you can't count the mass of the O in CO2 and H2O as coming frm the compound (It was burned in O2). but you can deduce it .
mass of C in 57.94 grams of CO2
massC=12/(32+12) *57.94= ....
then convert that to moles of C
mass H in 11.85g water: 2/(18)*11.85=.... then convert that to moles H
mass O in original y= 20.63-mass H - mass C
convert that to moles of O
now you have moles of each in the original Y.
find the ratio of them, that leads directly to the empiricalformula
A compound Y containg C, H and O only was burnt in a strem of pure oxygen.The carbon(iv)oxide and water produced were collected in pure weighed reading obtained are initial mass of Y=20.63mg mass of CO2 product=57.94mg mass of water=11.85mg.What is the emperical fomular of the compound Y? [H=1? C=12, O=16].?
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1 answer