Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

6. Chemical Reactions & Quantities

Percent Yield

GOB Chemistry

6. Chemical Reactions & Quantities

Percent Yield

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5:23 minutes

Problem 13

Textbook Question

The reaction of ethylene oxide with water to give ethylene glycol (automobile antifreeze) occurs in 96.0% actual yield. How many grams of ethylene glycol are formed by reaction of 35.0 g of ethylene oxide? (For ethylene oxide, MW = 44.0 amu; for ethylene glycol, MW = 62.0 amu.)

Verified step by step guidance

Step 1: Write the balanced chemical equation for the reaction. Ethylene oxide (C₂H₄O) reacts with water (H₂O) to form ethylene glycol (C₂H₆O₂). The balanced equation is: C₂H₄O + H₂O → C₂H₆O₂.

Step 2: Calculate the number of moles of ethylene oxide (C₂H₄O) using its molecular weight. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \). Substitute the given mass of ethylene oxide (35.0 g) and its molecular weight (44.0 g/mol).

Step 3: Use the stoichiometry of the reaction to determine the moles of ethylene glycol (C₂H₆O₂) produced. From the balanced equation, the molar ratio of ethylene oxide to ethylene glycol is 1:1. Therefore, the moles of ethylene glycol formed will be equal to the moles of ethylene oxide reacted.

Step 4: Convert the moles of ethylene glycol to grams using its molecular weight. Use the formula: \( \text{mass} = \text{moles} \times \text{molecular weight} \). Substitute the moles of ethylene glycol and its molecular weight (62.0 g/mol).

Step 5: Account for the actual yield of the reaction. Multiply the theoretical mass of ethylene glycol (calculated in Step 4) by the actual yield (96.0%) expressed as a decimal (0.960). This will give the actual mass of ethylene glycol produced.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the amount of product formed from a given amount of reactant by using molar ratios derived from the coefficients in the balanced equation.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is essential for converting between grams and moles, enabling the calculation of how many moles of a substance are present in a given mass, which is crucial for stoichiometric calculations.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100. It indicates how much of the expected product was actually obtained, which is important for assessing the success of a reaction and for practical applications in chemistry.

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Related Practice

Textbook Question

Consider the balanced chemical equation: A₂ + 2 B₂ → 2 AB₂. A reaction is performed with the initial amounts of A₂ and B₂ shown in part (a). The amount of product obtained is shown in part (b). Calculate the percent yield.

a. <IMAGE>

b. <IMAGE>

Multiple Choice

What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO₄?

Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)

Ammonia, NH₃, reacts with hypochlorite ion, OCl^–, to produce hydrazine, N₂H₄. How many grams of hydrazine are produced from 115.0 g NH₃ if the reaction has a 81.5% yield?

2 NH₃ + OCl^– → N₂H₄ + Cl^– + H₂O

The reduction of iron (III) oxide creates the following reaction:

Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g)

If the above reaction only went to 75% completion, how many moles of Fe₂O₃ were require to produce 0.850 moles of Fe?