Question

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

Answer 1

Mass of carbon present in 1760.36 g $CO_2$

$$=\frac{12}{44}×1760.36=586.786\ g$$

Then,

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

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

$$=\frac{14}{46}×920.10=280\ g$$

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

$$C_{15}N_6O_2$$