Question

Ethers (general formula \(\left.\mathrm{R}-\mathrm{O}-\mathrm{R}^{\prime}\right)\) have many important uses. Until recently, methyl tert-butyl ether (MTBE, below) was used as an octane booster and fuel additive for gasoline. It increases the oxygen content of the fuel, which reduces CO emissions. MTBE is synthesized by the catalyzed reaction of 2 -methylpropene with methanol. (a) Write a balanced equation for the synthesis of MTBE. (Hint: Alcohols add to alkenes similarly to the way water does.) (b) If the government required that auto fuel mixtures contain \(2.7 \%\) oxygen by mass to reduce CO emissions, how many grams of MTBE would have to be added to each \(100 . \mathrm{g}\) of gasoline? (c) How many liters of MTBE would be in each liter of fuel mixture? (The density of both gasoline and \(\mathrm{MTBE}\) is \(0.740 \mathrm{~g} / \mathrm{mL}\).) (d) How many liters of air \(\left(21 \% \mathrm{O}_{2}\right.\) by volume \()\) are needed at \(24^{\circ} \mathrm{C}\) and 1.00 atm to fully combust \(1.00 \mathrm{~L}\) of MTBE?

Short answer

The synthesis reaction is C₄H₈ + CH₃OH → C₅H₁₂O. Add 14.86 g of MTBE per 100 g of gasoline. This equates to 0.02008 L per liter of fuel mixture. For combustion, 1.352 moles of O₂ are needed, which requires 144.1 L of air.

Step-by-step solution

Write the chemical equation

MTBE is synthesized by the reaction of 2-methylpropene (\text{C}_{4}\text{H}_{8}) with methanol (\text{CH}_{3}\text{OH}). The balanced chemical equation is:\[\text{C}_{4}\text{H}_{8} + \text{CH}_{3}\text{OH} \rightarrow \text{C}_{5}\text{H}_{12}\text{O}\]

Calculate grams of MTBE needed

Given that each 100 g of gasoline needs to contain 2.7% oxygen by mass:1. Calculate the mass of O in 100 g of fuel:\[100 \text{ g} \times 0.027 = 2.7 \text{ g O}\]2. Calculate the mass of MTBE that would contain 2.7 g of oxygen. MTBE (C₅H₁₂O) has a molar mass of 88.15 g/mol, and one mole of it contains 16 g of O. Therefore:\[\text{Mass of MTBE} = \frac{2.7 \text{ g O} \times 88.15 \text{ g/mol}}{16 \text{ g/mol}} = 14.86 \text{ g}\]

Calculate the volume of MTBE in the fuel mixture

Given the density of MTBE is 0.740 g/mL, convert the mass of MTBE into volume:\[\text{Volume of MTBE} = \frac{14.86 \text{ g}}{0.740 \text{ g/mL}} \rightarrow 20.08 \text{ mL} \text{(or 0.02008 L)}\]Since this is for each 100 g of gasoline and both densities are the same, hence in 1 L of gasoline, the proportion remains the same.

Calculate the liters of air needed for complete combustion

First, establish the combustion reaction of MTBE:\[\text{C}_5 \text{H}_{12} \text{O} + 8 \text{O}_2 \rightarrow 5 \text{CO}_2 + 6 \text{H}_2 \text{O}\]1. Calculate the moles of MTBE in 1 L of fuel mixture:\[\text{Mass of MTBE in 1L} = 14.86 \text{ g}\]\[\text{Moles of MTBE} = \frac{14.86 \text{ g}}{88.15 \text{ g/mol}} = 0.169 \text{ moles}\]2. Find the moles of O needed for full combustion: \[\text{Moles of } \text{O}_2 \text{ needed} = 8 \times 0.169 = 1.352 \text{ moles}\]3. Calculate the volume of air required:\[\text{Volume of air} = 1.352 \text{ mol} \times \frac{22.4 \text{ L/mol}}{0.21} \rightarrow 144.1 \text{ L}\]Note: 22.4 L/mol is used for ideal gases at STP, and oxygen makes up 21% of air.

Key concepts

Chemical Equation

To fully understand the synthesis of MTBE, it’s crucial to know how to write chemical equations. Chemical equations represent reactions between reactants to form products.
In this exercise, MTBE is formed by the reaction of 2-methylpropene (C₄H₈) with methanol (CH₃OH). Writing the balanced equation for this reaction helps us understand the exact process at a molecular level:
\(\text{C}_{4}\text{H}_{8} + \text{CH}_{3}\text{OH} \rightarrow \text{C}_{5}\text{H}_{12}\text{O}\).
Balancing the equation ensures that the number of each type of atom on the reactants side is equal to the number on the products side. This obeys the Law of Conservation of Mass, which states that mass in an isolated system is neither created nor destroyed by chemical reactions.

Mass Calculation

Mass calculation plays a pivotal role when understanding the amount of product formed or reactants required in a chemical reaction.
In part (b) of the exercise, we needed to determine how much MTBE to add to gasoline to achieve 2.7% oxygen by mass. Here's how we approached it:

• First, calculate the mass of oxygen needed in 100 g of gasoline: \(100 \text{ g} \times 0.027 = 2.7 \text{ g O}\).
• Next, find the mass of MTBE that contains this amount of oxygen. Since MTBE (C₅H₁₂O) contains one mole (or 16 g) of oxygen per 88.15 g/mol of MTBE, we calculated: \(\text{Mass of MTBE} = \frac{2.7 \text{ g O} \times 88.15 \text{ g/mol}}{16 \text{ g/mol}} = 14.86 \text{ g}\).

This approach ensures we are considering the molar mass and proportion of oxygen in MTBE.

Density Conversion

Density conversion helps us transform between different units of mass and volume, particularly when substances have known densities.
For MTBE, knowing the density (0.740 g/mL) allowed us to convert from grams to milliliters, a frequent necessity in chemistry.

• To find the volume of MTBE that corresponds to the mass calculated in step 2 (14.86 g), we used the density conversion formula: \(\text{Volume of MTBE} = \frac{14.86 \text{ g}}{0.740 \text{ g/mL}} = 20.08 \text{ mL}\).
• Since gasoline and MTBE share the same density, converting grams directly to liters for ease in a 1 L mixture is an efficient approach. This means 0.02008 L of MTBE per liter of gasoline.

Understanding density helps bridge the gap between mass and volume, making mixed-phase calculations smoother.

Combustion Reaction

Combustion reactions are crucial in fields like energy production and environmental science. These reactions combine a fuel with an oxidant, typically oxygen, to produce heat and combustion products like CO₂ and H₂O.
In our exercise, we used the combustion reaction of MTBE to find the volume of air needed for its complete combustion:

• The equation for combustion is: \(\text{C}_5 \text{H}_{12} \text{O} + 8 \text{O}_2 \rightarrow 5 \text{CO}_2 + 6 \text{H}_2 \text{O}\).
• Calculating the moles of MTBE in 1 L (14.86 g / 88.15 g/mol = 0.169 mol) and the oxygen required (\(\text{Moles of O}_2 = 8 \times 0.169 = 1.352 \text{ mol}\), we used gas laws to find the volume of air (since O₂ is 21% of air): \(\text{Volume of air} = 1.352 \text{ mol} \times \frac{22.4 \text{ L/mol}}{0.21} = 144.1 \text{ L}\).

This ensures full combustion, considering air composition and ideal gas laws.