.33 , H =
6.67
1
= 6.67 , O =
53.33
16
= 3.33 .
• Ratio: Divide b y smallest (3.33): C = 1, H = 2, O = 1.
• Empirical form ula: CH
2
O.
Answ er : CH
2
O.
5.3 Example 3: Limiting Reagen t
Problem : 4 g H
2
reacts with 32 g O
2
to form w ater. Find the limiting reagen t and mass
of w ater (H
2
= 2 g/mol, O
2
= 32 g/mol, H
2
O = 18 g/mol).
Solution :
• Balanced e quation: 2 H
2
+ O
2
? 2 H
2
O.
• Moles: H
2
=
4
2
= 2 , O
2
=
32
32
= 1 .
• Mole rat io: 2 H
2
: 1 O
2
. H
2
needs 1 mol O
2
, a v ailable = 1 mol (su?icien t).
2
• O
2
needs 2 mol H
2
, a v ailable = 2 mol (su?icien t). Both fully react.
• W ater: 2 mol H
2
pro duces 2 mol H
2
O = 2×18 = 36 g.
Answ er : No limiting reagen t; 36 g H
2
O.
5.4 Example 4: Molarit y
Problem : Calculate molarit y of 4.9 g H
2
SO
4
(molar mass = 98 g/mol) in 500 mL
solution.
Solution :
• Moles: n =
4.9
98
= 0.05 mol.
• V olume = 0.5 L.
• Molarit y: M =
0.05
0.5
= 0.1 M.
Answ er : 0.1 M.
5.5 Example 5: P ercen t Yield
Problem : In a reaction, 50 g CaCO
3
(100 g/mol) pro duces 20 g CaO (56 g/mol). Find
% yield.
Solution :
• Equation: CaCO
3
? CaO+ CO
2
.
• Moles CaC O
3
:
50
100
= 0.5 mol.
• Theoretical CaO: 0.5 mol× 56 = 28 g.
• % yield:
20
28
×100 = 71.43% .
Answ er : 71.43%.
3