Question

Find the molecular weight of the following compounds- a) Ca( HCO3)2 b) (NH4)3PO4 c) Fe2(SO4)3 [ Ca=40, H=1, C=12, O=16, N= 14, P=31, Fe=56, S=32 ]​

Answer 1

Answer :-

→ Molecular mass of Ca(HCO₃)₂ is 162 u .

→ Molecular mass of (NH₄)₃PO₄ is 149 u

→ Molecular mass of Fe₂(SO₄)₃ is 400 u .

Explanation :-

The atomic masses of elements are given as :-

• Calcium = 40 u ; Hydrogen = 1 u

• Carbon = 12 u ; Oxygen = 16 u

• Nitrogen = 14 u ; Phosphorus = 31 u

• Iron = 56 u ; Sulphur = 32 u

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Number (a) :-

The given compound is Ca(HCO₃)₂ i.e. "Calcium bi-carbonate" .

Molecular weight will be :-

= 40 + (1 + 12 + 16 × 3)2

= 40 + (13 + 48)2

= 40 + (61)2

= 40 + 122

= 162 u

Number (b) :-

The given compound is (NH₄)₃PO₄ i.e. "Ammonium Phosphate" .

Molecular weight will be :-

= (14 + 1 × 4)3 + 31 + 16 × 4

= (14 + 4)3 + 31 + 64

= (18)3 + 95

= 54 + 95

= 149 u

Number (c) :-

The given compound is Fe₂(SO₄)₃ i.e. "Ferric Sulphate" .

Molecular weight will be :-

= 56 × 2 + (32 + 16 × 4)3

= 112 + (32 + 64)3

= 112 + (96)3

= 112 + 288

= 400 u

Answer 2

Answer :

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• Molecular weight (a) Ca(HCO₃)₂, (b) (NH₄)₃PO₄ and (c) Fe₂(SO₄)₃? of is 162u, 149u and 400u.

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Explanation :

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Given :

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• Atomic weight of Ca = 40, H = 1, C = 12, O = 16, N = 14, P = 31, Fe = 56 and S = 32.

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To Find :

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• Molecular weight of (a) Ca(HCO₃)₂, (b) (NH₄)₃PO₄ and (c) Fe₂(SO₄)₃?

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Solution :

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$\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(a)\::}}}}$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (1 + 12 + 16\:\times\:3)\:\times\:2$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (13 + 48)\:\times\:2$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (61 \:\times\:2)$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + 112$

$\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\purple{ Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 162}}}}\:\red{\clubsuit}$

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$\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(b)\::}}}}$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = (14 + 1\:\times\:4)\:\times\:3\:+(31 + 16\:\times\:4)$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = (14 + 4)\:\times\:3\:+ (31 + 64)$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 18\:\times\:3\:+ 95$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 54 + 95$

$\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\gray{ Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 149}}}}\:\clubsuit$

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$\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(c)\::}}}}$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 56\:\times\:2 + (32 + 16\:\times\:4)\:\times\:3$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + (32 + 64)\:\times\:3$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + (96\:\times\:3)$

$\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + 288$

$\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\pink{Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 400}}}}\:\green{\clubsuit}$

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• Therefore, molecular weight (a) Ca(HCO₃)₂, (b) (NH₄)₃PO₄ and (c) Fe₂(SO₄)₃? of is 162u, 149u and 400u.

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