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Chapter 9: Problem 59

A student runs a reaction to prepare \(40.0 \mathrm{~g}\) of aspirin and yet recovers only \(15.5 \mathrm{~g}\). What is the percent yield?

Short Answer

Expert verified
The percent yield of the aspirin reaction is 38.75%, calculated using the formula Percent Yield = \(\frac{Actual Yield}{Theoretical Yield}\) × 100 with an actual yield of 15.5 g and a theoretical yield of 40.0 g.

Step by step solution

01

Identify the Actual and Theoretical Yields

In this problem, the actual yield of aspirin is given as 15.5 g, while the theoretical yield of aspirin is given as 40.0 g.
02

Calculate the Percent Yield

Use the formula for calculating percent yield: Percent Yield = (Actual Yield / Theoretical Yield) × 100 Plug in the known values: Percent Yield = (15.5 g / 40.0 g) × 100
03

Solve for Percent Yield

Divide the actual yield by the theoretical yield and multiply by 100: Percent Yield = (0.3875) × 100
04

Report the Final Answer

Multiply the fraction by 100 to obtain the percent yield: Percent Yield = 38.75% The percent yield for this reaction is 38.75%.

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Most popular questions from this chapter

When \(490.0 \mathrm{mg}\) of iron reacts with excess bromine, a mixture of \(\mathrm{FeBr}_{2}\) and \(\mathrm{FeBr}_{3}\) is produced. It is determined that \(35.5 \%\) of the mixture is iron(II) (bromide). What is the total mass of \(\mathrm{FeBr}_{2} / \mathrm{FeBr}_{3}\) mixture produced?

Consider the balanced chemical equation $$ \begin{aligned} &\mathrm{Fe}(\mathrm{CO})_{5}(s)+2 \mathrm{PF}_{3}(l)+\mathrm{H}_{2}(g) \rightarrow \\ &\mathrm{Fe}(\mathrm{CO})_{2}\left(\mathrm{PF}_{3}\right)_{2} \mathrm{H}_{2}(s)+3 \mathrm{CO}(g) \end{aligned} $$ (a) How many grams of \(\mathrm{CO}\) could be produced from \(10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), excess \(\mathrm{Fe}(\mathrm{CO})_{5}\), and excess \(\mathrm{H}_{2} ?\) (b) How many grams of \(\mathrm{CO}\) could be produced from \(5.0\) moles of \(\mathrm{Fe}(\mathrm{CO})_{5}, 8.0\) moles of \(\mathrm{PF}_{3}\), and \(6.0\) moles of \(\mathrm{H}_{2}\) ? (c) How many moles of \(\mathrm{CO}\) could be produced from \(25.0 \mathrm{~g}\) of \(\mathrm{Fe}(\mathrm{CO})_{5}, 10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), and excess \(\mathrm{H}_{2} ?\) (d) The density of hydrogen gas at room temperature and atmospheric pressure is \(0.0820 \mathrm{~g} / \mathrm{L}\). When \(5.00 \mathrm{~L}\) of hydrogen gas at room temperature and atmospheric pressure is mixed with excess \(\mathrm{Fe}(\mathrm{CO})_{5}\) and excess \(\mathrm{PF}_{3}\), the mass of \(\mathrm{CO}\) collected is \(13.5 \mathrm{~g}\). What is the theoretical yield (in grams) and the percent yield of \(\mathrm{CO}\) ?

A \(2.230-\mathrm{g}\) sample of a solid is subjected to combustion analysis, yielding \(76.59 \% \mathrm{C}\) and \(6.39 \% \mathrm{H}\). It may also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(94 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

How many molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), would be in a tablet that contained \(250 \mathrm{mg}\) of aspirin? How many atoms of carbon would be in the aspirin in that tablet?

In a chemical reaction, the mass of the products is equal to the mass of the reactants consumed. Are the moles of product equal to the moles of reactant consumed? Explain your answer.
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