Solution Exercise 3

65 percent of the rock is fluorapatite and the mole proportion of F in fluorapatite is the ratio of the atomic weight of F to the molecular weight of fluorapatite, which is 19/504. So, the percent of F in the rock from fluorapatite would be:

[latex]\displaystyle 0.65\frac{19}{504}100\textrm{%}=2.45\textrm{%}[/latex]

5 percent of the rock is fluorite and the mole proportion of F in fluorite is 38/78, the ratio of the atomic weight of F to the molecular weight of fluorite. So, the percent of F in the rock from fluorite would be:

[latex]\displaystyle 0.05\frac{38}{78}100\textrm{%}=2.44\textrm{%}[/latex]

And the sum from these two minerals would be 4.89 percent. However, these two minerals comprise only 80 percent of the rock so that the total F in the rock would be:

(0.8) 4.89% = 3.9% F

Return to Exercise 3

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