Question from Acidic Environment

[steph_batten]

So stuck with this question. Please help!! 

The following data was collected for an organic compound
M (g/mol) : 88.1
Composition (by weight):
54.5% C
36.4% O
9.1% H
m.p.(degreeC) : -5.4
b.p.(degreeC) : 163.3
Universal Indicator: turns red

(b) Calculate the empirical formula of the compound
(c) Calculate the molecular formula of the compound
(d) Draw five possible isomers of this compound

[RuiAce]

So stuck with this question. Please help!! 

The following data was collected for an organic compound
M (g/mol) : 88.1
Composition (by weight):
54.5% C
36.4% O
9.1% H
m.p.(degreeC) : -5.4
b.p.(degreeC) : 163.3
Universal Indicator: turns red

(b) Calculate the empirical formula of the compound
(c) Calculate the molecular formula of the compound
(d) Draw five possible isomers of this compound

Compare the molar mass to the percentage composition to determine the molar mass of each atom that builds the compound

C: 88.1 * 54.5% = 48.0145
O: 88.1 * 36.4% = 32.0684
H: 88.1 * 9.1% = 8.0171

These molar masses reveal the total molar mass of carbon/oxygen/hydrogen atoms are present. Divide by the molar mass of the atoms themselves to therefore determine how many atoms are present. (All answers here will be approximates)

48.0145/12.01 = 4
32.0684/16.00 = 2
8.0171/1.008 = 8

Hence we anticipate that the atom ratio of C:O:H is 4:2:8, which simplifies to 2:1:4

So the empirical formula is given C₂H₄O

Now, the organic compound turns universal indicator red. Hence we should expect some kind of an acid.
The 4 carbons is self-explanatory.

What could possibly have 4 atoms, be an organic compound, an acid, and also coincidentally 2 oxygens?
The answer is a carboxylic acid (like acetic acid).
Here, I propose that the answer is butanoic acid.

Check: Butanoic acid has formula C₄H₇COOH.
12.01*4 + 1.008*8 + 16.00*2 = 88.104
Liquid at room temperature, which is between the m.p. and b.p.
Perfect

Scroll a bit down to find some isomers of butanoic acid.