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Schach [20]
3 years ago
6

Xenon (xe) of mass 5.08 g reacts with fluorine to form 9.49 g of a xenon fluoride compound. what is the empirical formula of thi

s compound?
Chemistry
1 answer:
jenyasd209 [6]3 years ago
8 0
We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.

  mass of Xenon (Xe) = 5.08 g
  mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g

Determine the number of moles of each of the element in the compound. 
    moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
   moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F

The empirical formula is therefore,
      Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
<em>     XeF₆</em>
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1- Alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2. XH2O. To find the value of X, you can heat the sample
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THE VALUE OF X IS 7 AND THE FORMULA OF THE HYDRATED SALT IS KAl(SO4)2.7H20

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Mass of water lost = 2.16 g

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The expression,

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X = molar mass of anhydrous salt * mass of water lost / mass of anhydrous salt * H20

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X = 557.28/ 85.32

X = 6.53

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