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

The plant needs \(2,169,330.75 \mathrm{g}\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(590,570.59 \mathrm{g}\) of \(\mathrm{NH}_{3}\) to manufacture 10,000 printed circuit boards with 100% yield.

Step-by-step solution

Calculate the mass of copper removed from each board

First, we need to determine the volume of the copper removed from each board. We know that the copper layer is 0.60 mm thick and the dimensions of each board are \(8.0 \times 16.0 \mathrm{cm}\). To calculate the volume of the copper layer, multiply the area of the board by its thickness and convert the thickness from mm to cm. Area: \(8.0 \times 16.0 \mathrm{cm^{2}} = 128 \mathrm{cm^{2}}\) Thickness: \(0.60 \mathrm{mm} = 0.060 \mathrm{cm}\) Copper volume per board: \(128 \mathrm{cm^{2}} \times 0.060 \mathrm{cm} = 7.68 \mathrm{cm^{3}}\) Since 80% of the copper is removed, we can calculate the volume of copper removed: Volume of copper removed: \(0.80 \times 7.68 \mathrm{cm^{3}} = 6.144 \mathrm{cm^{3}}\) Now, we can find the mass of copper removed from each board by using the density of copper: Mass of copper removed: \(6.144 \mathrm{cm^{3}} \times 8.96 \mathrm{g/cm^{3}} = 55.055 \mathrm{g}\)

Calculate the total mass of copper removed from all boards

Multiply the mass of copper removed from each board by the number of boards: Total mass of copper removed: \(55.055 \mathrm{g} \times 10,000 = 550,550 \mathrm{g}\)

Use stoichiometry to find the required masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

From the given reaction: \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\) For every mole of copper, one mole of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and four moles of \(\mathrm{NH}_{3}\) are required.

Calculate the moles of copper removed

Use the molar mass of copper (63.55 g/mol) to convert the total mass of copper removed into moles: Moles of copper removed: \(\frac{550,550 \mathrm{g}}{63.55 \mathrm{g/mol}} = 8,663.25 \mathrm{mol}\)

Calculate the required moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

For each mole of removed copper, we need one mole of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and four moles of \(\mathrm{NH}_{3}\): Moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) required: \(8,663.25 \mathrm{mol}\) Moles of \(\mathrm{NH}_{3}\) required: \(8,663.25 \mathrm{mol} \times 4 = 34,653 \mathrm{mol}\)

Calculate the required masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\)

Finally, we can use the molar masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) to convert the moles into masses: Mass of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) required: \(8,663.25 \mathrm{mol} \times (250.5 \mathrm{g/mol}) = 2,169,330.75 \mathrm{g}\) Mass of \(\mathrm{NH}_{3}\) required: \(34,653 \mathrm{mol} \times (17.03 \mathrm{g/mol}) = 590,570.59 \mathrm{g}\) So, the plant needs 2,169,330.75 g of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and 590,570.59 g of \(\mathrm{NH}_{3}\) to manufacture 10,000 printed circuit boards with 100% yield.

Key concepts

Chemical Reaction

Chemical reactions involve the transformation of substances through breaking and forming bonds. In the context of copper etching, a chemical reaction occurs when copper, in the presence of certain chemicals, transforms into other compounds. This specific reaction involves copper reacting with \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\), converting to a complex ion \(2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl} \). During chemical reactions, understanding the stoichiometry is crucial. It allows us to predict the amounts of reactants needed or products formed, based on the balanced chemical equation. This principle helps in determining the exact quantities of chemicals required for effective etching, as seen in this exercise. Each copper atom is involved directly in the reaction, leading to the removal of unwanted material from the circuit board.

Copper Etching

Copper etching is a vital process in electronics manufacturing, specifically in creating printed circuit boards (PCBs). It involves removing excess copper from a laminated board to achieve the desired circuit pattern.
Cuprous ammonia chloride complexes, among other reagents, are often used to facilitate this etching process. In the chemical reaction provided, the copper is dissolved, showcasing its transformation into a soluble complex.

• The goal is to remove about 80% of copper once the chemical process is complete.
• This precise etching is crucial because it defines the conductive paths that will carry electronic signals in the device.

Ensuring an accurate reaction requires careful calculation of reactants using stoichiometry, ensuring efficiency and precision in the production of PCBs.

Molar Mass

Molar mass is an essential concept in stoichiometry, as it allows us to convert mass to moles, facilitating chemical calculations. It is the mass of one mole of a substance, measured in grams per mole.
In the copper etching solution, determining the amount of copper removed involves using its molar mass of \(63.55 \mathrm{g/mol}\). This conversion is necessary for calculating how much of the chemical reagents, like \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\), are required to react with all the copper.

• The molar mass of copper guides the calculation of its total moles removed from all circuit boards.
• For chemistry-oriented manufacturing processes, precise molar mass usage ensures reaction efficiency and cost-effectiveness.

Understanding molar mass equips you to execute complex chemical equations accurately.

Printed Circuit Boards

Printed Circuit Boards (PCBs) serve as the backbone of modern electronic devices, hosting various electronic components in a compact and organized manner. The production starts by laminating a copper layer onto an insulating board.
The etching process, where excess copper is chemically removed, is essential to define the pattern of conductive paths designed to optimize electrical performance.

• The etched board leaves behind a precise network of copper tracks that form the circuitry.
• PCBs are tailored specifically for different electronic applications, requiring precision in both design and manufacturing.

With technological advancements, PCBs have become indispensable, making the understanding of processes like chemical etching crucial for production quality and efficiency.