Question

X forms an oxide X2O3 , 0.36 grams of X forms 0.56 grams of X2O3 .so that atomic weight of X is

Answer 1

let M be moles of X2O3
so in M moles of X2O3, you have
2 M of X2, and 3 M of O
so if mass X in 2M is .36, and mass O in 3M is .20, then
M=(.2/3)/18=.2/54=1/270 moles
but also
M=.18/atomicwtX

or atomicwtX=.18*270=48.6

Answer 2

Calculate % each element.
%X = (0.36/0.56)*100 = 64.3%
%O = (0.2/0.56)*100 = 35.7%

Take 100 g sample to give
64.3 g X
35.7 g O

Convert to mols.
64.3/at mass X = ?mols X.
35.7/16 = 2.23 mols O.
We know this is in the ratio of 2:3 so, letting X = atomic mass X
64.3/X = ?......*some # || = 2
35.7/16 = 2.23..*some # || = 3
What is some #? It is
2.23*# = 3. That # must be 3/2.23 = 1.34. The number multiplier for metal X must also be 1.34 so ? must be 2/1.34 = 1.49 so 64.3/X = 1.49 and X (atomic mass X) must be 64.3/1.49 = 43.2.
CHECK:
0.36g X/43.2 = 0.008333 mols X
0.2g O/16 = 0.0125 mols O
Divide both numbers by the smaller number to obtain:
0.00833/0.00833 = 1
0.0125/0.00833 = 1.50 and these are in the ratio of small whole numbers of 2:3 so formula is X2O3. The 43.2 number checks out for atomic mass X OK.
Bob Pursley obtained slightly higher in his work above which probably is the result of a typo or a rounding error. You can check it another way and see if the percentages agree.
X2O3 molar mass is (2*43.2)+3(16) = 134.4
%X = [2*(43.2)/134.4]*100 = 64.28 which rounds to 64.3 and
%O = [3*16/134.4]*100 = 35.71 which rounds to 35.7. Formula checks and percent checks; must be right.