Question

In the production of printed circuit boards for the electronics industry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching and the protective polymer is finally removed by solvents. One etching reaction is

$\mathbf{\text{Cu}}{\left({\text{NH}}_3\right)}_{\mathbf{4}}{\mathbf{\text{Cl}}}_{\mathbf{2}}\left(aq\right)\mathbf{+}\mathbf{4}{\mathbf{\text{NH}}}_{\mathbf{3}}\left(aq\right)\mathbf{+}\mathbf{\text{Cu}}\left(s\right)\mathbf{\to }\mathbf{2}\mathbf{\text{Cu}}{\left({\text{NH}}_3\right)}_{\mathbf{4}}\mathbf{\text{C}}\mathit{l}\left(aq\right)$

A plant needs to manufacture 10,000 printed circuit boards, each$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }\mathbf{16}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$in area. An average of$\mathbf{80}\mathbf{\%}$of the copper is removed from each board(density of copperrole="math" localid="1648626614637" $=\mathbf{8}\mathbf{.}\mathbf{96}\mathbf{}\raisebox{1ex}{$\mathbf{}\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{cm}}}^{\mathbf{3}}$}\right.$).What masses ofrole="math" localid="1648626916866" $\mathbf{\text{Cu}}\left({\text{NH}}_3\right){\mathbf{\text{4Cl}}}_{\mathbf{2}}$ and ${\mathbf{\text{NH}}}_{\mathbf{3}}$are needed to do this? Assume$\mathbf{100}\mathbf{\%}$ yield.

Short answer

Masses of $\mathbf{\text{Cu}}\left({\text{NH}}_3\right)\mathbf{4}{\mathbf{\text{Cl}}}_{\mathbf{2}}$and ${\mathbf{\text{NH}}}_{\mathbf{3}}$required to remove copper from each board are $\mathbf{175}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{g}$and $\mathbf{58}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$.

Step-by-step solution

 Step 1: Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determining the mass of copper coated 

$\mathbf{\text{Volumeofcoppercoated}}\mathbf{=}\left(\text{areaoftheboard}\right)\mathbf{\times }\left(\text{thickness}\right)$

$=\left(8.0\times 16.0{\mathrm{cm}}^2\right)\times \left(0.60mm\right)$

$=7.68{\mathrm{cm}}^3$

For 10,000 plates, the volume of copper coated $=7.68\times 1000{\text{cm}}^3$

$=7.68\times {10}^4{\mathrm{cm}}^3$

$\mathbf{\text{Massofcopper}}\mathbf{=}\mathbf{\left(}\mathbf{\text{density}}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\mathbf{\text{volume}}\mathbf{\right)}$

$=\left(8.96\right)\times \left(7.68\times {10}^4\right)$

$=68.81\times {10}^{4}g$

Determining the mass of copper removed

Amount of copper removed $=\left(80\%\right)\times \left(68.81\times {10}^{4}g\right)$

$=55.0502\times {10}^{4}g$

$\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathit{c}\mathit{o}\mathit{p}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{r}\mathit{e}\mathit{m}\mathit{o}\mathit{v}\mathit{e}\mathit{d}\mathbf{\text{=}}\frac{\left(\text{Mass}\right)}{\left(\text{Atomicmass}\right)}$

$=\frac{55.0502\times {10}^{4}g}{63.546}$

$=0.866\times {10}^{4}moles$

Determining the mass of  Cu(NH3)4Cl2

Thus, from the given reaction,

$\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\left(\mathrm{aq}\right)+4{\mathrm{NH}}_3\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)\to 2\mathrm{Cu}{\left(\mathrm{NH}3\right)}_4\mathrm{Cl}\left(\mathrm{aq}\right)$

For 1 mole of etching$\text{Cu}$, 1 mole of role="math" localid="1648624524768" $\mathrm{Cu}\left({\mathrm{NH}}_3\right)4{\mathrm{Cl}}_2$is needed

$\mathbf{\text{MassofCu}}\left({\text{NH}}_3\right)\mathbf{4}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{=}\mathbf{}\left(\text{Moles}\right)\mathbf{\times }\left(\text{Molarmass}\right)$

Moles $=0.866+4\left(17\right)+2\left(35.45\right)$

=$202.446\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Thus the required mass of$\mathrm{Cu}\left({\mathrm{NH}}_3\right)4{\mathrm{Cl}}_2=\left(202.446\right)\times \left(0.866\times {10}^4\right)$

$=175.31\times {10}^{4}g$

$=175.31\times {10}^{4}g$ of$\mathrm{Cu}\left({\mathrm{NH}}_3\right)4{\mathrm{Cl}}_2$

Determining the mass of Ammonia

For 1 mole of $\text{Cu}$, 4 moles of Ammonia is needed

And for etching,$0.866\times {10}^4$ of $\text{Cu}$,

$=4\times 0.866\times {10}^4$Moles of ${\text{NH}}_3$is needed.

role="math" localid="1648623953027" $\mathbf{\text{MassofAmmoiarequired=}}\left(\text{Moles}\right)\mathbf{\times }\left(\text{Molarmass}\right)$$=\left(4\times 0.866\times {10}^4\right)\times \left(17\right)$

$=58.88\times {10}^4$ Of${\text{NH}}_3$

Therefore, the masses of$\mathbf{\text{Cu}}\left({\text{NH}}_3\right)\mathbf{4}{\mathbf{\text{Cl}}}_{\mathbf{2}}$ and$\mathbf{175}\mathbf{.}\mathbf{31}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{g}$ areand$\mathbf{58}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{g}$ .