9) .mass of sample = 12.10 grams
mass of CO2 = 30.20 grams
molar mass of CO2 = 44 gram/mole
% by mass of C= 12/44 xmass of CO2/mass of sample x 100
= 12/44x 30.20/12.10 x100 = 68.069%
% by mass of H = 2/18 x mass of H2O/mass of sample x 100
= 2/18 x 14.80/12.10 x 100= 13.59%
% by mass of H= 13.59%
% by mass of O= 100 -[68.069 +13.59]
% by mass of O = 18.341%
Element % by mass atomic wt Relative number simple ratio
C 68.069 12 68.069/12 = 5.672 5.672/1.146= 4.94 = 5.0..
H 13.59 1.0 13.59/1.0 = 13.59 13.59/1.146 = 11.85= 12
O 18.341 16 18.341/16 = 1.146 1.146/1.146 = 1.0
Emperical formula = C5H12O.
티15-) Certam amount ofpure hydrogen chloride psisdissolved=water and-ted with-AgNO witine, lfthe mass of precipitate formed from...