Find the Empirical Formula!

[Modyasoo1]

0.529gm Alumminum and 0.471gm Oxygen Al=27 O=16




0.979gm Sodium 1.365gm Sulphur 1.021gm Oxygen

Na=23 S=32 O=16

i need help please

[Dark*Phantom]

Answer to quesion 1
If the mass of oxygen (O) = 0.471 gm
Mass of aluminium (Al) = 0.529 gm
Molar mass of oxygen = 16gm
Molar mass of aluminium = 27 gm
No. of moles in oxygen = mass / molar mass = 0.471/16 = 0.029 moles
No. of moles in aluminium = mass / molar mass = 0.529/27 = 0.019 moles
Ratio between aluminium and oxygen = 0.19:0.29 = 1:1.5
= 2:3

The empirical formula is Al2O3

Answer to quesion 2
If the mass of sodium (Na) = 0.979 gm
Mass of oxygen (O) = 1.021 gm
Mass of sulphur (S) = 1.365 gm
Molar mass of sodium = 23 gm
Molar mass of oxygen = 16 gm
Molar mass of aluminium = 27 gm
No. of moles in sodium = mass / molar mass = 0.979/23 = 0.043 moles
No. of moles in sulphur = mass / molar mass = 1.365/32 = 0.043 moles
No. of moles in oxygen = mass / molar mass = 1.021/16 = 0.064 moles
Ratio between sodium, sulphur and oxygen = 0.043 : 0.043 : 0.064 = 1 : 1 : 1.5 = 2 : 2 : 3

The empirical formula is Na2S2O3