Question

Copper is an excellent electrical conductor widely used in making electric circuits. In producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant polymer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired copper circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One reaction used to remove the exposed copper from the circuit board is $$ \begin{aligned} \mathrm{Cu}(s)+\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q) & \longrightarrow \\ & 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q) \end{aligned} $$ A plant needs to produce 5000 circuit boards, each with a surface area measuring 2.0 in. \(\times 3.0\) in. The boards are covered with a \(0.65-\mathrm{mm}\) layer of copper. In subsequent processing, \(85 \%\) of the copper is removed. Copper has a density of \(8.96 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed to produce the circuit boards, assuming that the reaction used gives a \(97 \%\) yield.

Short answer

To produce 5,000 circuit boards, 542,282 g of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) and 85,142 g of \(\mathrm{NH}_{3}\) are needed.

Step-by-step solution

Calculate the volume of copper used for a single circuit board

To find the volume of copper for a single circuit board, first convert the surface area to cm² and then multiply by the thickness in cm. Area of a single circuit board = \(2.0 \,\text{in} \times 3.0\, \text{in} \) Convert inches to cm: 1 in = 2.54 cm Area \(= 2.0 \,\text{in}\times 3.0\, \text{in} \times (2.54 \,\text{cm/in})^2 = 30.84\, \text{cm}^2\) Thickness of copper = 0.65 mm = 0.065 cm Volume of copper = Area × Thickness = \(30.84\, \text{cm}^2 \times 0.065\, \text{cm} = 2.005 \,\text{cm}^3\)

Convert the volume of copper to mass using copper's density

To calculate the mass of copper used on one circuit board, use copper's density: Density = \(\frac{\text{Mass}}{\text{Volume}}\) Mass of copper = Density × Volume Mass \(= 8.96\, \frac{\text{g}}{\text{cm}^3} \times 2.005\, \text{cm}^3 = 17.972\, \text{g}\)

Calculate the mass of copper removed from each board during the chemical process

According to the problem, 85% of the copper is removed during the chemical process. So, we need to find 85% of the total mass of copper. Mass of copper removed/board = percentage removed × mass of the total copper \(= 0.85 \times 17.972\, \text{g} = 15.276 \,\text{g}\)

Calculate the mass of copper reacted for all 5,000 boards

To find the total mass of copper reacted, multiply the mass removed by the number of boards. Total mass of copper reacted = mass of copper removed/board × number of boards \(= 15.276\, \text{g/board} \times 5000\, \text{boards} = 76,380\, \text{g}\)

Calculate the moles of copper reacted using its molar mass

To find the moles of copper, use the molar mass of copper: Moles of copper = \(\frac{\text{Mass of copper}}{\text{Molar mass of copper}}\) Molar mass of copper = \(\mathrm{63.55\, g/mol}\) Moles of copper = \(\frac{76,380\, \text{g}}{63.55\, \text{g/mol}} = 1202\, \text{mol}\)

Calculate the moles of Cu(NH3)4Cl2 and NH3 using stoichiometry and the balanced chemical reaction equation.

Using the balanced equation: \(\mathrm{Cu}(s) + \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q) + 4 \mathrm{NH}_{3}(a q) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\) Moles of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) = Moles of \(\mathrm{Cu}\) reacted:\(= 1202\, \text{mol}\) Moles of \(\mathrm{NH}_{3}\) = \(\frac{4 \,\text{mol }\mathrm{NH}_{3}}{1\, \text{mol} \, \mathrm{Cu}} \times 1202\, \text{mol \(\mathrm{Cu}\)} = 4808\, \text{mol}\)

Calculate the masses of Cu(NH3)4Cl2 and NH3 required, considering the reaction's 97% yield.

To find the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) required, we need to account for the reaction's \(97\%\) yield: Molar mass of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) = \(\mathrm{437.14\, g/mol}\) Mass of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) \(= \frac{\text{Moles × Molar mass}}{\text{Yield}}\) \(= \frac{1202\, \text{mol} \times 437.14\, \text{g/mol}}{0.97}\) \(= 542,282 \, \text{g}\) Mass of \(\mathrm{NH}_{3}\) \(= \frac{\text{Moles × Molar mass}}{\text{Yield}}\) \(= \frac{4808\, \text{mol} \times 17.03\, \text{g/mol}}{0.97}\) \(= 85,142 \, \text{g}\) So, to produce the 5,000 circuit boards, 542,282 g of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) and 85,142 g of \(\mathrm{NH}_{3}\) are needed.

Key concepts

Chemical Reactions in Electronics

In the production of printed circuit boards (PCBs), which are central to modern electronics, chemical reactions are employed to refine the board's surface and establish precise paths for electricity to flow. One such reaction involves the use of copper in a complex with ammonia and chloride ions to selectively remove copper from areas of the circuit board not protected by a polymer.

Understanding this process is critical for engineers and technicians in the electronics industry because it determines the accuracy and functionality of the final product. The reaction not only requires careful control to ensure the desired pattern is correctly etched onto the board but also embodies the intricate relationship between chemistry and electronic engineering. Precise stoichiometry, which is the calculation of the reactants and products in a chemical reaction, is essential to execute this process efficiently and cost-effectively.

Mole Concept and Calculations

The mole concept is fundamental to stoichiometry and chemistry at large. It allows chemists to measure substances based on the number of particles they contain rather than their mass. This is crucial in the PCB manufacturing process, as the amount of reactants needed to remove copper is based on mole calculations.

For instance, in our exercise, to determine the quantities of reactants required, we must first calculate the number of moles of copper to be etched away. From there, stoichiometric ratios - derived from the balanced chemical equation - inform us of the corresponding moles of other reactants needed. Utilizing the molar mass of these substances then enables us to convert moles back to grams, which is a more practical unit for measuring and handling chemicals in an industrial setting.

Yield Percentage in Chemical Reactions

Yield percentage is a measure of a chemical reaction's efficiency. It compares the actual amount of product obtained to the theoretical amount that could have been produced under perfect conditions. In circuit board production, as in all chemical manufacturing, not all reactions proceed to completion, and the yield is often less than 100%.

For example, the exercise demonstrates that the reaction to remove copper from the PCB has a 97% yield, meaning that we must account for this inefficiency when calculating reactant masses. The yield factor is applied after we've determined the theoretical mass of reactants based on stoichiometry. Adjusting for the yield ensures that we supply enough material to account for the reaction's inefficiencies, thereby guaranteeing that enough copper is etched away to meet production goals. It is an essential consideration for both cost control and environmental impact in the manufacturing process.