black smokers are found in the depths of the oceans. thinking that the coditions in these smokers might be conducive to the formation of organic compounds, two chemists in the germany found the following reaction could occur in similar conditions: 2CH3SH+CO-->CH3COSCH3+H2S
If you being with 10.0g of CH3SH and excess CO
A.) what is the thoretical yield of CH3COSCH3?
B.) if 8.65g of CH3COSCH3 IS ISOLATED, WHAT IS ITS PERCENT YIELD?
Use the same concepts for the last problem you posted. If you have trouble post your work and/or tell us in detail what you don't understand about the next step.
To find the theoretical yield of CH3COSCH3 and the percent yield, we need to follow these steps:
Step 1: Calculate the molar mass of CH3SH (M1) and CH3COSCH3 (M2).
The molar mass of CH3SH (M1) is:
M1 = (12.01 g/mol × 1) + (1.0078 g/mol × 3) + (32.06 g/mol × 1) + (1.0078 g/mol × 1)
= 62.16 g/mol
The molar mass of CH3COSCH3 (M2) is:
M2 = (12.01 g/mol × 2) + (32.06 g/mol × 1) + (1.0078 g/mol × 4) + (12.01 g/mol × 1) + (1.0078 g/mol × 3)
= 90.16 g/mol
Step 2: Convert the given mass of CH3SH to moles.
Mass of CH3SH = 10.0 g
Moles of CH3SH = Mass of CH3SH / Molar mass of CH3SH
= 10.0 g / 62.16 g/mol
= 0.161 moles
Step 3: Use the balanced chemical equation to write a mole ratio between CH3SH and CH3COSCH3.
From the balanced equation:
2 CH3SH + CO -> CH3COSCH3 + H2S
The mole ratio between CH3SH and CH3COSCH3 is:
2 moles of CH3SH : 1 mole of CH3COSCH3
Step 4: Calculate the theoretical yield of CH3COSCH3.
The theoretical yield can be calculated using the mole ratio and the moles of CH3SH:
Theoretical yield of CH3COSCH3 = Moles of CH3SH × (Molar mass of CH3COSCH3 / Molar mass of CH3SH) × (1 mole of CH3COSCH3 / 2 moles of CH3SH)
= 0.161 moles × (90.16 g/mol / 62.16 g/mol) × (1/2)
= 0.238 moles
Step 5: Calculate the percent yield.
The percent yield can be calculated using the formula:
Percent Yield = (Actual yield / Theoretical yield) × 100
Given: Actual yield = 8.65 g
Theoretical yield = 0.238 moles × 90.16 g/mol = 21.53 g
Percent Yield = (8.65 g / 21.53 g) × 100
≈ 40.15%
Therefore, the answers are:
A.) The theoretical yield of CH3COSCH3 is 0.238 moles or 21.53 g.
B.) The percent yield of CH3COSCH3 is approximately 40.15%.