C3H4
C2H4
C4H10
C2H6O
I need them in four significant figures
C2H4
C4H10
C2H6O
I need them in four significant figures

Divide atomic weight of carbon by molecular weight of the compound and multiply it by 100.
C3H4: Mass % C = 36.033 x 100 / 40.0641 = 89.93 (Propadiene)
C2H4: Mass % C = 24.022 x 100 / 28.0533 = 85.62 (Ethylene)
C4H10: Mass % C = 48.044 x 100 / 58.1226 = 82.65 (Butane)
C2H6O: Mass % C = 24.022 x 100 / 46.0687 = 52.14 (Dimethyl Ether)
C3H4: Mass % C = 36.033 x 100 / 40.0641 = 89.93 (Propadiene)
C2H4: Mass % C = 24.022 x 100 / 28.0533 = 85.62 (Ethylene)
C4H10: Mass % C = 48.044 x 100 / 58.1226 = 82.65 (Butane)
C2H6O: Mass % C = 24.022 x 100 / 46.0687 = 52.14 (Dimethyl Ether)

To calculate the mass % composition of any element in any compound, we have to first find the molecular wt of the compound and then the weight of that element in that compound
In C3H4, the molecular weight is, 3x12 + 4x1 = 40
Wt of C in this compound is 3x12=36
So mass % of C = (36/40)x100 = 90%
In C2H4, molecular wt = (2x12) + ( 4 x 1) = 28
Mass % of C = (12/28)x100= 42.85%
Similarly, you can calculate the others
In C3H4, the molecular weight is, 3x12 + 4x1 = 40
Wt of C in this compound is 3x12=36
So mass % of C = (36/40)x100 = 90%
In C2H4, molecular wt = (2x12) + ( 4 x 1) = 28
Mass % of C = (12/28)x100= 42.85%
Similarly, you can calculate the others

BM Carbon (C) 12
BM Hydrogen (H) 1
BM Oxygen (O) 16
C3H4 : 12 X 3 + 1 X 4 = 40 ==> % C = 36/40 = 90%
C2H4 : 12 X 2 + 1 X 4 = 28 ==> % C = 24/28 = 85.7%
C4H10 : 12 X 4 + 1 X 10 = 58 ==> % C = 48/58 = 82.7%
C2H6O : 12 X 2 + 1 X 6 + 16 X 1 = 46 ==> % C = 24/46 = 52.1%
BM Hydrogen (H) 1
BM Oxygen (O) 16
C3H4 : 12 X 3 + 1 X 4 = 40 ==> % C = 36/40 = 90%
C2H4 : 12 X 2 + 1 X 4 = 28 ==> % C = 24/28 = 85.7%
C4H10 : 12 X 4 + 1 X 10 = 58 ==> % C = 48/58 = 82.7%
C2H6O : 12 X 2 + 1 X 6 + 16 X 1 = 46 ==> % C = 24/46 = 52.1%