Question

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

Answer 1

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm


=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


X (+) + O2 (g) ---> X2O          or   


2 X(2+) + O2(g) ----> X2O2 = 2XO     or


4X(3+) + 3O2(g) ---> 2X2O3


 
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X


Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol


That does not correspond to any of the metal with valence 1+


So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.