[jayjay112]

Hi here is the question and anser but can someone explain the calculation in (i) and (ii) where
is the 59.57 coming from etc???

A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal.
(Need table of atomic weights!)

(i) In lower fluoride, find weight of metal that reacts with 1 mole F atoms = 59.57x(19/40.43) = 27.99 g
(The atomic weight of the metal must be an integer multiple of this value)
(ii)In higher fluoride, find how much F reacts with this much (27.99 g) M = 50.44x(27.99/49.56) g = 28.51 g = 1.5 moles of F ... but F is univalent so 55.98 g M = 3 moles of F (or some integer multiple of this)

No known MF9; Cd only forms CdF2  so metal is Fe

Thank you

[cliverlong]

but can someone explain the calculation in (i) and (ii) where
is the 59.57 coming from etc???[/b]

59.57 + 40.43 = 100.00

Ok, but why is this useful?

Tell us what you know about moles, mass and rmm/formula mass and we can go from there.

If you know nothing about moles, mass and rmm/formula mass  then you have no hope of tackling this question and should be tackling a simpler question.

Clive