A sample of liquid consisting of only C,H and O and having a mass of 0.5438g was burned in pure oxygen and 1.039g of CO2 and 0.6369g H2O were obtained. what is the empirical formula of the compound?
Convert g CO2 to g C.
Convert g H2O to g H.
0.5438g - g C - g H = g O
Convert grams of each to mols and find the ratio for the empirical formula. Post your work if you get stuck.
To determine the empirical formula of the compound, you need to find the ratio of the number of atoms of each element present in the compound. Let's break down the given information step by step:
1. Start by calculating the number of moles of CO2 and H2O produced:
- The molar mass of CO2 (carbon dioxide) is 12.01 g/mol for carbon + 2 * 16.00 g/mol for oxygen.
- The molar mass of H2O (water) is 2 * 1.01 g/mol for hydrogen + 16.00 g/mol for oxygen.
- Convert the mass of CO2 and H2O to moles using their respective molar masses.
Moles of CO2 = 1.039g / (12.01g/mol + 2 * 16.00g/mol) = 0.0244 mol
Moles of H2O = 0.6369g / (2 * 1.01g/mol + 16.00g/mol) = 0.0356 mol
2. Next, determine the number of moles of carbon (C), hydrogen (H), and oxygen (O) in the original compound:
- The mole ratio between CO2 and C is 1:1, meaning that the number of moles of carbon is also 0.0244 mol.
- The mole ratio between H2O and H is 2:2, meaning that the number of moles of hydrogen is also 0.0356 mol.
- The mole ratio between CO2 and O is 1:2, meaning that the number of moles of oxygen can be calculated by multiplying the moles of CO2 by 2: 0.0244 mol * 2 = 0.0488 mol.
3. Convert the number of moles of each element to the simplest, whole-number mole ratio:
- Divide the number of moles of each element by the smallest value among them (0.0244 mol). This will give you the mole ratio relative to carbon.
C: 0.0244 mol / 0.0244 mol = 1
H: 0.0356 mol / 0.0244 mol = 1.459
O: 0.0488 mol / 0.0244 mol = 2
4. Round the mole ratios to the nearest whole number:
C: 1
H: 1.459 (approximately 1)
O: 2
The empirical formula of the compound is CH2O, as the mole ratio of carbon, hydrogen, and oxygen can be simplified to 1:1:2.