An organic compound have the percentage C=68.85%’ H=4.91% and rest of oxygen. Calculate its molecular formula if V. D. is 61.An organic compound have the percentage C=68.85%’ H=4.91% and rest of oxygen. Calculate its molecular formula if V. D. is 61.
Answers
Answer:
Explanation:An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86.
%O= 100−(69.77+11.63)=18.60
Simple atomic ratio C:H:O=
12
69.77
:
1
11.63
:
16
18.6
=5.88:11.63:1.16=5:10:1
The empirical formula is C
5
H
10
O
Empirical formula mass =5×12+10×1+1×16=86. It is equal to molecular formula mass. Hence, molecular formula is same as empirical formula.
It does not reduce Tollens reagent. Hence, it is not an aldehyde
It forms an addition compound with sodium hydrogensulphite.
Hence, it contains carbonyl group.
It give positive iodoform test.
Hence, it is a methyl ketone
On vigorous oxidation it gives ethanoic and propanoic acid.
Hence, it is 2-pentanone.
CH
3
COCH
2
CH
2
CH
3
O
Vigorous
CH
3
CH
2
COOH+CH
3
COOH
Answer:
ANSWER
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86.
%O= 100−(69.77+11.63)=18.60
Simple atomic ratio C:H:O=
12
69.77
:
1
11.63
:
16
18.6
=5.88:11.63:1.16=5:10:1
The empirical formula is C
5
H
10
O
Empirical formula mass =5×12+10×1+1×16=86. It is equal to molecular formula mass. Hence, molecular formula is same as empirical formula.
It does not reduce Tollens reagent. Hence, it is not an aldehyde
It forms an addition compound with sodium hydrogensulphite.
Hence, it contains carbonyl group.
It give positive iodoform test.
Hence, it is a methyl ketone
On vigorous oxidation it gives ethanoic and propanoic acid.
Hence, it is 2-pentanone.
CH
3
COCH
2
CH
2
CH
3
O
Vigorous
CH
3
CH
2
COOH+CH
3
COOH