How many grams of `CaWO_4` would contain the same mass of tungsten that is present in 569 g of `FeWO_4`? (W = 184)

Explanation

Let the mass of `CaWO_(4)` be w g. As given
mass of W in w g of `CaWO_(4)`= mass of W in 569 g of `FeWO_(4)`
Moles of W in `CaWO_(4) xx` at. Wt. of W
= moles of W in `FeWO_(4) xx` wt of W.
As both `CaWO_(4)` and `FeWO_(4)` contain 1 atom of W each.
`therefore` moles of `CaWO_(4) xx` at wt. of W = moles of `FeWO_(4) xx` at. wt of W
`w/288 xx 184 = 569/304 xx 184`
w= 539.05 g