Snippet #1821470

Gracie glanced at the girl next to her out of the corner of her eye. "Nothing interesting," she said. She reached into her pocket, to maybe find something to play with to pass some time, and found a knife. "Hey, is it true you aren't supposed bring weapons to school? I wasn't sure if my parents were being serious or not," Gracie leaned over and asked the girl. Use your common sense, Gracie. You've had to go under cover in a school before. You know weapon's aren't aloud in any school, her mind told her. "Miss Smith, since your so interested in what the class is learning, perhaps yo can answer this ... simple question. A compound containing the elements C, H, N, and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H. Determine the mass in grams of C in the 1.2359 g sample of the compound." Gracie looked down for a second then looked back up. "0.6117 grams of Carbon." Working for the government meant she had to stdy some subjects early. Chemistry was a major one. Some kids were asleep and some were looking at her in a weird way. "How did you get that answer," the teacher asked, a puzzled look on her face. "The same way anyone else would, I thought. I can elaborate if you want. Well first I'll start by saying that the only thing necessary for a combustion reaction is oxygen and something that will burn, or be oxidized. There's no rule stating that a combustion reaction can only produce CO2 and H2O. This is known as rusting and it's a very slow combustion reaction, not like the types you're usually used to like clothes catching on fire or wood burning. Notice that no water or carbon dioxide was formed in the above reaction yet it is still a combustion reaction because we have oxygen reacting with another substance and the substance being oxidized." Gracie walked to the front of the class and began to write on the board.

0.6117 g C x 1 mol C/12.01 g C=0.05093 mol C

0.3564 g N x 1 mol N/14.01 g N=0.02544 mol N

0.2030 g O x 1 mol O/16.00 g O=0.01269 mol O

0.0648 g H x 1 mol/1.01 g H=0.0642 mol H

C: 0.05093 mol/0.01269 mol=4.013
N: 0.02544 mol/0.01269 mol=2.005
O: 0.01269 mol/0.01269 mol=1.000
H: 0.0642 mol/0.01269 mol=5.06

Empirical formula: C4H5N2O

"See, I converted everything into moles for you, and figred the empirical formula." Gracie walked back to her desk and sat down. "You know, if you want to challange me, your going to have to think harder," she said with a smirk.