Final answer:
The compound in question is a hydrocarbon composed of only carbon and hydrogen. By using the molar masses of carbon dioxide and water, we can calculate the moles of carbon and hydrogen in the compound and determine its empirical formula. The molecular formula of the compound is C5H12.
Step-by-step explanation:
The compound described in the question contains only carbon and hydrogen. To determine its molecular formula, we need to calculate the empirical formula first. We know that 0.2801 g of the compound gave 0.9482 g of carbon dioxide (CO2) on complete combustion.
The molar mass of CO2 is 44.01 g/mol (12.01 g/mol for carbon and 32.00 g/mol for oxygen).
Using the molar mass of CO2, we can calculate the moles of carbon dioxide produced: 0.9482 g ÷ 44.01 g/mol = 0.0216 mol CO2.
Since one mole of CO2 contains one mole of carbon, the moles of carbon in the compound is also 0.0216.
Now, we can calculate the moles of hydrogen in the compound. We know that water (H2O) is produced during combustion, and the molar mass of water is 18.02 g/mol (2.02 g/mol for hydrogen and 16.00 g/mol for oxygen).
The moles of water produced can be calculated using the mass of water: 0.9482 g CO2 × (2.02 g/mol / 44.01 g/mol) = 0.0434 mol H2O.
Since one mole of water contains two moles of hydrogen, the moles of hydrogen in the compound is 2 × 0.0434 mol = 0.0868 mol.
Now we have the moles of carbon (0.0216 mol) and moles of hydrogen (0.0868 mol) in the compound. We can find the empirical formula by dividing each value by their smallest common divisor:
C: 0.0216 mol ÷ 0.0216 mol = 1
H: 0.0868 mol ÷ 0.0216 mol = 4
Therefore, the empirical formula of the compound is CH4, which is methane.
Since the molar mass of the compound is given as 78.1 g/mol, and the empirical formula of methane (CH4) has a molar mass of 16.04 g/mol, we can calculate the molecular formula:
78.1 g/mol ÷ 16.04 g/mol = 4.87
Rounding the value to the nearest whole number, the molecular formula of the compound is C5H12.