Question

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If the total mass of oxygen present in CuSO4.5H2O is 160 g, then calculate the total number of electrons present in CuSO4.5H2O

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Answer 1

Molar mass of CuSO4.5H2O=1Cu+1S+4O+10H+5O=63.5+32.1+4×16+10×1+5×16=249.6g

Oxygen and copper will be present in 9:1 molar ratio always.

63.5g of Cu144g of O =3.782g of Cux g of O

Hence, grams of oxygen in this sample=63.5144×3.782=8.576gram

Answer 2

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Given Substance: CuSO₄. 5 H₂O

Given Mass: 160 g

Atomic Mass of constituting elements are:

Cu = 63.5 g

S = 32 g

O = 16 g

H = 1 g

Hence Molar Mass of CuSO₄. 5 H₂O is:

$\tt \orange{ \longrightarrow 63.5 + 32 + 4(16) + 5 ( 2(1) + 16 )}$

$\tt \orange{ \longrightarrow 63.5 + 32 + 64 + 5 ( 18 )}$

$\boxed{ \tt249.5 g}$

Calculating the number of moles of the given compound, we get:

$\tt \red{ \longrightarrow \: n =\frac{Given Mass}{Molar Mass} }$

$\tt \red{ \longrightarrow n = \frac{160 g}{249.5 g} }$

$\boxed{ \tt{ n = 0.64 \: moles}}$

Now let us calculate the number of molecules in 0.64 moles of CuSO₄.5H₂O. We know that, 1 mole of any compound has Avogadro number of molecules. Hence 0.64 moles would have:

$\tt \purple{ \longrightarrow1 \: mole : Nₐ :: 0.64 \: mole = ?}$

$\tt \purple{ \longrightarrow x = 0.64 × Nₐ}$

$\tt \purple { \longrightarrow 0.64 × 6.023 × 10²³}$

$\boxed{ \tt{3.85 × 10²³ \: molecules}}$

Let us calculate the number of electrons present in 1 molecule of CuSO₄.5H₂O. We know that, number of electrons present in an element is equal to it's atomic number. Hence we get:

$\tt \pink{ Cu = 1 × 29 = 29 \: electrons}$

$\tt \pink{ \longrightarrow \: S = 1 × 16 = 16 \: electrons}$

$\tt \pink{ \longrightarrow \: O = 9 × 8 = 72 \: electrons}$

$\tt \pink{ \longrightarrow H = 10 × 1 = 10 \: electrons}$

Total number of electrons in 1 molecule of CuSO₄.5H₂O is:

$\boxed{ \tt (29 + 16 + 72 + 10) = 127 electrons.}$

Therefore, if 1 molecule contains 127 electrons, then 3.85 × 10²³ molecules would contain:

$\tt \longrightarrow x = \dfrac{127 \times 3.85 \times 10^{23}}{1}\\\\\\ \boxed{ \tt{ x = 4.8895 \times 10^{25} \: \text{electrons}}}$

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