Answer
a) $\omega (Ca)=39.84\%$
b) $\omega (P)=18.53\%$
Work Step by Step
a) $\omega (Ca)=\frac{10\times Ar(Ca)}{Mr(Ca_{10}(PO_{4})_{6}(OH)_{2})} = \frac{10\times 40}{1004}=39.84\%$
b) $\omega (P)=\frac{6\times Ar(P)}{Mr(Ca_{10}(PO_{4})_{6}(OH)_{2})} = \frac{6\times 31}{1004}=18.53\%$
