Answer :
c=54.54%=54.54/12=4.545 4.545/2.2725=2
H=9.09%=9.09/1=9.09 9.09/2.2725=4
O=36.36%=36.36/16=2.2725 2.2725/2.2725=1
so,
the compound is C2H4O (CH3CHO)
H=9.09%=9.09/1=9.09 9.09/2.2725=4
O=36.36%=36.36/16=2.2725 2.2725/2.2725=1
so,
the compound is C2H4O (CH3CHO)
Question : A compound on analysis gave the following percentage composition by mass.
1, Hydroge = 9.09%
2, Oxygen = 35.35%
3,Carbon = 54.55%
5,Molar mass of compound = 88.88%
Find out molecular formula of compound
Answer:
Step 1 : = H + O + C
= 9.09+36.36+54.55
= 100
Step 2: = H = 9.09% 9.09g
O = 36.36% 36.36g
C =54.55% 54.55g
( here we change % into g)
Step 3: Convertion g into mol
Mole of H = 9.09÷1.008
=9.17 mol
Mole of O = 36.36÷16
= 2.272 mol
Mole of C = 54.55÷12.000
= 4.545 mol
Step 4 : H = 9.017 ÷ 2
= 4.5085
O = 2.272 ÷ 2
= 1.136
C = 4.545 ÷ 2
= 0.2725
therefore
4.5085 : 1.126 : 0.2725
Ratio = 4 : 1 : 2
emeeical formula = H4 OC2
Step 5 : emerical into molecular formula
Molecular formula = CH3 - CHO
hope it help you friend ^_^ .