Answer:
The answers are
C = 0.0895 moles
1. 0.02975 moles of N
2. 0.0299 moles O
Explanation:
Mass of sample = 2.18 g
Combustion yields 3.94 g of CO₂ and 1.89 g of H₂O
1.23 g of sample contains 0.235 g of N
3.94 g of CO₂ which contains one mole of C and two moles of O
Molar mass of CO₂ = 44 mass fraction of C = 12/44 = 0.273, Therefore mass of carbon in 2.18 g = 0.273×3.94 = 1.075 g
number of moles of C in sample = mass/(molar mass) = 1.075 g/(12.01 g/mol) = 0.0895 moles of C
(1) Mass of nitrogen, in 1.23 g of sample = 0.235
therefore mass in 2.18 g =0.235×2.18/1.23 = 0.4165 g
= 0.02975 moles of N
(2) To find the number of moles of oxygen in the sample we note that total mass of sample = 2.18 g containing only nitrogen, hydrogen, oxygen and carbon
Mass of hydrogen can be fround from 2/18×1.89 = 0.21 g
Mass of oxygen therefore = 2.18 -(0.21+0.4165+1.075) = 0.4785
number of moles = 0.475/16 = 0.0299 moles