a 5.15g sample of a hydrocarbon is burned in oxygen, producing 15.6g of carbon dioxide and 8.45g of water. Assuming an excess of oxygen, what is the empirical formula of the hydrocarbon?
how do i set this problem up and how do i solve it?
To determine the empirical formula of the hydrocarbon, you need to calculate the ratio of the elements in the compound. This can be done by following these steps:
1. Convert the mass of each element to moles.
- Carbon dioxide: CO2 (1 carbon atom, 2 oxygen atoms)
- Water: H2O (2 hydrogen atoms, 1 oxygen atom)
Mass of carbon = 15.6 g CO2 × (1 mol CO2 / 44.01 g CO2) × (1 mol C / 1 mol CO2) × (12.01 g C / 1 mol C) = X g C
Mass of hydrogen = 8.45 g H2O × (1 mol H2O / 18.015 g H2O) × (2 mol H / 1 mol H2O) × (1.01 g H / 1 mol H) = Y g H
Mass of oxygen = 15.6 g CO2 × (1 mol CO2 / 44.01 g CO2) × (2 mol O / 1 mol CO2) × (16.00 g O / 1 mol O) + 8.45 g H2O × (1 mol H2O / 18.015 g H2O) × (1 mol O / 1 mol H2O) × (16.00 g O / 1 mol O) = Z g O
2. Convert the masses of each element to moles.
Moles of carbon (C) = X g C / 12.01 g/mol
Moles of hydrogen (H) = Y g H / 1.01 g/mol
Moles of oxygen (O) = Z g O / 16.00 g/mol
3. Determine the smallest whole-number ratio.
Divide the moles of each element by the smallest number of moles calculated in step 2.
4. Write the empirical formula.
Use the obtained ratio as subscripts to write the empirical formula. Round any fractional subscripts to the nearest whole number.
Now, let's solve it step by step:
1. Calculate the moles of each element:
Moles of carbon = 15.6 g / 44.01 g/mol = 0.3541 mol C
Moles of hydrogen = 8.45 g / 18.015 g/mol = 0.4693 mol H
Moles of oxygen = (15.6 g × 2 + 8.45 g) / 44.01 g/mol + 18.015 g/mol = 0.5183 mol O
2. Find the smallest number of moles (0.3541 mol C in this case).
3. Divide the number of moles of each element by the smallest number:
Moles of carbon = 0.3541 mol C / 0.3541 mol C = 1
Moles of hydrogen = 0.4693 mol H / 0.3541 mol C ≈ 1.326
Moles of oxygen = 0.5183 mol O / 0.3541 mol C ≈ 1.465
4. Round the fractional subscripts to the nearest whole number:
The ratio becomes approximately C1H1.326O1.465. Since we need whole numbers, let's multiply everything by 2 to get whole numbers:
C2H2.652O2.93
Thus, the empirical formula of the hydrocarbon is C2H2.65O2.93.