Answer
is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.
<span>
Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] =
3s(Ca₃(PO₄)₂) =
3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.
s = ⁵√(Ksp ÷ 108).
The empirical and molecular formulas will be
and
respectively.
<h3>Empirical and molecular formula</h3>
The compound contains C, H, and O.
C = 61.15/12 = 5.0958
H = 5.3/1 = 5.3
O = 31.55/16 = 1.9719
Divide by the smallest
C = 2.6
H = 2.7
O = 1
Thus, the empirical formula is 
Empirical formula mass = (12x5) + (1x5) + 16x2 = 97
n = 152.15/97 = 2
The molecular formula is 
More on molecular and empirical formulas can be found here: brainly.com/question/14425592
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