**Calculate the molar mass of #"X"#**

1 mol of #"X"_3"O"_4# contains 3 mol of #"X"# and 4 mol of #"O"#.

Assume that you have 1 mol of #"X"_3"O"_4#.

Then

#"Mass of O" = 4 color(red)(cancel(color(black)("mol O"))) × "16.00 g O"/(1 color(red)(cancel(color(black)("mol O")))) = "64.00 g O"#

and

#"Mass of X"_3"O"_4 = 64.00 color(red)(cancel(color(black)("g O"))) × ("100 g X"_3"O"_4)/(27.6 color(red)(cancel(color(black)("g O")))) = "232 g"#

∴ #"Mass of X" = "Mass of X"_3"O"_4 - "Mass of O" = "232 g - 64.00 g" = "168 g"#

This is the mass of 3 mol of #"X"#.

∴ #"Molar mass of X" = "168 g"/"3 mol" = "56.0 g/mol"#

**Calculate the empirical formula of the second oxide**

Assume that you have 100 g of the second oxide.

Then it contains 30 g of #"O"# and 70 g of #"X"#.

#"Moles of X" = 70 color(red)(cancel(color(black)("g X"))) × "1 mol X"/(56.0 color(red)(cancel(color(black)( "g X")))) = "1.25 mol X"#

#"Moles of O" = 30 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "1.88 mol O"#

From this point on, I like to summarize the calculations in a table.

#bbul("Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×2"color(white)(mm)"Integers")#

#color(white)(mm)"X" color(white)(XXXmm)70 color(white)(Xmm)1.25
color(white)(mmm)1color(white)(mmmm)2color(white)(mmmmml)2#

#color(white)(mm)"O"color(white)(XXXmm)30 color(white)(Xmm)1.88color(white)(mmm)1.50color(white)(mmll)3.00color(white)(mmmll)3#

The empirical formula is #"X"_2"O"_3#.