Reacting Mass Quesion(F4)

Question

13.70 g of an oxide of metal X (relative atomic mass = 207.0) reacts with excess hydrogen to produce 1.44 g of water.What is the empirical formula of the oxide?

Answer 1

2H2 + O2 --> 2H2O

1.44g water has 1.44/18 = 0.08 mole

∴Hydrogen is excess
∵Oxygen is the limited reactant

By the formula, 2 mole of water is produced by 1 mole of oxygen
∵The mole of oxygen is 0.08/2 = 0.04 mole
∵The mass in the oxide is 0.04*32 = 1.28 g

The remaining mass is the mass of the metal X = 13.7-1.28 = 12.42 g
∵The mole of the metal X is 12.42/207 = 0.06 mole

By mole ratio
The relative ratio of the oxygen is 0.04/0.04 = 1
The relative ratio of the metal X is 0.06/0.4 = 1.5

To make the number be integer, 1*2 = 2
1.5*2 = 3
.
∵The empirical formula of the oxide is X3O2

2007-04-19 10:39:54 補充：
SOrry計mole ratio 果陣計錯左By mole ratioThe relative ratio of the oxygen atom is 0.08/0.06 = 1.3333The relative ratio of the metal X atom is 0.06/0.06 = 1To make the number be integer, 1.333*3 = 41*3 = 3∵The empirical formula of the oxide is X3O4

Answer 2

Molecular mass of H2O = 2x1 + 16 = 18 g mol-1
Fraction by mass of O in H2O = 16/18

Mass of O in 13.70 g of the oxide of X
= Mass of O in 1.44 g of H2O
= 1.44 x (16/18)
= 1.28 g
Mass of X in 13.70 g of the oxide of X
= 13.70 - 1.28
= 12.42 g

In the oxide of X, mole ratio X : O
= (12.42/207) : (1.28/16)
= 0.06 : 0.08
= 3 : 4

Empirical formula = X3O4

2007-04-18 22:30:49 補充：
Checking :Suppose it is only known that 13.7 g of X3O4 is used.X3O4 4H2 → 3X 4H2ONo. of moles of X3O4 used = 13.7/685 = 0.02 molNo. of moles of H2O formed = 0.02 x 4 = 0.08 molMass of H2O formed = 0.08 x 18 = 1.44 gThis is consistent with the data given.