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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.

How Many Hydrogen Atoms Are in 0.1488 g of H₃PO₄? | Verla