/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 136 The compound ibuprofen, used in ... [FREE SOLUTION] | 91影视

91影视

The compound ibuprofen, used in some pain relievers, has the molecular formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\). If \(0.250 \mathrm{~g}\) of ibuprofen is burned in oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced?

Short Answer

Expert verified
The combustion of 0.250 g ibuprofen in oxygen will produce 0.690 g of \(CO_{2}\) and 0.196 g of \(H_{2}O\).

Step by step solution

01

Write the balanced chemical equation for ibuprofen combustion

First, we need to write the balanced chemical equation for the combustion of ibuprofen. When ibuprofen (C鈧佲們H鈧佲倛O鈧) burns in oxygen (O鈧), it produces carbon dioxide (CO鈧) and water (H鈧侽). The balanced chemical equation is: \[C_{13}H_{18}O_{2} + 17O_{2} \rightarrow 13CO_{2} + 9H_{2}O\]
02

Calculate the molar mass of ibuprofen, CO鈧, and H鈧侽

In order to perform stoichiometry calculations, we need the molar masses of ibuprofen, CO鈧, and H鈧侽: Ibuprofen: Molar mass = (13 脳 12.01 g/mol C) + (18 脳 1.01 g/mol H) + (2 脳 16.00 g/mol O) = 206.28 g/mol CO鈧: Molar mass = (1 脳 12.01 g/mol C) + (2 脳 16.00 g/mol O) = 44.01 g/mol H鈧侽: Molar mass = (2 脳 1.01 g/mol H) + (1 脳 16.00 g/mol O) = 18.02 g/mol
03

Use stoichiometry to find the moles and grams of CO鈧 and H鈧侽 produced

Given 0.250 g of ibuprofen, we can now use stoichiometry to calculate the grams of CO鈧 and H鈧侽 produced: 1. Determine the moles of ibuprofen used: moles of ibuprofen = grams of ibuprofen / molar mass of ibuprofen moles of ibuprofen = 0.250 g / 206.28 g/mol = 0.00121 mol ibuprofen 2. Determine the moles of CO鈧 and H鈧侽 produced: From the balanced chemical equation, 1 mol ibuprofen forms 13 mol CO鈧 and 9 mol H鈧侽. moles of CO鈧 = 0.00121 mol ibuprofen 脳 (13 mol CO鈧 / 1 mol ibuprofen) = 0.0157 mol CO鈧 moles of H鈧侽 = 0.00121 mol ibuprofen 脳 (9 mol H鈧侽 / 1 mol ibuprofen) = 0.0109 mol H鈧侽 3. Convert the moles of CO鈧 and H鈧侽 to grams: grams of CO鈧 = 0.0157 mol CO鈧 脳 44.01 g/mol = 0.690 g CO鈧 grams of H鈧侽 = 0.0109 mol H鈧侽 脳 18.02 g/mol = 0.196 g H鈧侽
04

Provide the answer with units

The combustion of 0.250 g ibuprofen in oxygen will produce 0.690 g of CO鈧 and 0.196 g of H鈧侽.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Balanced Chemical Equation
In the context of chemical reactions, a balanced chemical equation is essential because it represents the conservation of mass. The number of atoms for each element involved in a reaction remains constant throughout the process. For the combustion of ibuprofen, the balanced equation is: \[C_{13}H_{18}O_{2} + 17O_{2} \rightarrow 13CO_{2} + 9H_{2}O\] This equation signifies that one molecule of ibuprofen combines with 17 molecules of oxygen to produce 13 molecules of carbon dioxide and 9 molecules of water. To ensure the equation is balanced, count the atoms of each element on both sides of the equation. All elements鈥攃arbon, hydrogen, and oxygen鈥攎atch up, confirming mass conservation. A balanced equation not only allows for predicting the products of a reaction accurately but is also crucial for further stoichiometric calculations.
Molar Mass Calculation
The concept of molar mass is pivotal when converting between grams and moles in chemistry. Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). Calculating molar mass involves adding the atomic masses of all the atoms in the molecule.

For ibuprofen (\(C_{13}H_{18}O_{2}\)):

- Carbon (C): 13 atoms \(\times\) 12.01 g/mol = 156.13 g/mol - Hydrogen (H): 18 atoms \(\times\) 1.01 g/mol = 18.18 g/mol - Oxygen (O): 2 atoms \(\times\) 16.00 g/mol = 32.00 g/mol Summing these gives a molar mass of 206.28 g/mol for ibuprofen.

For \(CO_{2}\):

- Carbon (C): 1 atom \(\times\) 12.01 g/mol = 12.01 g/mol - Oxygen (O): 2 atoms \(\times\) 16.00 g/mol = 32.00 g/mol Total: 44.01 g/mol

For \(H_{2}O\):

- Hydrogen (H): 2 atoms \(\times\) 1.01 g/mol = 2.02 g/mol - Oxygen (O): 1 atom \(\times\) 16.00 g/mol = 16.00 g/mol Total: 18.02 g/molThese molar masses are essential for converting between mass and amount in moles during stoichiometric calculations.
Combustion Reaction
Combustion reactions are a class of chemical reactions where a substance reacts with oxygen gas, often liberating energy in the form of heat and light. These reactions typically involve hydrocarbons and produce carbon dioxide and water as primary products. In the combustion of ibuprofen, the hydrocarbon compound reacts with oxygen to produce: - \(CO_{2}\), representing the carbon atoms in ibuprofen being oxidized. - \(H_{2}O\), representing the hydrogen atoms combining with oxygen.These reactions are exothermic, meaning they release energy. This energy release is one of the reasons combustion reactions are critical in engines and heaters. Understanding the combustion process, especially in organic molecules like ibuprofen, allows chemists to predict the type and amount of products generated, which is crucial in industrial applications and environmental considerations.
Conversion of Moles to Grams
Converting moles to grams is a fundamental step in stoichiometry, which helps in quantifying chemical reactions. To perform this conversion, one uses the molar mass of the substance as a conversion factor. This process helps in determining how much of a substance is involved or produced in a reaction.First, compute the number of moles from the given mass using the formula:\[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]Once the moles are known, it's straightforward to convert back to mass:\[\text{grams} = \text{moles} \times \text{molar mass (g/mol)}\]In our exercise, given 0.250 g of ibuprofen, we calculated 0.00121 moles. With the balanced equation, we then found the moles of \(CO_{2}\) and \(H_{2}O\) would be 0.0157 and 0.0109 respectively. Finally, using their molar masses, we converted these moles into grams. This tidy progression from moles to grams enables chemists to predict and measure reaction yields accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Chlorine \(\left(\mathrm{Cl}_{2}\right)\) and fluorine \(\left(\mathrm{F}_{2}\right)\) react to form \(\mathrm{ClF}_{3}\). A reaction vessel contains \(10.00 \mathrm{~g} \mathrm{Cl}_{2}\) and \(10.00 \mathrm{~g} \mathrm{~F}_{2}\). (Hint: Refer to Problem 9.82.) (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(12.50 \mathrm{~g}\) of \(\mathrm{ClF}_{3}\) is recovered. What is the percent yield of the reaction?

A compound used as an insecticide that contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{Cl}\) is subjected to combustion analysis, yielding \(55.55 \% \mathrm{C}\) and \(3.15 \% \mathrm{H}\). (a) What is the empirical formula of this compound? (b) What is the molecular formula if its empirical formula is \(1 / 3\) the mass of its actual formula?

Consider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction

Consider the unbalanced chemical equation \(\mathrm{S}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Balance the equation. (b) If you react \(4.80 \mathrm{~g}\) of sulfur with \(16.20 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), how many grams of \(\mathrm{SO}_{2}\) can theoretically be produced?

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?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.