Answer to Question #331841 in General Chemistry for jjjjk

Question

Exactly 1.387 g of compound Z was combusted to form 3.499 g CO₂ mo sand 0.614 g H₂O. Compound Z contains only carbon, hydrogen, and oxygen. Determine the compound's molecular formula if the molar mass of compound Z is 122.118 g/mole.

Answer 1

С_xH_yO_z + O₂ → xCO₂ + x/yH₂O

n(CO₂) = m(CO₂)/Mr(CO₂) = 3.499 g / 44 g/mol = 0.0795 moll

It containd 0.0795 moll of C

m(C) = n(C)*Mr(C) = 0.0795 moll * 12 g/moll = 0.954 g

n( H₂O) = m(H₂O)/Mr(H₂O) = 0.614 g / 18 g/mol = 0.0341 moll

It containd 2*0.0795 moll of H = 0.0682 moll

m(H) = n(H)*Mr(H) = 0.0682 moll * 1 g/moll = 0.0682 g

m(O) = m(С_xH_yO) - m(C) - m(H) = 1.387 g - 0.954 g - 0.0682 g = 0.3648 g

n(O) = m(O)/Mr(O) = 0.3648 g / 16 g/mol = 0.0228 moll

ratio n(C):n(H):n(O) = 0.0795 : 0.0682 : 0.0228 = 3.5:3:1 = 7:6:2

Molecular formula is С₇H₆O₂