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Method 1 :

Let CₘHₙ be the molecular formula of X.

Each mole of any gas contains a volume of 24000 cm³.
No. of moles of CₘHₙ = (120 cm³) / (24000 cm³/mol) = 0.005 mol
Molar mass of CₘHₙ = (0.290 g) / (0.005 mol) = 58.0 g/mol

No. of moles of CₘHₙ = (0.135 g) / (58.0 g/mol) = 0.00233 mol
No. of moles of CO₂ = (0.410 g) / (44.0 g/mol) = 0.00932 mol
No. of moles of H₂O = (0.209 g) / (18.0 g/mol) = 0.0116 mol

CₘHₙ + [m + (n/4)]O₂ → mCO₂ + (n/2)H₂O

Mole ratio CₘHₙ : CO₂ = 1 : m
1 : m = 0.00233 : 0.00932
M = 4

Mole ratio CₘHₙ : H₂O = 1 : (n/2)
1 : (n/2) = 0.00233 : 0.0116
n = 10

Molecular formula = C₄H₁₀


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Method 2 :

Each mole of any gas contains a volume of 24000 cm³.
No. of moles of the gas = (120 cm³) / (24000 cm³/mol) = 0.005 mol
Molar mass of the gas = (0.290 g) / (0.005 mol) = 58.0 g/mol

Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Each mole of CO₂ contains 1 mole of C.
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Each mole of H₂O contains 2 moles of H.

Moles of C in X = Moles of C in CO₂ formed = [(0.410 g) / (44.0 g/mol)] × 1 = 0.00932 mol
Moles of H in X = Mass of H in H₂O formed = [(0.209 g) / (18.0 g/mol)] × 2 = 0.0232 mol

Mole ratio C : H = 0.00932 : 0.0232 = 1 : 2.49 ≈ 2 : 5

Let (C₂H₅)ₓ be the molecular formula of X.

Molecular mass of X (in g/mol) :
(12.0×2 + 1.0×5) x = 58.0
29x = 58.0
x = 2

Molecular formula = C₄H₁₀

Supposing the molar volume at room temperature to be 24.0 L:
(0.290 g) / (0.120 L / (24 L/mol)) = 58.0 g X/mol
(0.135 g total) / (58.0 g X/mol) = 0.0023276 mol X

(0.410 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) / (0.0023276 mol X) = 4.00
(0.209 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) / (0.0023276 mol X) = 9.97
Round to the nearest whole numbers to find the molecular formula:
C4H10

The coefficients of the molecular formula have the common factor 2, so divide the molecular formula by 2 to find the empirical formula:
C2H5