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Question Number 178166 by Spillover last updated on 13/Oct/22

8.5g of hydrated copper (ii)  sulphate CuSO_4 .xH_2 O was heated  to dryness.if  4.0g of anhydrous  copper (ii) sulphate[CuSO_4 ] obtain  Find number of molecule of water of  crystallization

$$\mathrm{8}.\mathrm{5g}\:\mathrm{of}\:\mathrm{hydrated}\:\mathrm{copper}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{sulphate}\:\mathrm{CuSO}_{\mathrm{4}} .\mathrm{xH}_{\mathrm{2}} \mathrm{O}\:\mathrm{was}\:\mathrm{heated} \\ $$$$\mathrm{to}\:\mathrm{dryness}.\mathrm{if}\:\:\mathrm{4}.\mathrm{0g}\:\mathrm{of}\:\mathrm{anhydrous} \\ $$$$\mathrm{copper}\:\left(\mathrm{ii}\right)\:\mathrm{sulphate}\left[\mathrm{CuSO}_{\mathrm{4}} \right]\:\mathrm{obtain} \\ $$$$\mathrm{Find}\:\mathrm{number}\:\mathrm{of}\:\mathrm{molecule}\:\mathrm{of}\:\mathrm{water}\:\mathrm{of} \\ $$$$\mathrm{crystallization} \\ $$

Answered by MikeH last updated on 14/Oct/22

  mass of H_2 O = 4.5 g  Empirical Formula               CuSO_4                            H_2 O  moles:    ((4.0)/(64+32+4(16)))        ((4.5)/(18))               = 0.025                           0.25  mole ratio:    ((0.025)/(0.025))                  ((0.25)/(0.025))                   =       1                           10  ⇒ CuSO_4 .10 H_2 O

$$\:\:\mathrm{mass}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \mathrm{O}\:=\:\mathrm{4}.\mathrm{5}\:\mathrm{g} \\ $$$$\mathrm{Empirical}\:\mathrm{Formula} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{CuSO}_{\mathrm{4}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\mathrm{moles}:\:\:\:\:\frac{\mathrm{4}.\mathrm{0}}{\mathrm{64}+\mathrm{32}+\mathrm{4}\left(\mathrm{16}\right)}\:\:\:\:\:\:\:\:\frac{\mathrm{4}.\mathrm{5}}{\mathrm{18}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0}.\mathrm{025}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}.\mathrm{25} \\ $$$$\mathrm{mole}\:\mathrm{ratio}:\:\:\:\:\frac{\mathrm{0}.\mathrm{025}}{\mathrm{0}.\mathrm{025}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{0}.\mathrm{25}}{\mathrm{0}.\mathrm{025}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10} \\ $$$$\Rightarrow\:\mathrm{CuSO}_{\mathrm{4}} .\mathrm{10}\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$

Commented by Spillover last updated on 16/Oct/22

thanks

$${thanks} \\ $$

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