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 starting material is 1,3,5-tris(4-nitrophenyl)benzene.

Step 1 : Reduction
Sn/HCl converts each $$\;p\text{-}NO_2\;$$ group to $$\;p\text{-}NH_2\;$$.
Product: 1,3,5-tris(4-aminophenyl)benzene.

Step 2 : Diazotisation
Excess $$NaNO_2/HCl$$ at $$0^{\circ}\text{C}$$ changes every $$-NH_2$$ to a diazonium chloride group $$-N_2^+Cl^-$$.
Compound P: 1,3,5-tris(4-diazonium-chloride-phenyl)benzene.

Step 3 : Hydrolysis
Warming P with water replaces each $$-N_2^+Cl^-$$ by $$-OH$$, giving Q = 1,3,5-tris(4-hydroxyphenyl)benzene.

Step 4 : Bromination
In aqueous medium the phenolic ring gives electrophilic substitution at both ortho positions.
Each phenol ring of Q picks up two bromine atoms $$\;(2\times79)\;$$ and loses two hydrogens $$\;(2\times1)$$.
Mass increase per ring $$= 2\,(79-1)=158$$.
For three rings, total increase $$=3\times158=474\;\text{g mol}^{-1}$$, exactly the value given. Thus R contains six bromine atoms.

Step 5 : Azo coupling
Compound P reacts with excess phenol in basic medium. Each diazonium chloride group couples with a phenoxide ion at its para position:

$$\text{Ar}-N_2^+Cl^- + C_6H_5O^- \rightarrow \text{Ar}-N=N-C_6H_4OH + Cl^- + H^+$$

Mass change for one diazonium site:
• Phenol adds $$C_6H_5OH$$: $$6\times12 + 6\times1 + 16 = 94$$.
• A proton is lost during coupling: $$94-1 = 93$$.
• Chloride ion $$Cl^-$$ of mass $$35.5$$ is lost.
Net gain per site $$=93-35.5 = 57.5\;\text{g mol}^{-1}$$.
Given total gain $$172.5\;\text{g mol}^{-1}$$, the reaction involves three identical sites $$\;(172.5/57.5 = 3)$$, confirming that every diazonium group has coupled.

Compound S is therefore
1,3,5-tris[4-(p-hydroxyphenylazo)phenyl]benzene.

Step 6 : Counting atoms in one molecule of S

Carbons:
• Central benzene ring: $$6$$ C.
• Three inner phenyl rings: $$3\times6 = 18$$ C.
• Three outer phenol rings: $$3\times6 = 18$$ C.
Total carbon atoms $$= 6 + 18 + 18 = 42$$.

Heteroatoms (all atoms other than C and H):
• Azo nitrogens: each $$-N=N-$$ has $$2$$ N; for three links $$2\times3 = 6$$ N.
• Phenolic oxygens: one $$O$$ on each outer ring $$\Rightarrow 3$$ O.
Total heteroatoms $$= 6 + 3 = 9$$.

Step 7 : Required total
Carbon atoms $$+$$ heteroatoms $$= 42 + 9 = 51$$.

Hence, the total number of carbon atoms and heteroatoms present in one molecule of S is 51.