When the element A is burned in an excess of oxygen, the oxide A2O3(s) is formed. 0.5386 g of element A is treated with oxygen and 0.711 g of A2O3 are formed. Identify element A.

Answer 1

.....4A + 3O2 ==> 2A2O3
..0.5386...x.....0.711
x = 0.711 - 0.5386 = 0.1724 g O2.

Convert g O2 to g A2O3. Let A = atomic mass A and O = atomic mass O.
1724 x (2*molar mass A2O3/3 molar mass O) = 0.711. Then
0.1724 x (4A+6O)/6O) = 0.711
Substitute and solve for A. My answer is in the low 70s. Look up the atomic mass on the periodic element to identify the element.

Here is another way.
You have 0.1724 g O2. mols O2 = g/molar mass = 0.1724/32 = ?
Convert mols O2 to mols A2O3. That's ? x (2 mol A2O3/3 mols O2) = ? x 2/3 = ?
Since mols = g/molar mass then molar mass = g/mols = ?. I get approx 198.
If molar mass A2O3 is about 198 (you will have a slightly different number). Subtract 48 for 3 oxygen atoms at 16 each and that will give you 2*atomic mass A. Divide by 2 and look it up on the periodic table.