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:

$Given :</p><p></p><p>\:</p><p></p><p>Atomic weight of Ca = 40, H = 1, C = 12, O = 16, N = 14, P = 31, Fe = 56 and S = 32.</p><p></p><p>\:</p><p></p><p>To Find :</p><p></p><p>\:</p><p></p><p>Molecular weight of (a) Ca(HCO₃)₂, (b) (NH₄)₃PO₄ and (c) Fe₂(SO₄)₃?</p><p></p><p>\:</p><p></p><p>Solution :</p><p></p><p>\:</p><p></p><p>\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(a)\::}}}}♣Calculationfor(a):</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (1 + 12 + 16\:\times\:3)\:\times\:2\end{gathered}→Molecularweight[Ca(HCO3)2]=40+(1+12+16×3)×2</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (13 + 48)\:\times\:2\end{gathered}→Molecularweight[Ca(HCO3)2]=40+(13+48)×2</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + (61 \:\times\:2)\end{gathered}→Molecularweight[Ca(HCO3)2]=40+(61×2)</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 40 + 112\end{gathered}→Molecularweight[Ca(HCO3)2]=40+112</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\purple{ Molecular\:weight_{[Ca(HCO_{3})_{2}]} = 162}}}}\:\red{\clubsuit}\end{gathered}→Molecularweight[Ca(HCO3)2]=162♣</p><p></p><p>\:</p><p></p><p>\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(b)\::}}}}♣Calculationfor(b):</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = (14 + 1\:\times\:4)\:\times\:3\:+(31 + 16\:\times\:4)\end{gathered}→Molecularweight[(NH4)3PO4]=(14+1×4)×3+(31+16×4)</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = (14 + 4)\:\times\:3\:+ (31 + 64)\end{gathered}→Molecularweight[(NH4)3PO4]=(14+4)×3+(31+64)</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 18\:\times\:3\:+ 95\end{gathered}→Molecularweight[(NH4)3PO4]=18×3+95</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 54 + 95\end{gathered}→Molecularweight[(NH4)3PO4]=54+95</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\gray{ Molecular\:weight_{[(NH_{4})_{3}PO_{4}]} = 149}}}}\:\clubsuit\end{gathered}→Molecularweight[(NH4)3PO4]=149♣</p><p></p><p>\:</p><p></p><p>\footnotesize\underline{\clubsuit{\textit{\textbf{\:Calculation\:for\:(c)\::}}}}♣Calculationfor(c):</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 56\:\times\:2 + (32 + 16\:\times\:4)\:\times\:3\end{gathered}→Molecularweight[Fe2(SO4)3]=56×2+(32+16×4)×3</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + (32 + 64)\:\times\:3\end{gathered}→Molecularweight[Fe2(SO4)3]=112+(32+64)×3</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + (96\:\times\:3)\end{gathered}→Molecularweight[Fe2(SO4)3]=112+(96×3)</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\sf Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 112 + 288\end{gathered}→Molecularweight[Fe2(SO4)3]=112+288</p><p></p><p>\begin{gathered}\\ \footnotesize\rightarrow \:\underline{\boxed{\bf{\pink{Molecular\:weight_{[Fe_{2}(SO_{4})_{3}]} = 400}}}}\:\green{\clubsuit}\end{gathered}→Molecularweight[Fe2(SO4)3]=400♣</p><p></p><p>\:</p><p></p><p>Therefore, molecular weight (a) Ca(HCO₃)₂, (b) (NH₄)₃PO₄ and (c) Fe₂(SO₄)₃? of is 162u, 149u and 400u.$

Answer 2

Answer:

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 ]