Question

identify the number of atoms found in each chemical below:
al₂(co₃)₃ 4 mg₃(po₄)₂
? aluminum atom(s) ? magnesium atom(s)
? carbon atom(s) ? phosphorus atom(s)
? oxygen atom(s) ? oxygen atom(s)

Explanation:

Step1: Analyze \(Al_2(CO_3)_3\)

• For aluminum (\(Al\)): The subscript is \(2\), so there are \(2\) aluminum atoms.
• For carbon (\(C\)): Inside the carbonate \((CO_3)\) group, the subscript of \(C\) is \(1\), and there are \(3\) carbonate groups (\(3\) outside the parentheses). So \(1\times3 = 3\) carbon atoms.
• For oxygen (\(O\)): Inside the carbonate group, the subscript of \(O\) is \(3\), and there are \(3\) carbonate groups. So \(3\times3=9\) oxygen atoms.

Step2: Analyze \(4Mg_3(PO_4)_2\)

• For magnesium (\(Mg\)): The subscript is \(3\), and there are \(4\) formula units. So \(3\times4 = 12\) magnesium atoms.
• For phosphorus (\(P\)): Inside the phosphate \((PO_4)\) group, the subscript of \(P\) is \(1\), there are \(2\) phosphate groups per formula unit, and \(4\) formula units. So \(1\times2\times4=8\) phosphorus atoms.
• For oxygen (\(O\)): Inside the phosphate group, the subscript of \(O\) is \(4\), there are \(2\) phosphate groups per formula unit, and \(4\) formula units. So \(4\times2\times4 = 32\) oxygen atoms.

Answer:

• \(Al_2(CO_3)_3\): \(2\) aluminum atom(s), \(3\) carbon atom(s), \(9\) oxygen atom(s)
• \(4Mg_3(PO_4)_2\): \(12\) magnesium atom(s), \(8\) phosphorus atom(s), \(32\) oxygen atom(s)