chem empirical formula 一問

Question

8.38g of the oxide of metal Y requires 2.88g of carbon monoxide for complete reduction. What is the empirical formula of the oxide of metal Y? (RAM of Y=40)

Please include detailed steps for the calculation

Answer 1

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 ?