An organic compound on analysis gave C=57.8% and H=3.6% and rest of oxygen.The vapour density of the compound was found to be 83.Find the molecular formula of the compound?
Take a 100 g sample to give you 57.8g C, 3.6g H and 38.6 g O.
Convert to mols.
57.8/12 = ?
3.6/1 = ?
38.6/16 = ?
Now find the ratio of CHO to each other with the lowest being no smaller than 1.00. The easy way to do this is to divide the smaller number by itself, then divide the other numbers by the same small number. In this case, you don't get whole numbers so
multiply all of those resultant numbers by 2, 3, 4, 5, etc until you come out with numbers that can be rounded to whole numbers. You will find that number to be 4.
Post your work you get stuck. I didn't so the molecular formula. There is more than one definition for vapor density and I don't know which you are using.
One formula that ik of is molecular mass/Empertical formula mass=n
n is the subscript you need for your formula.
To find the molecular formula of the compound, we need to gather all the available information and use it to determine the empirical formula and then the molecular formula.
1. Start by assuming you have 100 grams of the compound.
- Given: C = 57.8%, H = 3.6%, and O = rest
- Therefore, C = 57.8 grams, H = 3.6 grams, and O = ((100 - C - H) grams)
2. Convert the masses of each element to moles using their atomic masses.
- Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol
- Moles of C: 57.8 g / 12.01 g/mol
- Moles of H: 3.6 g / 1.008 g/mol
- Moles of O: ((100 - C - H) g) / 16.00 g/mol
3. Determine the mole ratios between the elements.
- Divide the moles of each element by the smallest number of moles obtained in Step 2.
- This will give you the simplest whole number ratio of the elements.
4. Use the empirical formula mass to calculate the empirical formula.
- Empirical formula mass = (amount of C in grams) + (amount of H in grams) + (amount of O in grams)
- Convert the moles obtained in Step 3 to grams by multiplying them by their respective atomic masses.
- Calculate the empirical formula mass.
5. Find the molecular formula using the empirical formula and the vapor density.
- Molecular formula mass = (vapor density x empirical formula mass) / 22.4
- Divide the molecular formula mass by the empirical formula mass to find the number of empirical formula units in the molecular formula.
Now, let's calculate the values and find the molecular formula.
Given: C = 57.8%, H = 3.6%, and rest O
We have assumed 100 grams of the compound.
1. C = 57.8 grams, H = 3.6 grams, O = (100 - C - H) grams
C = 57.8 grams, H = 3.6 grams, O = 100 - 57.8 - 3.6 = 38.6 grams
2. Moles of C = C / atomic mass(C)
Moles of C = 57.8 g / 12.01 g/mol = 4.81 mol
Moles of H = H / atomic mass(H)
Moles of H = 3.6 g / 1.008 g/mol = 3.57 mol
Moles of O = O / atomic mass(O)
Moles of O = 38.6 g / 16.00 g/mol = 2.41 mol
3. Divide by the smallest number of moles (2.41 mol) to get the simplest ratio:
C: 4.81 mol / 2.41 mol = 2
H: 3.57 mol / 2.41 mol = 1.48
O: 2.41 mol / 2.41 mol = 1
The simplest empirical formula ratio is C2H1.48O1, which can be approximated to C2H3O.
4. Calculate the empirical formula mass:
Empirical formula mass = (amount of C in grams) + (amount of H in grams) + (amount of O in grams)
Empirical formula mass = (2 moles x atomic mass(C)) + (1.48 moles x atomic mass(H)) + (1 mole x atomic mass(O))
Empirical formula mass = (2 x 12.01) + (1.48 x 1.008) + (1 x 16.00)
Empirical formula mass = 24.02 + 1.49 + 16.00
Empirical formula mass = 41.51 g/mol
5. Find the molecular formula using the empirical formula and vapor density:
Molecular formula mass = (vapor density x empirical formula mass) / 22.4
Molecular formula mass = (83 x 41.51) / 22.4
Molecular formula mass = 1537.33 / 22.4
Molecular formula mass = 68.75 g/mol
Divide the molecular formula mass by the empirical formula mass to find the number of empirical formula units in the molecular formula:
Molecular formula mass / Empirical formula mass = 68.75 g/mol / 41.51 g/mol
Molecular formula mass / Empirical formula mass = 1.657
The closest whole number approximation for 1.657 is 2.
Hence, the molecular formula of the compound is (C2H3O) x 2, which can be simplified to C4H6O2.