Answer to Question #96359 in General Chemistry for ABIGAIL

Question #96359
Complete combustion of a 970.97-gram sample of caffeine-containing carbon nitrogen and oxygen produces 1760.36g of CO2 450.375 of H20 and 920.10 of NO2. WHat is the empirical formula
1
Expert's answer
2019-10-11T10:29:57-0400

Mass of carbon present in 1760.36 g "CO_2"

"=\\frac{12}{44}\u00d71760.36=586.786\\ g"

Then,

Moles of carbon = "\\frac{586.786}{12}=48.9"

Mass of nitrogen present in 920.10 g of "NO_2\\\\"


"=\\frac{14}{46}\u00d7920.10=280\\ g"


Moles of nitrogen = "\\frac{280}{14}=20"


Mass of oxygen = total mass of caffeine - mass of carbon - mass of nitrogen

"=104.184\\ g"

Moles of oxygen = "\\frac{104.184}{16}=6.511"

So,

Ratio of carbon to nitrogen to oxygen


"=48.9:20:6.511\\\\ or\\\\ 7.5:3:1\\\\or\\\\15:6:2"

So empirical formula is

"C_{15}N_6O_2"


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