✔ 最佳答案
Let YmOn be the required empirical formula.
Molar mass of CO = 12 + 16 = 28 g mol-1
Molar mass of O = 16 g mol-1
YmOn + nCO → mY + nCO2
Mole ratio YmOn : CO = 1 : n
No. of moles of CO = mass/(molar mass) = 2.88/28 = 0.103 mol
No. of moles of YmOn = (0.103/n) mol
1 mol of YmOn contains n mol of O atoms.
No. of moles of O in YmOn = (0.103/n) x n = 0.103 mol
Mass of O in YmOn = mol x (molar mass) = 0.103 x 16 = 1.65 g
In 8.38 g of YmOn :
Mass of O = 1.65 g
Mass of Y = 8.38 - 1.65 = 6.73 g
Mole ratio Y : O = (6.73/40) : 0.103 = 5 : 3
Empirical formula = Y5O3
2007-08-18 01:07:38 補充:
Actually, what is this oxide ? Is it a hypothetical one ?