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#.
