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A student weighs out 3.23 g of Fe(NO3)3 • ?H2O and heats it. The remaining anhydrous salt has a mass of 1.93 g. Calculate the empirical formula for the hydrate.

Respuesta :

Answer:

The empirical formula is : [tex]Fe(NO_3)_3.9H_2O[/tex]

Explanation:

Let us assume the moles of hydration in ferric nitrate is '[tex]x[/tex]'

Thus the formula of the compound becomes :

[tex]Fe(NO_3)_3.xH_2O[/tex]

Now when its heated , the reaction proceeds like  :

[tex]Fe(NO_3)_3.xH_2O[/tex]⇒[tex]Fe(NO_3)_3+xH_2O[/tex]

Given ,

  • The mass of anhydrated [tex]Fe(NO_3)_3=1.93g[/tex]

Molar mass of [tex]Fe,N,O=56,14,16[/tex] respectively.

The mass of 1 mol of [tex]Fe(NO_3)_3 = 56+(14+16*3)*3=56+62*3=242g[/tex]

The number of moles = [tex]\frac{1.93}{242} \\=0.008mol[/tex]

Therefore the number of hydrated moles must also be 0.008

Given ,

  • The mass of hydrated [tex]Fe(NO_3)_3=3.23g[/tex]

[tex]0.008=\frac{3.23}{242+18x} \\242+18x=\frac{3.23}{0.008}\\242+18x=403.75\\18x=161.75\\x=9[/tex]

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