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Ch.3 - Chemical Reactions and Reaction Stoichiometry

Chapter 3, Problem 54c

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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Hey everyone. So here we ask, calculate the empirical molecular formula of loose scene, which has a molar mass of 131.17 g per mole and has a mass percent composition of 54.94% carbon 9.99% hydrogen 10.68% nitrogen in 24.39% oxygen. And we know that the empirical formula gonna give us the relative number of atoms in the molecular formula can give us the actual number of atoms. So we're gonna assume we have 100 g of the compound and now we can convert our percentages into grams. The 54th 0.94 grams of carbon 9.99. And supplies region 10. grams of nitrogen in 24.39 grams of oxygen. And now we're going to convert the masses of each into malls using the molar masses for one ball of carbon. We have 12.11 g of carbon. When we get 4.57 malls in one mold of hydrogen have 1.8 g. And we get 9.91 malls in one more of nitrogen, we have 14 0.7 g. We're going to get 0.76 malls in one more of oxygen. 15 0.999 grams. We're going to get 1.52 malls. So now we need to divide each by the smallest number of moles to get the small toll number ratios over carbon, we have 4.57 by 0.76. We're going to get six for hydrogen with 9.910 point 76. We're gonna get 13 for nitrogen 0.76 divided by 0.76. To give us one and for oxygen you have 1.52 by 0.76. Would you give us two? Our empirical formula? We're gonna have C six age 13 in 02. Now we need to figure out what numbers multiply the subscript by to get the whole number ratios for the molecular formula. They're going to do this by using the formula X. It was a Mueller mass. What about the empirical mass? Our empirical mass or C six? Age 13 in is six. 12.11 g plus 13 times 1.8 g plus 14.7 g plus two times 15.999 g. And it's gonna give us 100 30. 1 0.17 g. We're gonna have eggs equals 100 31.17. About 131.17 which is one. So now we need to multiply the subscript and empirical formula by one. Our molecular formula. It's gonna be the same as an empirical formula. They're going to have C six age 13 N. 02. Thanks for watching my video and I hope it was helpful
Related Practice
Textbook Question

Determine the empirical and molecular formulas of each of the following substances: (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60% Na, and has a molar mass of 169 g/mol.

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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

(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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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?

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Textbook Question

(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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