# How many subatomic particles are present in ONE molecule of H_3PO_4?

Well, we specify the isotopomer ${\text{^1H_3""^31P}}^{16} {O}_{4}$.....and I gets $\text{48 protons, 48 electrons, and 48 neutrons......}$
For $H , Z = 1$; $P , Z = 15$; $O , Z = 8$, and the $\text{atomic number, Z}$ BY DEFINITION specifies the number of nuclear protons present in each element.
And so we gots $3 \times 1 + 15 + 4 \times 8 = 48 \cdot \text{protons}$, i.e. 48 massive, positively charged, nucular particles. And there are also 48 electrons, i.e. 48 negatively charged of negligible mass, that are conceived to whizz about the nuclear cores.
As regards neutrons, NEUTRALLY charged, massive, nucular particles, ""^1H contributes NO neutrons. ""^31P contributes 16 neutrons, and oxygen contributes $4 \times 8$ neutrons, i.e. 48 neutrons in total.