Question

A sample of hydrated copper (II) sulfate weighs 124.8g. The sample has been determined to contain 31.8g of copper (II) ions and 48.0g of sulfate ions.
A) How many molecules if water of crystallization are present in the sample?
B) Deduce the chemical formula of hydrated coppee (II) sulfate.
Please provide the formulas as well. Thank you.

Answer 1

Explanation:

We Have :-

$A \: sample \: of \: hydrated \: copper \: (II) \: sulfate \: of \: weight \: 124.8g. \\ \\ \\ The \: sample \: has \: been \: determined \: to \: contain \: :- \\ \\ \\ </p><p>31.8g \: of \: copper \: (II) \: ions \\ \\ \\ </p><p>48.0g \: of \: sulfate \: ions.$

To Find :-

$A) \: How \: many \: molecules \: of \: water \: of \: crystallization \: are \: present \: in \: the \: sample \: ? \\ \\ \\ </p><p>B) \: Deduce \: the \: chemical \: formula \: of \: hydrated \: copper \: (II) \: sulfate.$

Solution :-

$First \: we \: need \: to \: calculate \: the \: mass \: of \: water \: of \: molecules \\ \\ \\ mass \: of \: sample - mass \: of \: Cu \: ions + mass \: of \: SO_{4} \: ions \\ \\ \\ 124.8 \: g - 31.8 \: g + 48.0 \: g \\ \\ \\ = 45 \: g \\ \\ \\ Moles \: of \: Cu \: = \dfrac{31.8}{64} \\ \\ \\ = 0.496875 \\ \\ \\ Moles \: of \: SO_{4} = \dfrac{48}{96} \\ \\ \\ = 0.5 \\ \\ \\ Moles \: of \: water = \dfrac{45}{18} \\ \\ \\ = 2.5 \\ \\ \\ So \: the \: ratio \: is \: 0.496875:0.5:2.5 \\ \\ \\ Dividing \: by \: the \: lowest \: ratio \\ \\ \\ We \: get \: 1:1:5 \\ \\ \\ We \: have \: 5 \: molecules \: of \: water \: of \: crystallisation \\ \\ \\ so \: chemical \: fomula \: will \: be \: CuSO_{4}.5H_{2}O$

Answer 2

Answer:

1 : 1 : 5

formula is :- CuSO4 . 5 H20