"empirical formulas"
combustion of a 0.255g compound conatining only C,H and O, yields 0.561g CO2 and 0.306g H2O. what is the empirical formula of the compound?
Convert g CO2 to g C.
Convert g 2P to g H.
0.255 - g C - g H = g O.
Convert g C, H, O, to mols.
Find the ratio.
Post your work if you get stuck.
umm... i need some help, sorry but how do you convert them?? my chem teacher isnt the best :(
I made a typo. It should read covert g H2O to g H.
g CO2 = 0.561/molar mass CO2 = ?
1 mol C in 1 mol CO2, then g C = mols CO2 x molar mass C = about 0.153g C.
Same process for the others (but note there are two mol H in 1 mol H2O).
To determine the empirical formula of a compound, we need to find the ratio of the different elements present in the compound based on the given mass of each element.
1. Start by finding the moles of CO2 and H2O produced:
- Moles of CO2 = mass of CO2 / molar mass of CO2
- Moles of H2O = mass of H2O / molar mass of H2O
The molar mass of CO2 (carbon dioxide) is calculated by adding the atomic masses of carbon (12.01 g/mol) and oxygen (16.00 g/mol) together. Similarly, the molar mass of H2O (water) is calculated by adding the atomic masses of hydrogen (1.01 g/mol) and oxygen (16.00 g/mol).
2. Next, determine the moles of carbon (C), hydrogen (H), and oxygen (O) in the original compound.
- Moles of carbon = 1 * moles of CO2 (since each CO2 molecule contains one carbon atom)
- Moles of hydrogen = 2 * moles of H2O (since each H2O molecule contains two hydrogen atoms)
- Moles of oxygen = 2 * moles of CO2 + 1 * moles of H2O (since each CO2 molecule contains two oxygen atoms, and each H2O molecule contains one oxygen atom)
3. Calculate the mass of each element in the compound:
- Mass of carbon = moles of carbon * atomic mass of carbon
- Mass of hydrogen = moles of hydrogen * atomic mass of hydrogen
- Mass of oxygen = moles of oxygen * atomic mass of oxygen
4. Finally, calculate the ratio of the elements to find the empirical formula of the compound:
- Divide the mass of each element by the smallest mass among the three elements. If necessary, round the ratios to the nearest whole number to obtain whole number subscripts.
The resulting ratios will give you the empirical formula of the compound.
Let's perform the calculations:
1. Moles of CO2:
Moles of CO2 = 0.561g / (12.01g/mol + 16.00g/mol) = 0.025 mol
Moles of H2O:
Moles of H2O = 0.306g / (2*(1.01g/mol) + 16.00g/mol) = 0.017 mol
2. Moles of carbon (C), hydrogen (H), and oxygen (O) in the original compound:
Moles of carbon (C) = 1 * moles of CO2 = 0.025 mol
Moles of hydrogen (H) = 2 * moles of H2O = 2 * 0.017 mol = 0.034 mol
Moles of oxygen (O) = 2 * moles of CO2 + 1 * moles of H2O = 2 * 0.025 mol + 1 * 0.017 mol = 0.067 mol
3. Mass of each element in the compound:
Mass of carbon (C) = moles of carbon * atomic mass of carbon = 0.025 mol * 12.01 g/mol = 0.301 g
Mass of hydrogen (H) = moles of hydrogen * atomic mass of hydrogen = 0.034 mol * 1.01 g/mol = 0.034 g
Mass of oxygen (O) = moles of oxygen * atomic mass of oxygen = 0.067 mol * 16.00 g/mol = 1.072 g
4. Ratios of elements:
Since the smallest mass is 0.034 g, we divide each mass by this value and round to obtain whole number subscripts:
Carbon (C): 0.301 g / 0.034 g ≈ 8.85 ≈ 9
Hydrogen (H): 0.034 g / 0.034 g = 1
Oxygen (O): 1.072 g / 0.034 g = 31.53 ≈ 32
Therefore, the empirical formula of the compound is C9H1O32.