(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 CO 2 and 4.083 mg of H 2 O. What is the empirical formula for nicotine? If nicotine has a molar mass of 160 ± 5 g/mol, what is its molecular formula?

Hey everyone, we're asked to determine the empirical and molecular formula of the following unknown compound. First let's go ahead and determine the number of grams we have for each atom Starting off with carbon, we can take our 0.6581 g of carbon dioxide Using carbon dioxides, Mueller mass of 44.01 g. We can get some moles of carbon dioxide And per one mole of carbon dioxide. We can see that we have one mole of carbon And using carbons atomic mass, we know that per one mole of carbon. We have 12.01 g of carbon. This will get us to a value of 0. g of carbon. Moving on to hydrogen, we can go ahead and take our 0.112, two grams of water. And using waters Mueller mass, we have 18.01 g of water per one mole of water and per one mole of water, we see that we have two mole of hydrogen using hydrogen atomic mass. We know that per one mole we have 1.01 g of hydrogen. So when we calculate this out, we end up with a total of 0.01246g of hydrogen. And now to find our nitrogen We can take our total number of grams of our compound, which is 0.227g. And subtract the values we just calculated which is 0.1794 g of carbon and 0. grams of hydrogen. And when we calculate this out, we end up with a total of 0.03514 g of nitrogen. Now that we have our grams of carbon, hydrogen and nitrogen, let's go ahead and convert this into moles. Starting off with our carbon, we have 0.1794 g of carbon and we know that we have 12.01g of carbon per one mole of carbon. This will get us a total of 0. mole of carbon. Looking at our hydrogen, We have 0.01246g of hydrogen And we know that we have 1.01 g of hydrogen per one mole of hydrogen. So when we calculate this out, we end up with 0. more of hydrogen. Next looking at our nitrogen, we have 0.03514 g of nitrogen. And we know that we have 14.01 g of nitrogen per one mole of nitrogen. So when we calculate this out, we end up with 0.00258 mole of nitrogen. Now, in order to get our empirical formula will have to divide each of our values by the least amount of moles we have present. In this case, it's going to be our nitrogen. So dividing each 1x0. We're going to end up with six carbon five hydrogen and one nitrogen. So our empirical formula Is going to be C6 H five N. Now in order to get the molecular formula will have to calculate the molar mass of our empirical formula. And when we do we find that it's 91.11 g per mole. So to get our molecular formula, We're going to take the molar mass of our unknown compound, which they told us in our questions them as 273. g per mole. And we'll divide this by our empirical formulas Mueller mass which is 91.11 g per mole. This will get us to a total of three. This means our molecular formula is going to be C six H 5 n. Multiplied by three. So we end up with a molecular formula of c. 18, Age 15 N three. And these are our final answers. So I hope this made sense. And let us know if you have any questions

Ch.3 - Chemical Reactions and Reaction Stoichiometry

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Brown 14th Edition

Ch.3 - Chemical Reactions and Reaction Stoichiometry

Problem 56b

Chapter 3, Problem 56b

(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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Determine the moles of carbon in CO2 and the moles of hydrogen in H2O. Use the molar mass of CO2 (44.01 g/mol) and H2O (18.02 g/mol) to convert the mass of CO2 and H2O to moles. Remember that each mole of CO2 contains one mole of carbon and each mole of H2O contains two moles of hydrogen.

Calculate the mass of carbon and hydrogen in the sample by multiplying the moles of carbon and hydrogen by their respective atomic masses (12.01 g/mol for C and 1.008 g/mol for H).

Subtract the total mass of carbon and hydrogen from the initial mass of the nicotine sample to find the mass of nitrogen. Use the atomic mass of nitrogen (14.01 g/mol) to convert this mass to moles of nitrogen.

Find the simplest whole number ratio of moles of C, H, and N to determine the empirical formula. Divide the moles of each element by the smallest number of moles calculated among C, H, and N.

To find the molecular formula, divide the molar mass of nicotine (given as 160 ± 5 g/mol) by the molar mass of the empirical formula unit. Multiply the subscripts in the empirical formula by this ratio to get the molecular formula.

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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 elemental composition of organic compounds. In this process, a sample is burned in excess oxygen, producing carbon dioxide (CO2) and water (H2O). The masses of these products are measured to calculate the amounts of carbon, hydrogen, and other elements in the original sample, which is essential for determining the empirical formula.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It is derived from the moles of each element calculated from the combustion products. For nicotine, the empirical formula can be determined by converting the masses of CO2 and H2O produced into moles of carbon and hydrogen, and then finding the ratio of these moles to establish the formula.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule of a compound. It can be derived from the empirical formula by comparing the molar mass of the compound to the molar mass of the empirical formula. If the molar mass of nicotine is known, the molecular formula can be calculated by determining how many times the empirical formula fits into the molar mass.

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Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as Na2CO3 # xH2O, where x is the number of moles of H2O per mole of Na2CO3. When a 2.558-g sample of washing soda is heated at 125 C, all the water of hydration is lost, leaving 0.948 g of Na2CO3. What is the value of x?

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(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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Key Concepts

Combustion Analysis

Empirical Formula

Molecular Formula

(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?