Medicinal “ether" is produced when ethyl alcohol is treated with an acid. How many grams of medicinal “ether" (CH3CH2OCH2CH3) would you produce with 50.0 g of ethyl alcohol (CH3CH2OH)
if the percent yield is 85%?

34.3 g

Explanation:

Step 1: Write the balanced equation

2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O

Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH

The molar mass of CH₃CH₂OH is 46.07 g/mol.

50.0 g × 1 mol/46.07 g = 1.09 mol

Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced

The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.

Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced

The percent yield of the reaction is 85%.

0.545 mol × 85% = 0.463 mol

Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃

The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.

0.463 mol × 74.12 g/mol = 34.3 g

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