A trinitro compound, 1,3,5-tris-(4-nitrophenyl)benzene, on complete reaction with an excess of Sn/HCl gives a major product, which on treatment with an excess of NaNO/HCl at 0°C provides P as the product. P, upon treatment with excess of HO at room temperature, gives the product Q. Bromination of Q in aqueous medium furnishes the product R. The compound P upon treatment with an excess of phenol under basic conditions gives the product S.

The molar mass difference between compounds Q and R is 474 g mol⁻¹ and between compounds P and S is 172.5 g mol⁻¹.

The total number of carbon atoms and heteroatoms present in one molecule of S is:
[Use: Molar mass (in g mol⁻¹): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5. Atoms other than C and H are considered as heteroatoms.]

The correct answer is 51.

Step-by-Step Reaction Sequence and Analysis:

1. Starting Material:
The starting compound is 1,3,5-tris-(4-nitrophenyl)benzene. It consists of a central benzene ring attached to three 4-nitrophenyl groups at the 1, 3, and 5 positions. Its molecular formula is $C_{24}{H}_{15}{N}_3{O}_6$.

2. Reduction with Sn/HCl:
Treating the trinitro compound with an excess of Sn/HCl reduces all three nitro groups (-NO₂) to primary amino groups (-NH₂). The product formed is 1,3,5-tris-(4-aminophenyl)benzene.

3. Diazotization to form Compound P:
Reaction of 1,3,5-tris-(4-aminophenyl)benzene with an excess of NaNO₂/HCl at 0°C converts all three amino groups into diazonium chloride groups (-N₂⁺Cl^-).
Thus, Compound P is 1,3,5-tris-(4-diazoniophenyl)benzene tri-chloride, which has 3 diazonium salt groups.
Molecular formula of P: $C_{24}{H}_{15}{N}_6{\mathrm{Cl}}_3$.

4. Formation of Compound Q (Hydrolysis):
When P is treated with excess water (H₂O) at room temperature, all three diazonium groups are hydrolyzed to phenolic hydroxyl groups (-OH).
Thus, Compound Q is 1,3,5-tris-(4-hydroxyphenyl)benzene, having 3 phenol groups.
Molecular formula of Q: $C_{24}{H}_{18}{O}_3$.

5. Bromination to form Compound R:
Bromination of Q in aqueous medium leads to polybromination of the activated aromatic rings containing the -OH groups. Each phenolic ring has two ortho-positions relative to the -OH group that get brominated.
Since there are 3 such rings, a total of 6 bromine atoms are introduced (2 Br per phenolic ring).
Molecular formula of R: $C_{24}{H}_{12}{O}_3{\mathrm{Br}}_6$.
Verification of molar mass difference between Q and R:
Each bromination replaces a -H with -Br, causing a mass increase of (80 - 1) = 79 g mol^-1 per site.
For 6 sites: $6\times 79=474$ g mol^-1, which matches the given difference of 474 g mol^-1.

6. Formation of Compound S (Azo Coupling):
Compound P (tris-diazonium salt) reacts with an excess of phenol under basic conditions to undergo an azo coupling reaction at the para-position of each phenol molecule.
Since P has 3 diazonium groups, it couples with 3 phenol molecules to form an tris-azo dye.
Structure of Compound S: Central 1,3,5-triphenylbenzene core attached via 3 azo linkages (-N=N-) to 3 phenol rings at their para-positions.

Verification of Molar Mass Difference between P and S:
Compound P: $C_{24}{H}_{15}{N}_6{\mathrm{Cl}}_3$ (Molar Mass = $24\times 12+15\times 1+6\times 14+3\times 35.5=493.5$ g mol^-1)
Compound S: $C_{42}{H}_{30}{N}_6{O}_3$ (Molar Mass = $42\times 12+30\times 1+6\times 14+3\times 16=666$ g mol^-1)
Mass Difference = $666-493.5=172.5$ g mol^-1, which perfectly matches the given value.

7. Calculating Carbon and Heteroatoms in Compound S:

Molecular formula of Compound S is $C_{42}{H}_{30}{N}_6{O}_3$.

• Number of Carbon (C) atoms = 42
• Heteroatoms are atoms other than C and H, which are Nitrogen (N) and Oxygen (O).
• Number of Nitrogen (N) atoms = 6
• Number of Oxygen (O) atoms = 3
• Total number of heteroatoms = $6+3=9$

Therefore, the total number of carbon atoms and heteroatoms present in one molecule of S is:

$\text{Total}=42\text{ (Carbon atoms)}+9\text{ (Heteroatoms)}=51$