Skip to main content
Ch.3 - Chemical Reactions and Reaction Stoichiometry

Chapter 3, Problem 54b

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.

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
1316
views
Was this helpful?

Video transcript

Hey everyone. So here we have to calculate the empirical and molecular formula of magnesium like tea With the multi master 202.47 g per mole. And it has a mass percent composition of % 12.01 magnesium 4.98% hydrogen 35.6% carbon, 47.42% oxygen. We know the empirical formula going to give us the relative number of atoms and I'm molecular formula. Gonna give us our actual number of Adams. So we're gonna assume we have 100 g of the compound. We're going to convert our percentages into grams. You have 12.01 magnesium 4.98 grams of hydrogen, 35.6 grams of carbon And 47.42 brands of oxygen. So now we're going to convert the masses of each into malls using the molar masses in one ball of magnesium, We have 24 .305 grams of magnesium. And this gives us 0.49 malls of magnesium. In one month of hydrogen. You have 1.008 grams of hydrogen. Just give us 4.94 malls of hydrogen and one more of carbon. You have 12.011 g of carbon. This gives us 2.98 malls of carbon and one more of oxygen. We have 15.999 brands of oxygen. Give us 2.96 malls of oxygen. And now we need to divide each by the smallest number of moles. To get the smallest whole number ratios for magnesium 0.49 By 0.49. Let's give us one for hydrogen. We have 4.94. What about 0.49 And give us for carbon Left. By 0.49. Just give us six. And for oxygen That's 2.96. 0.49. Let's give us six. So far empirical formula. We have M. G. C. Six. Age 10 oh six. Now we need to find what number two most by the subsequent spy. To get the whole number ratios for the molecular formula. We're gonna do this by using the equation X. It was smaller mass by the empirical mass. Our empirical mask of mg. C. Six. H. 1006. It's 24.30.5 grams by sex. 12.011 g plus Times 1.008g plus six times 15 .999 g. I'm gonna get 20.2 0.47 grams. You're gonna have X. He goes to go to 0.47. What about 20.2 0.47. Could you give us 1? And so now it needs most by the subscript in the empirical formula. By one of the molecular formula. It's gonna be the same as our empirical formula. We're gonna have M. G. These sets age 10 oh six. 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: (b) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass and has a molar mass of 195 g/mol.

1294
views
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.

1528
views
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.

1440
views
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.

1228
views
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?

1691
views
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?

3617
views
1
comments