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ChemistryGeneral Chemistry

How Many Hydrogen Atoms Are in 0.1488 g of Phosphoric Acid?

The 0.1488 g sample contains 1.518 × 10⁻³ mol H₃PO₄. Multiplying by three hydrogen atoms per molecule and Avogadro’s constant gives 2.743 × 10²¹ H atoms.

Question

How many hydrogen atoms are in 0.1488 g0.1488\,\mathrm g of phosphoric acid, H3PO4\mathrm{H_3PO_4}?

Answer

There are 2.743×10212.743\times10^{21} hydrogen atoms.

First find the molar mass of phosphoric acid:

M(H3PO4)=3(1.008)+30.974+4(15.999)=97.994 g mol−1.M(\mathrm{H_3PO_4})=3(1.008)+30.974+4(15.999)=97.994\,\mathrm{g\,mol^{-1}}.

Convert the sample mass to moles of H3PO4\mathrm{H_3PO_4}:

0.1488 g97.994 g mol−1=1.518×10−3 mol H3PO4.\frac{0.1488\,\mathrm g}{97.994\,\mathrm{g\,mol^{-1}}} =1.518\times10^{-3}\,\mathrm{mol\ H_3PO_4}.

Each H3PO4\mathrm{H_3PO_4} molecule contains three hydrogen atoms, so

(1.518×10−3 mol H3PO4)(3 mol H atoms1 mol H3PO4)(6.02214076×1023 atoms mol−1)=2.743×1021 H atoms.(1.518\times10^{-3}\,\mathrm{mol\ H_3PO_4}) \left(\frac{3\,\mathrm{mol\ H\ atoms}}{1\,\mathrm{mol\ H_3PO_4}}\right) (6.02214076\times10^{23}\,\mathrm{atoms\,mol^{-1}}) =\mathbf{2.743\times10^{21}\ H\ atoms}.

The factor of 3 is essential: multiplying only by Avogadro's constant would count H3PO4\mathrm{H_3PO_4} molecules, not hydrogen atoms.

Evidence boundary

The calculation assumes a pure, neutral, anhydrous H₃PO₄ sample and uses average atomic weights plus the exact Avogadro constant. It counts all H atoms represented by the formula; it is not a count of dissociated H⁺ ions in solution or of a particular hydrogen isotope. The result is rounded to four significant figures.

Sources

These references support the concepts and methods used in the explanation above.