Find a molecule knowing its contents

Knowing that $CH_4$ has an atomic mass of 16.04 g/mol, and that the atomic mass of our $Fe_xO_y$ is 10 times that, we can multiply $16.04 * 10$ to determine the atomic mass of our molecule to be $\approx160.4$ g/mol.

Knowing that the atomic masses of Fe and O are 55.845 u and 15.999 u, respectively, an equation can be made to solve for the variables indicating how many of each atom (Fe and O) will be in the molecule:

Where x = # Fe atoms, y = # O atoms:

$(55.845)x + (15.999)y = 160.4$

The equation above can be used to check our work.

It's also given that 70% of the molecule's total atomic mass (which we know to be 160.4 g/mol) will be made up of iron, and 30% will be made up of oxygen. This is enough information to determine the exact masses of each element that will make up the molecule. From this information we can make the equations:

$(0.7)x = 160.4$

$(0.3)y = 160.4$

Where x = mass of Fe in $Fe_xO_y$ , and y = mass of O in $Fe_xO_y$ , solving for x and y gives us:

$x=160.4(0.7)=112.28$ g/mol of $Fe$ in $Fe_xO_y$

$y=160.4(0.3)=48.120$ g/mol of $O$ in $Fe_xO_y$

Finally, since we know the atomic mass of each element, we can divide the total atomic mass of each element (the ones we solved for above) by their individual atomic masses to determine how many of each atom make up the molecule.

$112.28/55.845\approx2$

$48.120/15.999\approx3$

Therefore, there must be $x=2$ $Fe$ atoms and $y=3$ $O$ atoms, giving us the molecular formula $**Fe_2O_3**$ .

Mr(CH4)=16 ∴Mr(FexOy)=160 The molar mass of Fex is 70%×160=112 The molar mass of Oy is 48 ∴The amount of iron in this molecule is (112/56)=2 The amount of oxygen is 3 ∴The molecule is Fe2O3 This is the normal solution