(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO 2 and 1.37 mg of H 2 O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Hi everyone today, we have a problem asking us to determine the empirical formula of a hydrocarbon. If combustion analysis revealed the following composition, 23.093g of carbon dioxide and 4.701g of water. So our first step here is going to be to calculate the molar mass of carbon dioxide. So we have carbon Which has a molar mass of 12 and there's only one of them, So that equals 12. And then we have oxygen, which has a molar mass of 16, and we have two of them, Which equals 32. So 32 plus equals 44. And now we want to start out with our .09, 3 grams of carbon dioxide. and we are going to divide by the Molar mass. So 40 for grams of carbon dioxide Over one Under mole of carbon dioxide. And then we're going to multiply by the multiple ratio to get moles of carbon. So we're going to have one mole of carbon over one mole of carbon dioxide. So our grams of carbon dioxide are going to cancel out and our moles of carbon dioxide are going to cancel out And that equals 0. moles of carbon. Now we need to do the same thing with ox with water. So we have hydrogen with a molar mass of one and we have two of them. So that equals two. And then we have oxygen with a molar mass of 16 And we have one of them. So that equals one and equal 16 And then two plus equals 18. And now we want to start out with our four .701g of water. And we're going to divide by our molar mass so 18 g of water. And that will give us moles of water. And then we're gonna multiply by our multiple ratio. So we have two moles of hydrogen For every one mole of water. So our grams of water is going to cancel out and our moles of water are going to cancel out. And that is going to give us 0. moles of hydrogen. And now we need to divide the two by each other. So 0.52 divided by 0.52 is obviously one. So we have a 1-1 ratio which gives us an empirical formula of C. H. And that is our final answer. Thank you for watching. Bye.

Ch.3 - Chemical Reactions and Reaction Stoichiometry

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Brown 14th Edition

Ch.3 - Chemical Reactions and Reaction Stoichiometry

Problem 55a

Chapter 3, Problem 55a

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO₂ and 1.37 mg of H₂O. If the compound contains only carbon and hydrogen, what is its empirical formula?

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Convert the mass of CO2 to moles of carbon: Use the molar mass of CO2 (44.01 g/mol) to find the moles of CO2, then use the ratio of carbon in CO2 (1 mole of C per mole of CO2) to find the moles of carbon.

Convert the mass of H2O to moles of hydrogen: Use the molar mass of H2O (18.02 g/mol) to find the moles of H2O, then use the ratio of hydrogen in H2O (2 moles of H per mole of H2O) to find the moles of hydrogen.

Determine the simplest whole number ratio of moles of carbon to moles of hydrogen: Divide the moles of each element by the smallest number of moles calculated in the previous steps.

Write the empirical formula: Use the whole number ratio from the previous step to write the empirical formula of the compound.

Verify the empirical formula: Ensure that the empirical formula is consistent with the given data and the assumption that the compound contains only carbon and hydrogen.

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Key Concepts

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

Combustion Analysis

Combustion analysis is a technique used to determine the composition of organic compounds by burning them in excess oxygen. The products of combustion, typically carbon dioxide (CO2) and water (H2O), are measured to calculate the amounts of carbon and hydrogen in the original compound. This method is essential for deriving empirical formulas, as it provides the necessary data to relate the mass of the products back to the elements in the compound.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is derived from the mass of each element obtained from combustion analysis and is crucial for understanding the basic composition of the substance. For example, if a compound contains carbon and hydrogen, the empirical formula will indicate the ratio of carbon atoms to hydrogen atoms, which can be determined from the moles of each element calculated from their respective masses.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles (atoms, molecules, etc.) it contains. One mole of any substance contains Avogadro's number of entities, approximately 6.022 x 10^23. In combustion analysis, converting the mass of CO2 and H2O produced into moles allows for the determination of the moles of carbon and hydrogen in the original compound, which is essential for calculating the empirical formula.

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Textbook Question

Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol.

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Textbook Question

Determine the empirical and molecular formulas of each of the following substances: (b) Cadaverine, a foul-smelling substance produced by the action of bacteria on meat, contains 58.55% C, 13.81% H, and 27.40% N by mass; its molar mass is 102.2 g/mol.

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Textbook Question

Determine the empirical and molecular formulas of each of the following substances: (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains 59.0% C, 7.1% H, 26.2% O, and 7.7% N by mass; its molar mass is about 180 u.

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Textbook Question

(b) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO₂ and 0.1159 g of H₂O. What is the empirical formula for menthol? If menthol has a molar mass of 156 g/mol, what is its molecular formula?

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Textbook Question

(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of CO₂ and 2.58 mg of H₂O. What is the empirical formula of the compound?

(b) Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg of CO₂ and 4.083 mg of H₂O. What is the empirical formula for nicotine? If nicotine has a molar mass of 160 ± 5 g/mol, what is its molecular formula?

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(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Verified Solution

Key Concepts

Combustion Analysis

Empirical Formula

Mole Concept

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO₂ and 1.37 mg of H₂O. If the compound contains only carbon and hydrogen, what is its empirical formula?