What is the molecular formula of this compound?
An organic compound contains 54.55% C and 9.90% H. The molar mass is close to 90. What is the molecular formula?
Where I have gotten so far:
CxHyOz = 58g
C=12.001
H=1.0079
O=15.999
If 100% = 58
40% = 23.2
6.67% = 3.8744
Which leaves 53.33% (100%  40%  6.67%) = 30.9314
So that means there are 2 Carbon, 4 Hydrogen and 2 Oxygen.
How do I finish this off?
An organic compound contains 54.55% C and 9.90% H. The molar mass is close to 90. What is the molecular formula?
Where I have gotten so far:
CxHyOz = 58g
C=12.001
H=1.0079
O=15.999
If 100% = 58
40% = 23.2
6.67% = 3.8744
Which leaves 53.33% (100%  40%  6.67%) = 30.9314
So that means there are 2 Carbon, 4 Hydrogen and 2 Oxygen.
How do I finish this off?

Sorry, but I do not follow your method of calculating at all  where does: CxHyOz = 58g come from
I think that below is a fairly standardised method of working out these problems:
C = 54.55
H = 9.90
O = 100  ( 54.55+9.90) = 35.55
Divide each by atomic mass:
C = 54.55 /12.011 = 4.542
H = 9.90/1.0079 = 9.822
O = 35.55/15.999 = 2.222
Divide by smallest:
C = 2.04
H = 4.42
O = 2.222/2.222 = 1
With respect to other answers  I do not think that you can round this to H = 4
If you round H to 4.5, then multiply through by 2 to remove decimal fraction
C = 4
H = 9
O = 2
Empirical formula = C4H9O2 molar mass 89.1135 g/mol  which is close to 90 as required
empirical and molecular formula = C4H9O2
I think that below is a fairly standardised method of working out these problems:
C = 54.55
H = 9.90
O = 100  ( 54.55+9.90) = 35.55
Divide each by atomic mass:
C = 54.55 /12.011 = 4.542
H = 9.90/1.0079 = 9.822
O = 35.55/15.999 = 2.222
Divide by smallest:
C = 2.04
H = 4.42
O = 2.222/2.222 = 1
With respect to other answers  I do not think that you can round this to H = 4
If you round H to 4.5, then multiply through by 2 to remove decimal fraction
C = 4
H = 9
O = 2
Empirical formula = C4H9O2 molar mass 89.1135 g/mol  which is close to 90 as required
empirical and molecular formula = C4H9O2

Assuming the remaining percentage is oxygen, we have the following. Assume a 100 gram sample so that each percentage can be expressed as a mass, and convert the masses to moles. Then divide each number of moles by the smallest number of moles to get integers. These become the subscripts in the empirical formula.
54.55% C
9.90% H
35.55% O
54.55 g C x (1 mol C / 12.01 g C = 4.542 / 2.22 = 2
9.90 g H x (1 mol H / 1.01 g H = 9.90 mol H / 2.22 = 4
35.55 g O x (1 mol O / 16.0 g O = 2.22 mol O / 2.22 = 1
The empirical formula is C2H4O, and its empirical molar mass is 44.05
If the actual molar mass is 90 g/mol, then there must be twice as many of each element, and so the molecular formula is C4H8O2.
54.55% C
9.90% H
35.55% O
54.55 g C x (1 mol C / 12.01 g C = 4.542 / 2.22 = 2
9.90 g H x (1 mol H / 1.01 g H = 9.90 mol H / 2.22 = 4
35.55 g O x (1 mol O / 16.0 g O = 2.22 mol O / 2.22 = 1
The empirical formula is C2H4O, and its empirical molar mass is 44.05
If the actual molar mass is 90 g/mol, then there must be twice as many of each element, and so the molecular formula is C4H8O2.