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Chapter 9: Problem 61

According to his prelaboratory theoretical yield calculations, a student's experiment should have produced \(1.44 \mathrm{g}\) of magnesium oxide. When he weighed his product after reaction, only \(1.23 \mathrm{g}\) of magnesium oxide was present. What is the student's percent yield?

Short Answer

Expert verified
The student's percent yield of magnesium oxide in the experiment is approximately 85.42%.

Step by step solution

01

Identify the given values

The problem provides us with the theoretical yield (1.44 g) and the actual yield (1.23 g) of magnesium oxide.
02

Apply the percent yield formula

To determine the percent yield, we will use the formula: Percent yield = \(\frac{Actual\: Yield}{Theoretical\: Yield} \cdot 100\)
03

Plug in the values and calculate the result

Insert the given theoretical yield (\(1.44\:g\)) and actual yield (\(1.23\:g\)) into the formula: Percent yield = \(\frac{1.23\:g}{1.44\:g} \cdot 100\) Now, we will perform the calculations: Percent yield = \(\frac{1.23}{1.44}\cdot 100 \approx 85.42\%\)
04

Interpret the result

The student's percent yield of magnesium oxide in the experiment is approximately 85.42%. This means that the student was able to produce approximately 85.42% of the magnesium oxide that was expected according to the prelaboratory theoretical yield calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
In chemistry, the theoretical yield refers to the maximum amount of product that can be formed from a given amount of reactants, assuming perfect conditions with no losses. It is based entirely on stoichiometric calculations derived from balanced chemical equations.
Calculating the theoretical yield involves the following steps:
  • First, you need the balanced chemical equation for the reaction. This gives you the molar ratio of reactants to products.
  • Next, determine the amount of reactants available. Often given in grams, convert this to moles using the molar mass.
  • Use the stoichiometric ratios from the balanced equation to calculate the moles of product that should form.
  • Finally, convert the moles of product to grams, as desired, to find the theoretical yield.
This metric is vital as a baseline for evaluating how efficient an actual reaction was through percent yield.
Actual Yield
The actual yield is simply the amount of product that is actually obtained from a chemical reaction in the laboratory. Unlike the theoretical yield, actual yield can be less than expected due to:
  • Loss of material during handling, transfer, or due to experimental errors.
  • Side reactions that consume reactants or produce other products.
  • Incomplete reactions where not all reactants convert into products.
It鈥檚 important to weigh the final product carefully using an accurate balance. This measured yield is used along with the theoretical yield to calculate percent yield, showing the reaction's efficiency.
Magnesium Oxide
Magnesium oxide is a chemical compound often represented by the formula MgO. It forms as a result of magnesium reacting with oxygen, typically through heating: Mg + O2 鈫� MgO. Known for its high melting point and stability, it is used in various applications from refractory materials to supplements. Understanding the formation of magnesium oxide in experiments aids in learning reaction principles like yield and reactions of metals with non-metals. In the classroom, it serves as a straightforward model for calculating theoretical and actual yields.
Chemical Reactions
Chemical reactions describe the process through which substances (reactants) are transformed into different substances (products). These transformations are fundamental to both research and industry.
Key aspects of chemical reactions to consider include:
  • Reactants - starting materials in a reaction.
  • Products - substances formed from the reaction.
  • Catalysts - substances that can increase reaction rate without being consumed.
  • Stoichiometry - quantitative relationships which ensure the mass balance in reactions.
Each reaction must be balanced, meaning, the mass and the number of atoms before and after the reaction should be equal. Learning about chemical reactions helps us understand the theoretical yield and predict product formation.
Yield Calculation
Yield calculation is essential in determining the efficiency of a chemical reaction. The percent yield shows how much of the theoretical yield was achieved as actual product in the laboratory, calculated as:\(\text{Percent yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\).This formula highlights:
  • Comparison between what was ideally possible (theoretical) and what truly happened (actual).
  • The efficiency of a reaction, important for optimizing processes in science and industry.
A high percent yield indicates a reaction where most reactants are converted to products. In contrast, lower yields suggest losses that could arise from experimental errors, incomplete reactions, or side processes. Understanding yield calculation is paramount for chemists in evaluating reaction conditions and improving future experiments.

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Most popular questions from this chapter

Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. $$\mathrm{PbO}(s)+\mathrm{C}(s) \rightarrow \mathrm{Pb}(l)+\mathrm{CO}(g)$$ Calculate the expected yield of lead if \(50.0 \mathrm{kg}\) of lead oxide is heated with \(50.0 \mathrm{kg}\) of carbon.

For each of the following unbalanced chemical equations, suppose \(1.00 \mathrm{g}\) of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected. a. \(\mathrm{UO}_{2}(s)+\mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HNO}_{3}(a q)\) c. \(\mathrm{Zn}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{B}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{3} \mathrm{OH}(l) \rightarrow \mathrm{B}\left(\mathrm{OCH}_{3}\right)_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)

The traditional method of analysis for the amount of chloride ion present in a sample was to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible effectively to remove all chloride ion from the sample. $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s) $$ Suppose a \(1.054-\mathrm{g}\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Alkali metal hydroxides are sometimes used to "scrub" excess carbon dioxide from the air in closed spaces (such as submarines and spacecraft). For example, lithium hydroxide reacts with carbon dioxide according to the unbalanced chemical equation $$\mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Suppose a lithium hydroxide canister contains \(155 \mathrm{g}\) of \(\operatorname{LiOH}(s) .\) What mass of \(\mathrm{CO}_{2}(g)\) will the canister be able to absorb? If it is found that after 24 hours of use the canister has absorbed \(102 \mathrm{g}\) of carbondioxide, what percentage of its capacity has been reached?

Silicon carbide, \(\mathrm{SiC},\) is one of the hardest materials known. Surpassed in hardness only by diamond, it is sometimes known commercially as carborundum. Silicon carbide is used primarily as an abrasive for sandpaper and is manufactured by heating common sand (silicon dioxide, \(\mathrm{SiO}_{2}\) ) with carbon in a furnace. $$\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{CO}(g)+\mathrm{SiC}(s)$$ What mass of silicon carbide should result when 1.0 kg of pure sand is heated with an excess of carbon?
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