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.
