If only 1.31 grams of water are actually produced in the lab, what is the percent yield?
To calculate the percent yield, you need two pieces of information: the actual yield (the amount of water produced) and the theoretical yield (the expected amount of water based on the reaction you're performing).
In this case, you already know the actual yield, which is 1.31 grams of water. However, you need to determine the theoretical yield. To do this, you would typically start with the balanced chemical equation for the reaction. For example, if the chemical equation is:
A + B → C + H2O
You need to determine the stoichiometry of the reaction, which refers to the mole ratios between the reactants and products. For example, let's say you discover that for every 2 moles of A, you produce 3 moles of water (H2O).
With this information, you can calculate the theoretical yield. Start by converting the given mass (1.31 grams) to moles using the molar mass of water. The molar mass of water (H2O) is approximately 18.015 g/mol. So, you divide the mass by the molar mass:
1.31 grams / 18.015 g/mol ≈ 0.073 moles of water
Next, use the stoichiometry of the reaction to determine the theoretical yield. Since you found that for every 2 moles of A, you produce 3 moles of water, you can set up a ratio:
2 moles of A : 3 moles of water
Now, divide the moles of water by the stoichiometric ratio:
0.073 moles of water / (3 moles of water / 2 moles of A) ≈ 0.0487 moles of A
Finally, use the molar mass of A to convert moles back to grams. Suppose the molar mass of A is 50 g/mol:
0.0487 moles of A × 50 g/mol ≈ 2.435 grams
Now that you have the theoretical yield of 2.435 grams and the actual yield of 1.31 grams, you can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Percent Yield = (1.31 g / 2.435 g) × 100 ≈ 53.8%
So, the percent yield in this case is approximately 53.8%.