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To determine the percentage of each element present in the salt Potassium Hydrogen Phosphate Pentahydrate (K2HPO4 * 5H₂O).

Chemistry

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The question asks to determine the percentage of each element present in the salt Potassium Hydrogen Phosphate Pentahydrate (K2HPO4⋅5H2O\text{K}_2\text{HPO}_4 \cdot 5\text{H}_2\text{O}).
The given atomic masses are: K=39\text{K}=39, P=31\text{P}=31, O=16\text{O}=16, H=1\text{H}=1.

  1. Calculate the molar mass of K2HPO4⋅5H2O\text{K}_2\text{HPO}_4 \cdot 5\text{H}_2\text{O}.

    The formula contains:

    • 2 atoms of Potassium (K)
    • 1 atom of Phosphorus (P)
    • 4 atoms of Oxygen (O) from HPO4\text{HPO}_4
    • 1 atom of Hydrogen (H) from HPO4\text{HPO}_4
    • 5 molecules of water (5H2O\text{5H}_2\text{O}), which means 5×2=105 \times 2 = 10 atoms of Hydrogen and 5×1=55 \times 1 = 5 atoms of Oxygen.

    Total number of atoms:

    • K: 2
    • P: 1
    • O: 4+5=94 + 5 = 9
    • H: 1+10=111 + 10 = 11

    Molar mass:

    M=(2×39)+(1×31)+(9×16)+(11×1)M = (2 \times 39) + (1 \times 31) + (9 \times 16) + (11 \times 1) M=78+31+144+11M = 78 + 31 + 144 + 11 M=264 g/molM = 264 \text{ g/mol}
  2. Calculate the mass contribution of each element.

    • Mass of K: 2×39=78 g2 \times 39 = 78 \text{ g}
    • Mass of P: 1×31=31 g1 \times 31 = 31 \text{ g}
    • Mass of O: 9×16=144 g9 \times 16 = 144 \text{ g}
    • Mass of H: 11×1=11 g11 \times 1 = 11 \text{ g}
  3. Calculate the percentage of each element.

    Percentage of an element = mass of elementmolar mass of compound×100%\frac{\text{mass of element}}{\text{molar mass of compound}} \times 100\%

    Percentage of K:

    %K=78 g264 g×100%\% \text{K} = \frac{78 \text{ g}}{264 \text{ g}} \times 100\% %K=29.545%\% \text{K} = 29.545\% %K≈29.55%\% \text{K} \approx 29.55\%

    Percentage of P:

    %P=31 g264 g×100%\% \text{P} = \frac{31 \text{ g}}{264 \text{ g}} \times 100\% %P=11.742%\% \text{P} = 11.742\% %P≈11.74%\% \text{P} \approx 11.74\%

    Percentage of O:

    %O=144 g264 g×100%\% \text{O} = \frac{144 \text{ g}}{264 \text{ g}} \times 100\% %O=54.545%\% \text{O} = 54.545\% %O≈54.55%\% \text{O} \approx 54.55\%

    Percentage of H:

    %H=11 g264 g×100%\% \text{H} = \frac{11 \text{ g}}{264 \text{ g}} \times 100\% %H=4.166%\% \text{H} = 4.166\% %H≈4.17%\% \text{H} \approx 4.17\%

    The percentages of the elements are:

    • Potassium (K): 29.55%\boxed{29.55\%}
    • Phosphorus (P): 11.74%\boxed{11.74\%}
    • Oxygen (O): 54.55%\boxed{54.55\%}
    • Hydrogen (H): 4.17%\boxed{4.17\%}
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