Problem 106 - Chemistry The Central Science

Question

The semiconductor gallium arsenide, GaAs, is used in high-speed integrated circuits, light-emitting diodes, and solar cells. Its density is 5.32 \(\mathrm{g} / \mathrm{m}^{3} .\) It can be made by reacting trimethylgallium, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Ga},\) with arsine gas, \(\mathrm{AsH}_{3}\) . The other product of the reaction is methane, \(\mathrm{CH}_{4}\) . (a) If you reacted 450.0 g of trimethylgallium with 300.0 \(\mathrm{g}\) of arsine, what mass of GaAs could you make? (b) Which reactant, if any, would be left over, and how many moles of the leftover reactant would remain? (c) One application of GaAs uses it as a thin film. If you take the mass of GaAs from part (a) and make a 40 -nm thin film from it, what area, in \(\mathrm{cm}^{2},\) would it cover? Recall that \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\)

Step-by-Step Solution

Answer

The mass of GaAs that could be made is 449.9 g. Arsine gas would be left over, and 0.738 moles of it would remain. The area covered by the 40-nm thin GaAs film would be 2.115 x 10^8 \(cm^2\).

Step 1: Write the balanced chemical equation

We have the following chemical reaction: Trimethylgallium + Arsine → Gallium Arsenide + Methane In chemical form: \((CH_3)_3Ga + AsH_3 \rightarrow GaAs + CH_4\) When we balance the equation, we get: \((CH_3)_3Ga + AsH_3 \rightarrow GaAs + 3CH_4\)

Step 2: Convert masses to moles

Convert the given masses of trimethylgallium (450.0 g) and arsine gas (300.0 g) into moles using their molar masses (144.65 g/mol for trimethylgallium and 77.95 g/mol for arsine). Moles of trimethylgallium = \(\frac{450.0\,g}{144.65\,g/mol} = 3.111\,mol\) Moles of arsine = \(\frac{300.0\,g}{77.95\,g/mol} = 3.849\,mol\)

Step 3: Identify the limiting and excess reactants

Determine the mole ratio between trimethylgallium and arsine in the balanced equation (1:1). R = \(\frac{3.111\,mol}{3.849\,mol}\) = 0.808 Since R < 1, trimethylgallium is the limiting reactant, and arsine is the excess reactant.

Step 4: Calculate the mass of GaAs produced

Use stoichiometry to determine the amount of GaAs produced from the limiting reactant (trimethylgallium): Moles of GaAs produced = Moles of trimethylgallium = 3.111 mol Now, convert moles of GaAs to mass using its molar mass (144.64 g/mol): Mass of GaAs = \(3.111\,mol \times 144.64\,g/mol = 449.9\,g\)

Step 5: Calculate the moles of leftover reactant

Calculate the moles of leftover arsine using stoichiometry: Moles of arsine consumed = Moles of GaAs produced = 3.111 mol Moles of leftover arsine = Moles of arsine - Moles of arsine consumed = 3.849 mol - 3.111 mol = 0.738 mol

Step 6: Calculate the area of 40-nm thin film

Determine the volume of GaAs produced using its mass (449.9 g) and density (5.32 g/cm³): Volume of GaAs = \(\frac{449.9\,g}{5.32\,g/cm^3} = 84.58\,cm^3\) Now, calculate the area of the 40-nm thin film: Thickness of the film = 40 nm = \(40 \times 10^{-9}\,m = 4 \times 10^{-7}\,cm\) Area of the thin film = \(\frac{Volume\,of\, GaAs}{Thickness\, of\, the\, film}\) Area = \(\frac{84.58\,cm^3}{4 \times 10^{-7}\, cm}\) = 2.115 x 10^8 \({cm^2}\) #Answer#: (a) The mass of GaAs that could be made is 449.9 g. (b) Arsine gas would be left over, and 0.738 moles of it would remain. (c) The area covered by the 40-nm thin GaAs film would be 2.115 x 10^8 \({cm^2}\).

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances into new materials. Here, trimethylgallium reacts with arsine gas. This reaction is essential in making gallium arsenide, a crucial semiconductor. A balanced chemical equation shows the proportion of reactants and products, ensuring mass conservation. In our case, the balanced equation is \((CH_3)_3Ga + AsH_3 \rightarrow GaAs + 3CH_4\). This equation tells us that one molecule of trimethylgallium reacts with one molecule of arsine to produce one molecule of gallium arsenide and three molecules of methane. Balancing equations ensures we understand the exact amounts needed for the reaction.

Mole Calculations

Mole calculations convert mass into moles, a unit that measures the amount of substance. This is done using the molar mass, which represents the mass of one mole of a substance in grams. Here, we convert 450.0 g of trimethylgallium into moles using its molar mass (144.65 g/mol), resulting in 3.111 mol. Similarly, we calculate moles for arsine, yielding 3.849 mol. These calculations are crucial for stoichiometry as they allow us to determine how many moles of each reactant are available for the reaction.

Limiting Reactant

The limiting reactant is the substance that is completely consumed in a reaction, dictating how much product is formed. In our exercise, we determine the limiting reactant by comparing the mole ratio of trimethylgallium to arsine. With a 1:1 ratio needed, we find \(R = \frac{3.111}{3.849} = 0.808\). Since this ratio is less than 1, trimethylgallium is limiting. This means it will run out first, limiting the amount of gallium arsenide produced to 3.111 moles.

Excess Reactant

The excess reactant is the substance that remains after the reaction is complete. Once we've identified trimethylgallium as the limiting reactant, arsine becomes the excess reactant. To find the leftover moles of arsine, we subtract the consumed amount (3.111 mol) from the initial amount (3.849 mol), giving us 0.738 mol of arsine left over. This leftover reactant remains unreacted because there's not enough limiting reactant to use it up completely.

Molar Mass

Molar mass is a fundamental concept in stoichiometry used to connect mass with moles. It describes how many grams are in one mole of a substance. For example, trimethylgallium has a molar mass of 144.65 g/mol. This means that 144.65 g of trimethylgallium equals 1 mole. Similarly, arsine's molar mass is 77.95 g/mol, and gallium arsenide's is 144.64 g/mol. Understanding molar mass allows us to perform conversions between mass and moles, which are crucial for balancing and proceeding with stoichiometric calculations.

Thin Films

Thin films are layers of material ranging from fractions of a nanometer to several micrometers in thickness. In this exercise, gallium arsenide is used to make a 40-nm thin film. Calculating the film’s area involves finding the volume using its mass (449.9 g) and density (5.32 g/cm³). We convert this mass into a volume (84.58 cm³) and then use the thin film’s thickness (converted to cm) to find the area formula: \(\frac{Volume}{Thickness}\). Thin films are essential in many technologies, providing solutions like coatings in semiconductors.

Semiconductors

Semiconductors like gallium arsenide (GaAs) are materials with electrical conductivity between conductors and insulators. They are fundamental in modern electronics, including high-speed integrated circuits, LEDs, and solar cells. Gallium arsenide is advantageous due to its efficiency in converting electricity into photonic signals, essential for high-frequency and optoelectronic applications. This makes GaAs prefered in devices requiring excellent performance and efficiency. Understanding its preparation through stoichiometric analysis ensures we can produce it accurately for various applications.