🔬 Solid State Chemistry

Molecular Formula Determination from Crystal Structures

Problem 1 Easy
A compound is formed by atoms A occupying CCP (cubic close packing) positions and atoms B occupying all the octahedral voids. Determine the molecular formula of the compound.
Answer: Molecular Formula: AB
Detailed Solution: Step 1: Determine number of atoms A in CCP
In CCP (also known as FCC), there are:
• Corner atoms: 8 × 1/8 = 1 atom
• Face center atoms: 6 × 1/2 = 3 atoms
Total A atoms = 1 + 3 = 4 per unit cell

Step 2: Determine number of octahedral voids
Number of octahedral voids in CCP/FCC = Number of atoms in the packing
Total octahedral voids = 4

Step 3: Determine number of B atoms
B atoms occupy ALL octahedral voids
Total B atoms = 4

Step 4: Calculate the ratio
A : B = 4 : 4 = 1 : 1

Molecular Formula: AB
Example: NaCl has this structure where Na+ ions form FCC and Cl- ions occupy all octahedral voids (or vice versa).
Problem 2 Easy
In a crystal structure, atoms X form HCP (hexagonal close packing) arrangement and atoms Y occupy all the tetrahedral voids. What is the formula of the compound?
Answer: Molecular Formula: XY2
Detailed Solution: Step 1: Number of atoms in HCP
For HCP unit cell:
• Corner atoms: 12 × 1/6 = 2 atoms
• Face center atoms: 2 × 1/2 = 1 atom
• Body center atoms: 3 × 1 = 3 atoms
Total X atoms = 2 + 1 + 3 = 6 per unit cell

Step 2: Number of tetrahedral voids
Number of tetrahedral voids = 2 × (Number of atoms in packing)
Total tetrahedral voids = 2 × 6 = 12

Step 3: Number of Y atoms
Y atoms occupy ALL tetrahedral voids
Total Y atoms = 12

Step 4: Calculate the ratio
X : Y = 6 : 12 = 1 : 2

Molecular Formula: XY2
Key Point: In any close packing (HCP or CCP), number of tetrahedral voids = 2n and octahedral voids = n, where n = number of atoms in packing.
Problem 3 Easy
A compound is formed by cation A forming FCC lattice and anion B occupying half of the octahedral voids. Determine the formula of the compound.
Answer: Molecular Formula: A2B
Detailed Solution: Step 1: Number of A atoms in FCC
• Corner atoms: 8 × 1/8 = 1
• Face centers: 6 × 1/2 = 3
Total A atoms = 4

Step 2: Number of octahedral voids
In FCC, octahedral voids = 4

Step 3: Number of B atoms
B occupies HALF of octahedral voids
Total B atoms = 1/2 × 4 = 2

Step 4: Calculate the ratio
A : B = 4 : 2 = 2 : 1

Molecular Formula: A2B
Example: Antifluorite structure like K2O where O2- forms FCC and K+ occupies all tetrahedral voids.
Problem 4 Medium
In a compound, atoms A form CCP arrangement. Atoms B occupy 1/4 of the tetrahedral voids and atoms C occupy half of the octahedral voids. Determine the formula of the compound.
Answer: Molecular Formula: A4B2C2 or simplified as A2BC
Detailed Solution: Step 1: Number of A atoms in CCP/FCC
Total A atoms = 4

Step 2: Number of voids
• Tetrahedral voids = 2 × 4 = 8
• Octahedral voids = 4

Step 3: Number of B atoms
B occupies 1/4 of tetrahedral voids
Total B atoms = 1/4 × 8 = 2

Step 4: Number of C atoms
C occupies 1/2 of octahedral voids
Total C atoms = 1/2 × 4 = 2

Step 5: Calculate the ratio
A : B : C = 4 : 2 : 2 = 2 : 1 : 1

Molecular Formula: A2BC
Problem 5 Medium
Oxide ions (O2-) form a CCP lattice and cations A occupy 1/8 of the tetrahedral voids, while cations B occupy half of the octahedral voids. Determine the formula of the compound.
Answer: Molecular Formula: AB2O4
Detailed Solution: Step 1: Number of O2- ions in CCP
Total O2- ions = 4

Step 2: Number of voids
• Tetrahedral voids = 2 × 4 = 8
• Octahedral voids = 4

Step 3: Number of A cations
A occupies 1/8 of tetrahedral voids
Total A ions = 1/8 × 8 = 1

Step 4: Number of B cations
B occupies 1/2 of octahedral voids
Total B ions = 1/2 × 4 = 2

Step 5: Calculate the ratio
A : B : O = 1 : 2 : 4

Molecular Formula: AB2O4
Example: Spinel structure like MgAl2O4 or ferrites like Fe3O4 (written as FeO·Fe2O3).
Problem 6 Medium
In a crystal, anions X form HCP lattice. Cations Y occupy 1/3 of the octahedral voids and cations Z occupy 1/3 of the tetrahedral voids. Find the simplest formula.
Answer: Molecular Formula: Y2Z4X6 or simplified as YZ2X3
Detailed Solution: Step 1: Number of X anions in HCP
Total X anions = 6

Step 2: Number of voids in HCP
• Octahedral voids = 6
• Tetrahedral voids = 2 × 6 = 12

Step 3: Number of Y cations
Y occupies 1/3 of octahedral voids
Total Y ions = 1/3 × 6 = 2

Step 4: Number of Z cations
Z occupies 1/3 of tetrahedral voids
Total Z ions = 1/3 × 12 = 4

Step 5: Calculate the ratio
Y : Z : X = 2 : 4 : 6 = 1 : 2 : 3

Molecular Formula: YZ2X3
Problem 7 Hard
A compound is formed by atoms A forming FCC lattice. Atoms B occupy all octahedral voids and atoms C occupy all tetrahedral voids. What is the formula of the compound?
Answer: Molecular Formula: AB4C8 or simplified as ABC2
Detailed Solution: Step 1: Number of A atoms in FCC
Total A atoms = 4

Step 2: Number of voids
• Octahedral voids = 4
• Tetrahedral voids = 2 × 4 = 8

Step 3: Number of B atoms
B occupies ALL octahedral voids
Total B atoms = 4

Step 4: Number of C atoms
C occupies ALL tetrahedral voids
Total C atoms = 8

Step 5: Calculate the ratio
A : B : C = 4 : 4 : 8 = 1 : 1 : 2

Molecular Formula: ABC2
Note: Such structures are rare as filling all voids creates a very crowded structure. This is mostly a theoretical example.
Problem 8 Hard
In a compound, O2- ions form CCP arrangement. Fe2+ ions occupy 1/8 of tetrahedral voids while Fe3+ ions occupy half of the octahedral voids. Determine the formula and name the compound.
Answer: Molecular Formula: Fe2+Fe3+2O4 or Fe3O4
Common Name: Magnetite (Inverse Spinel)
Detailed Solution: Step 1: Number of O2- ions
In CCP: O2- = 4

Step 2: Available voids
• Tetrahedral voids = 8
• Octahedral voids = 4

Step 3: Number of Fe2+ ions
Fe2+ in 1/8 of tetrahedral voids
Fe2+ = 1/8 × 8 = 1

Step 4: Number of Fe3+ ions
Fe3+ in 1/2 of octahedral voids
Fe3+ = 1/2 × 4 = 2

Step 5: Formula determination
Fe2+ : Fe3+ : O2- = 1 : 2 : 4

Formula: Fe2+Fe3+2O4 or Fe3O4
Important: Fe3O4 is also written as FeO·Fe2O3. This is the inverse spinel structure of magnetite, a naturally magnetic iron oxide.
Problem 9 Hard
In a crystal structure, atoms X form HCP lattice. 2/3 of the octahedral voids are occupied by atoms Y and 1/3 of tetrahedral voids are occupied by atoms Z. What is the formula of the compound?
Answer: Molecular Formula: X6Y4Z4 or simplified as X3Y2Z2
Detailed Solution: Step 1: Number of X atoms in HCP
Total X atoms = 6

Step 2: Number of voids in HCP
• Octahedral voids = 6
• Tetrahedral voids = 2 × 6 = 12

Step 3: Number of Y atoms
Y occupies 2/3 of octahedral voids
Total Y atoms = 2/3 × 6 = 4

Step 4: Number of Z atoms
Z occupies 1/3 of tetrahedral voids
Total Z atoms = 1/3 × 12 = 4

Step 5: Calculate the ratio
X : Y : Z = 6 : 4 : 4 = 3 : 2 : 2

Molecular Formula: X3Y2Z2
Problem 10 Hard
In a complex oxide, O2- ions are arranged in CCP. Metal ions M+ occupy 1/4 of the tetrahedral voids, M2+ occupy 1/8 of the tetrahedral voids, and M3+ occupy 1/4 of the octahedral voids. Determine the formula of this complex oxide.
Answer: Molecular Formula: M+2M2+M3+O4
Detailed Solution: Step 1: Number of O2- ions in CCP
Total O2- ions = 4

Step 2: Available voids
• Tetrahedral voids = 2 × 4 = 8
• Octahedral voids = 4

Step 3: Number of M+ ions
M+ occupies 1/4 of tetrahedral voids
M+ = 1/4 × 8 = 2

Step 4: Number of M2+ ions
M2+ occupies 1/8 of tetrahedral voids
M2+ = 1/8 × 8 = 1

Step 5: Number of M3+ ions
M3+ occupies 1/4 of octahedral voids
M3+ = 1/4 × 4 = 1

Step 6: Formula determination
M+ : M2+ : M3+ : O2- = 2 : 1 : 1 : 4

Molecular Formula: M+2M2+M3+O4
Charge Balance Check:
Positive charges: 2(+1) + 1(+2) + 1(+3) = +7
Negative charges: 4(-2) = -8
(Note: Charge balance may require adjustment in oxidation states for real compounds)
Problem 11 Easy
In a compound, atoms A form CCP lattice and atoms B occupy 1/4 of the tetrahedral voids. Determine the simplest formula of the compound.
Answer: Molecular Formula: A2B
Detailed Solution: Step 1: Number of A atoms in CCP
Total A atoms = 4

Step 2: Number of tetrahedral voids
Tetrahedral voids in CCP = 2 × 4 = 8

Step 3: Number of B atoms
B occupies 1/4 of tetrahedral voids
Total B atoms = 1/4 × 8 = 2

Step 4: Calculate the ratio
A : B = 4 : 2 = 2 : 1

Molecular Formula: A2B
Example: Similar to antifluorite structure where oxide ions form CCP and metal ions occupy tetrahedral voids (Li2O, Na2O).
Problem 12 Medium
Sulphide ions (S2-) form FCC lattice. Zinc ions (Zn2+) occupy half of the tetrahedral voids. What is the formula of this compound?
Answer: Molecular Formula: ZnS (Zinc Blende structure)
Detailed Solution: Step 1: Number of S2- ions in FCC
Total S2- ions = 4

Step 2: Number of tetrahedral voids
Tetrahedral voids = 2 × 4 = 8

Step 3: Number of Zn2+ ions
Zn2+ occupies 1/2 of tetrahedral voids
Total Zn2+ ions = 1/2 × 8 = 4

Step 4: Calculate the ratio
Zn : S = 4 : 4 = 1 : 1

Molecular Formula: ZnS
Important: This is the Zinc Blende (Sphalerite) structure. ZnS also exists in Wurtzite structure (HCP arrangement). Both are polymorphs of ZnS.
Problem 13 Medium
In a crystal, anions form HCP arrangement. Cations occupy 2/3 of the octahedral voids. What is the formula of the compound?
Answer: Molecular Formula: M2X3 (where M = metal, X = anion)
Detailed Solution: Step 1: Number of anions in HCP
Total anions = 6

Step 2: Number of octahedral voids
Octahedral voids in HCP = 6

Step 3: Number of cations
Cations occupy 2/3 of octahedral voids
Total cations = 2/3 × 6 = 4

Step 4: Calculate the ratio
Cation : Anion = 4 : 6 = 2 : 3

Molecular Formula: M2X3
Example: Corundum structure like Al2O3, Fe2O3, Cr2O3 where O2- forms HCP and metal ions occupy 2/3 of octahedral voids.
Problem 14 Medium
A compound has fluoride ions (F-) forming CCP structure. Calcium ions (Ca2+) occupy all the tetrahedral voids. Determine the formula of this compound.
Answer: Molecular Formula: CaF2 (Fluorite structure)
Detailed Solution: Step 1: Number of F- ions in CCP
Total F- ions = 4

Step 2: Number of tetrahedral voids
Tetrahedral voids = 2 × 4 = 8

Step 3: Number of Ca2+ ions
Ca2+ occupies ALL tetrahedral voids
Total Ca2+ ions = 8

Step 4: Calculate the ratio
Ca : F = 8 : 4 = 2 : 1

Wait! This gives Ca2F, which is incorrect!

Correction: In Fluorite structure, Ca2+ forms FCC and F- occupies tetrahedral voids
• Ca2+ in FCC = 4
• F- in all tetrahedral voids = 8
• Ratio: Ca : F = 4 : 8 = 1 : 2

Molecular Formula: CaF2
Important: Fluorite structure: Ca2+ forms FCC, F- in all tetrahedral voids. Antifluorite: reverse positions (e.g., Na2O).
Problem 15 Hard
In a mixed oxide, O2- ions form CCP lattice. Metal A+ ions occupy 1/8 of tetrahedral voids and metal B3+ ions occupy 1/2 of octahedral voids. Find the formula and determine the oxidation states satisfy charge neutrality.
Answer: Molecular Formula: A+B3+2O4 or AB2O4
Detailed Solution: Step 1: Number of O2- ions
Total O2- = 4

Step 2: Available voids
• Tetrahedral voids = 8
• Octahedral voids = 4

Step 3: Number of A+ ions
A+ in 1/8 of tetrahedral voids
A+ = 1/8 × 8 = 1

Step 4: Number of B3+ ions
B3+ in 1/2 of octahedral voids
B3+ = 1/2 × 4 = 2

Step 5: Formula and charge balance
A : B : O = 1 : 2 : 4
Formula: AB2O4

Charge Balance:
Positive: 1(+1) + 2(+3) = +7
Negative: 4(-2) = -8
Not balanced! Need adjustment in oxidation states.

If A is actually A2+:
Positive: 1(+2) + 2(+3) = +8
Negative: 4(-2) = -8 ✓ Balanced!

Correct Formula: A2+B23+O4
Example: MgAl2O4 (Spinel), ZnFe2O4 (Zinc ferrite) have this structure.
Problem 16 Hard
In a compound, atoms X are in HCP arrangement. Atoms Y occupy all octahedral voids and atoms Z occupy 1/4 of tetrahedral voids. What is the formula?
Answer: Molecular Formula: X2Y2Z
Detailed Solution: Step 1: Number of X atoms in HCP
Total X atoms = 6

Step 2: Number of voids
• Octahedral voids = 6
• Tetrahedral voids = 2 × 6 = 12

Step 3: Number of Y atoms
Y occupies ALL octahedral voids
Total Y atoms = 6

Step 4: Number of Z atoms
Z occupies 1/4 of tetrahedral voids
Total Z atoms = 1/4 × 12 = 3

Step 5: Calculate the ratio
X : Y : Z = 6 : 6 : 3 = 2 : 2 : 1

Molecular Formula: X2Y2Z
Problem 17 Hard
Oxide ions form CCP structure. Ni2+ occupies 1/8 of tetrahedral voids and 1/2 of octahedral voids. Find the formula of this nickel oxide.
Answer: Molecular Formula: Ni3O4
Detailed Solution: Step 1: Number of O2- ions
Total O2- = 4

Step 2: Available voids
• Tetrahedral voids = 8
• Octahedral voids = 4

Step 3: Ni2+ in tetrahedral voids
Ni2+ (tetrahedral) = 1/8 × 8 = 1

Step 4: Ni2+ in octahedral voids
Ni2+ (octahedral) = 1/2 × 4 = 2

Step 5: Total Ni2+ ions
Total Ni = 1 + 2 = 3

Step 6: Formula
Ni : O = 3 : 4

Molecular Formula: Ni3O4
Note: Ni3O4 can be written as NiO·Ni2O3 or Ni2+Ni23+O4, indicating mixed oxidation states.
Problem 18 Medium
In a compound, anions A form FCC lattice. Cations B occupy 1/6 of the octahedral voids. What is the simplest formula?
Answer: Molecular Formula: A6B or A6B (already simplified)
Detailed Solution: Step 1: Number of A anions in FCC
Total A anions = 4

Step 2: Number of octahedral voids
Octahedral voids = 4

Step 3: Number of B cations
B occupies 1/6 of octahedral voids
Total B cations = 1/6 × 4 = 2/3

Step 4: Get whole number ratio
To get whole numbers, multiply by 3:
A : B = 4 : 2/3 = 12 : 2 = 6 : 1

Molecular Formula: A6B
Note: Fractional occupation of voids requires multiplying to get integer formula. This represents the empirical formula per unit cell group.
Problem 19 Hard
In a ternary compound, O2- ions form HCP lattice. Li+ ions occupy all the tetrahedral voids and Co3+ ions occupy all the octahedral voids. Find the formula of this lithium cobalt oxide.
Answer: Molecular Formula: LiCoO2
Detailed Solution: Step 1: Number of O2- ions in HCP
Total O2- = 6

Step 2: Available voids
• Tetrahedral voids = 2 × 6 = 12
• Octahedral voids = 6

Step 3: Number of Li+ ions
Li+ in ALL tetrahedral voids
Li+ = 12

Step 4: Number of Co3+ ions
Co3+ in ALL octahedral voids
Co3+ = 6

Step 5: Calculate ratio
Li : Co : O = 12 : 6 : 6 = 2 : 1 : 1

Formula: Li2CoO (This doesn't match reality!)

Correction for actual LiCoO2 structure:
In reality, LiCoO2 has:
• O2- in CCP (not HCP)
• Li+ in half of octahedral voids
• Co3+ in remaining half of octahedral voids

This gives: Li : Co : O = 2 : 2 : 4 = 1 : 1 : 2
Formula: LiCoO2
Important: LiCoO2 is used in lithium-ion batteries. It has a layered structure with alternating Li and Co layers.
Problem 20 Hard
In a complex structure, atoms P form CCP arrangement. Atoms Q occupy 3/8 of tetrahedral voids and atoms R occupy 3/4 of octahedral voids. Determine the empirical formula.
Answer: Molecular Formula: P4Q3R3
Detailed Solution: Step 1: Number of P atoms in CCP
Total P atoms = 4

Step 2: Available voids
• Tetrahedral voids = 2 × 4 = 8
• Octahedral voids = 4

Step 3: Number of Q atoms
Q occupies 3/8 of tetrahedral voids
Q = 3/8 × 8 = 3

Step 4: Number of R atoms
R occupies 3/4 of octahedral voids
R = 3/4 × 4 = 3

Step 5: Calculate ratio
P : Q : R = 4 : 3 : 3

Molecular Formula: P4Q3R3
Summary Table of Key Formulas:
Structure Type Packing Voids Occupied Formula
Rock Salt FCC (anions) All octahedral AB
Zinc Blende FCC (anions) Half tetrahedral AB
Fluorite FCC (cations) All tetrahedral AB2
Antifluorite FCC (anions) All tetrahedral A2B