Answer:
Molecule = \[C{{a}_{3}}{{\left(
P{{O}_{4}} \right)}_{2}}\]
Mass of =\[Ca=3\times
40=120amu\]
Mass of Phosphorous = \[2\times
31=62amu\]
Mass of Oxygen = \[8\times
16=128amu\]
\[\therefore \]Molar mass of
\[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}=120+62+128=310amu\]
Mass % of Calcium = \[\frac{120}{310}\times
100=38.7%\]
Mass % of Phosphorous = \[\frac{62}{310}\times
100=20%\]
Mass % of Oxygen = \[\frac{128}{310}\times
100=41.3%\]
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