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 $$ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow $$ $$ 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q) $$ A plant needs to manufacture \(10,000\) printed circuit boards, each \(8.0 \times 16.0 \mathrm{cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{g} / \mathrm{cm}^{3}\) . What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume 100\(\%\) yield.

Short answer

To manufacture 10,000 printed circuit boards, the plant needs \(991,818.49 \mathrm{g}\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(590,979.83 \mathrm{g}\) of \(\mathrm{NH}_{3}\), assuming a 100% yield.

Step-by-step solution

Calculate the volume of copper removed from one board

First, we need to find the total volume of copper removed from one board. The area of each board is \(8.0 \times 16.0 \mathrm{cm}^{2}\), and the thickness of the copper layer is 0.60 mm or 0.06 cm. So, the volume of copper on each board is: \(Area \times Thickness = 8.0 \times 16.0 \mathrm{cm}^{2} \times 0.06 \mathrm{cm} = 7.68 \mathrm{cm}^{3}\) Since 80% of the copper is removed, we find the volume of removed copper: \(7.68 \mathrm{cm}^{3} \times 0.80 = 6.144 \mathrm{cm}^{3}\)

Find the total volume of copper needed to be removed from 10000 boards

Now, we find the total volume of copper needed to be removed from 10000 boards by multiplying the volume of copper removed from one board by the number of boards (10000): \(6.144 \mathrm{cm}^{3/board} \times 10000 \mathrm{boards} = 61440 \mathrm{cm}^{3}\)

Calculate the mass of copper required for etching

We are given the density of copper, which is \(8.96 \mathrm{g} / \mathrm{cm}^{3}\). Using the density formula, we can find the mass of copper required for etching: \(Density = \frac{Mass}{Volume} \Rightarrow Mass = Density \times Volume\) \(Mass _{Cu} = 8.96 \mathrm{g/cm^{3}} \times 61440 \mathrm{cm}^{3} = 550758.4 \mathrm{g}\)

Use stoichiometry to determine the mass of Cu(NH3)4Cl2 and NH3

In the given reaction, one mole of Cu(NH3)4Cl2 reacts with 4 moles of NH3 to produce 2 moles of Cu from 1 mole of Cu(s). Using stoichiometry and the molar mass of Cu, Cu(NH3)4Cl2, and NH3, we can find the required masses for etching. Moles of Cu(s) needed: \(moles _{Cu} = \frac{550758.4 \mathrm{g}}{63.55 \mathrm{g/mol}} = 8669.73 \mathrm{moles}\) Moles of Cu(NH3)4Cl2 required: \(moles _{Cu(NH3)4Cl2} = \frac{1}{2} \times moles _{Cu} = 4334.86 \mathrm{moles}\) Mass of Cu(NH3)4Cl2 required: \(mass _{Cu(NH3)4Cl2} = 4334.86 \mathrm{moles} \times 228.78 \mathrm{g/mol}(molar\ mass) = 991818.49 \mathrm{g}\) Moles of NH3 required: \(moles _{NH3} = 4 \times moles _{Cu} = 34678.92 \mathrm{moles}\) Mass of NH3 required: \(mass _{NH3} = 34678.92 \mathrm{moles} \times 17.03 \mathrm{g/mol}(molar\ mass) = 590979.83 \mathrm{g}\) So, the plant needs 991818.49 g of Cu(NH3)4Cl2 and 590979.83 g of NH3 to manufacture 10000 printed circuit boards assuming 100% yield.

Key concepts

Copper Etching

Copper etching is a chemical process used extensively in the manufacturing of printed circuit boards (PCBs). During this process, unwanted copper is selectively removed from copper-clad laminates to form the desired circuit path. This is done by protecting certain areas with a polymer resist, while the exposed copper areas are removed by chemical reactions.

Chemical etching is highly valued because it allows for the precision removal of copper without the need for physical tools, which can create precise and intricate circuit patterns. Etching typically involves a chemical reaction with the copper, which transforms it into a soluble compound that can be washed away easily.

Printed Circuit Boards

Printed Circuit Boards, or PCBs, are a cornerstone in modern electronics, providing a platform to mechanically support and electrically connect electronic components using conductive pathways. These boards are typically made of a non-conductive substrate laminated with copper.

The process of creating a PCB involves several steps, starting with designing the circuit layout, printing the design onto a copper-clad board, and then etching away the unwanted copper. The result is a functional and durable board that is essential for various electronic applications, from smartphones to home appliances.

Chemical Reactions

Chemical reactions are the backbone of copper etching processes. In the example from the exercise, the etching compound Cu(NH₃)₄Cl₂ reacts with NH₃ and copper (Cu) to form a different complex. This reaction exemplifies how chemical etching processes rely on specific reactions to remove materials selectively.

Understanding reactions like these involves stoichiometry, which refers to the quantitative relationship between reactants and products in a chemical equation. It's important to balance the equation to ensure that all atoms are accounted for, which is crucial when scaling up reactions for large-scale production, like in the mass manufacturing of PCBs.

Density Calculations

Density calculations are essential for determining how much copper can be removed from the circuit boards during etching. The density of copper is a known quantity, which allows us to convert the volume of copper to be removed into a mass.
Using the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), one can determine the mass of copper by multiplying the total volume of copper to be etched by its density. This step is necessary to assess how much of each reactant is needed for the chemical reactions that occur during etching, ensuring precise material use and cost efficiency.