How many atoms of oxygen are in one formula unit of $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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How many atoms of oxygen are in one formula unit of $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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Answer 1 · 2016-05-10T05:06:29

Well, clearly there are 12. Why?

Explanation:

$A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ formally derives from an aluminum trication, $A l^{3 +}$ $Al^(3+)$ , and a sulfate dianion, $S O_{4}^{2 -}$ $SO_4^(2-)$ . Now ionic substances form to balance charges, and the only way to achieve neutrality for these net charges is to take 6 positive charges, i.e. $(2 \times A l^{3 +})$ $(2xxAl^(3+))$ , and 6 negative charges $(3 \times S O_{4}^{2 -})$ $(3xxSO_4^(2-))$ .

If you count up all the atoms that carry these charges, there are $(2 \times A l + 3 \times S + 3 \times 4 \times O)$ $(2xxAl+3xxS+3xx4xxO)$ . Is this clear?

How would I form a salt from $C a^{2 +}$ $Ca^(2+)$ and phosphate anion, $P O_{4}^{3 -}$ $PO_4^(3-)$ ? What composition would give electrical neutrality?

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How many atoms of oxygen are in one formula unit of $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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How many atoms of oxygen are in one formula unit of $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?