5.00 g of a certain Compound known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide water Use this information to find the molecular formula of .

Answer

Combustion Analysis: Finding Molecular Formula of a Compound

Finding the Molecular Formula from Combustion Data

We are given the complete combustion of 5.00 g of an unknown compound X, which contains carbon, hydrogen, and oxygen. The combustion produces:

Product	Mass
Carbon Dioxide (CO₂)	9.57 g
Water (H₂O)	5.87 g

Step 1: Determine Moles of Carbon from CO₂

Molar mass of CO₂ = 44.01 g/mol

mol C = 9.57 g / 44.01 g/mol ≈ 0.2175 mol

Each mole of CO₂ gives 1 mole of carbon ⇒ 0.2175 mol C

Step 2: Determine Moles of Hydrogen from H₂O

Molar mass of H₂O = 18.02 g/mol

mol H₂O = 5.87 g / 18.02 g/mol ≈ 0.3259 mol

Each mole of H₂O gives 2 moles of hydrogen:

mol H = 0.3259 × 2 ≈ 0.6518 mol H

Step 3: Determine Mass of C and H in the Compound

mass C = 0.2175 mol × 12.01 g/mol ≈ 2.611 g

mass H = 0.6518 mol × 1.008 g/mol ≈ 0.657 g

Step 4: Determine Mass and Moles of Oxygen

Total mass of compound = 5.00 g

mass O = 5.00 − (2.611 + 0.657) = 1.732 g

mol O = 1.732 / 16.00 ≈ 0.1083 mol

Step 5: Find the Empirical Formula

Moles of each element:

C: 0.2175
H: 0.6518
O: 0.1083

Divide all by smallest (≈ 0.1083):

C: 0.2175 / 0.1083 ≈ 2.01 ≈ 2
H: 0.6518 / 0.1083 ≈ 6.02 ≈ 6
O: 0.1083 / 0.1083 ≈ 1.00

Empirical formula = C₂H₆O

Step 6: Determine the Molecular Formula

Molar mass of empirical formula (C₂H₆O):

(2×12.01) + (6×1.008) + (16.00) = 46.07 g/mol

Given molecular mass = 46.0 g/mol ⇒ same as empirical formula

✅ Final Molecular Formula of Compound X: C₂H₆O

This formula corresponds to compounds such as ethanol or dimethyl ether, both with the same molecular formula but different structures (isomers).

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