Question #45a3b

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Question #45a3b

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Answer 1 · 2014-12-17T01:46:49

The molecule $A l {(P O_{4})}_{3}$ $Al(PO_4)_3$ contains the elements Aluminum, Phosphorus and Oxygen.

For this molecule the Aluminum has one (1) atom.
The Phosphate ion $(P O_{4})$ $(PO_4)$ has three units which means the subscript 3 outside of the parenthesis gets distributed to each value inside thus there are three (3) Phosphorus atoms and twelve (12) Oxygen atoms.

Al = 1
P = 3
O = 16

This means that the molecule $A l {(P O_{4})}_{3}$ $Al(PO_4)_3$ contains a total of 16 atoms (1 +3+12 = 16).

Technically however, there is no molecule of $A l {(P O_{4})}_{3}$ $Al(PO_4)_3$ since Aluminum has an ionic charge of +3, $A l^{+} 3$ $Al^+3$ and Phosphate has a charge of -3, $P O_{4}^{- 3}$ $PO_4^-3$ . This means that the actual molecule is $A l P O_{4}$ $AlPO_4$ .

Therefore the atom count is

Al = 1
P = 1
O = 4

For a total of 6 atoms in the molecule.

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Question #45a3b

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