An organic compound x contains 40% carbon, 6.7% hydrogen the third compound oxygen. If the relative molecular mass of the compound is 60, calculate (i) the empirical formular of x (ii) the molecular formular of x
Take a 100 g sample which give you
40 g C
6.7 g H
100-40-6.7 = 53.3 g O
Convert these g to mols. mol = g/atomic mass.
40/12 = about 3.3 mols C
6.7/1 = 6.7 mols H
53.3/16 = about 3.3
Now find the ratio of these three elements to each other. The easy way to do that is to divide the smallest number by itself, then divide the other numbers by the same small number; i.e.,
C = 3.3/3.3 = 1.0
H = 6.7/3.3 = 1.988 whch rounds to 2.0
O = 3.3/3.3 = 1.0
NOTE. If you do those mols more accurately it comes out a little better. So the empirical formula is
C1H2O for an empirical mass of 1*12 + 2*1 + 1*16 = about 30
empirical mass x number = molar mass = 60
30*number = 60
number = 2
Therefore, the molecular formula is
(CH2O)2 or C2H4O2
Source very helpful
To determine the empirical formula and molecular formula of compound X, we need to follow a step-by-step process.
(i) Calculating the empirical formula:
1. Determine the mass of each element:
- Carbon (C): 40% of the total mass of compound x
- Hydrogen (H): 6.7% of the total mass of compound x
- Oxygen (O): Subtract the total mass of C and H from the relative molecular mass of compound x (100% - (40% + 6.7%))
2. Convert the percentage values to grams:
- Mass of C = 40% of total mass of x = 40g
- Mass of H = 6.7% of total mass of x = 6.7g
- Mass of O = (100% - (40% + 6.7%)) of total mass of x = 53.3g
3. Calculate the moles of each element using their respective molar masses (g/mol):
- Molar mass of C = 12 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of O = 16 g/mol
- Moles of C = Mass of C / Molar mass of C = 40g / 12 g/mol = 3.33 mol
- Moles of H = Mass of H / Molar mass of H = 6.7g / 1 g/mol = 6.7 mol
- Moles of O = Mass of O / Molar mass of O = 53.3g / 16 g/mol = 3.33 mol
4. Determine the simplest whole number ratio:
- Divide the number of moles of each element by the smallest number of moles (3.33 mol) to obtain the simplest whole number ratio:
- C = 3.33 mol / 3.33 mol = 1
- H = 6.7 mol / 3.33 mol ≈ 2
- O = 3.33 mol / 3.33 mol = 1
5. Write the empirical formula using the determined ratios:
- Empirical formula of compound X: CH2O
(ii) Calculating the molecular formula:
1. Determine the empirical formula mass:
- Empirical formula mass = (Mass of C atom x number of C atoms) + (Mass of H atom x number of H atoms) + (Mass of O atom x number of O atoms)
- Empirical formula mass = (12 g/mol x 1) + (1 g/mol x 2) + (16 g/mol x 1) = 30 g/mol
2. Determine the ratio between the relative molecular mass and the empirical formula mass:
- Ratio = Relative molecular mass / Empirical formula mass
- Ratio = 60 g/mol / 30 g/mol = 2
3. Multiply the subscripts in the empirical formula by the ratio obtained:
- Molecular formula of compound X = (C1H2O1) x 2
- Molecular formula of compound X = C2H4O2
Therefore, the (i) empirical formula of compound X is CH2O, and the (ii) molecular formula of compound X is C2H4O2.
To calculate the empirical formula of compound X, we need to determine the simplest ratio of atoms present in the compound.
(i) Calculating the empirical formula:
1. Start by assuming we have a 100g sample of compound X. This assumption will make the percent composition of the compound equal to the mass of each element in grams.
- Carbon: 40% of 100g = 40g
- Hydrogen: 6.7% of 100g = 6.7g
- Oxygen: The remaining mass after subtracting the masses of carbon and hydrogen from the total mass:
= Total mass - (Mass of carbon + Mass of hydrogen)
= 100g - (40g + 6.7g)
= 53.3g
2. Convert the masses of each element to moles using their atomic masses from the periodic table.
- Carbon: Atomic mass of carbon (C) = 12 g/mol
Moles of carbon = Mass of carbon / Atomic mass of carbon = 40g / 12 g/mol = 3.33 mol
- Hydrogen: Atomic mass of hydrogen (H) = 1 g/mol
Moles of hydrogen = Mass of hydrogen / Atomic mass of hydrogen = 6.7g / 1 g/mol = 6.7 mol
- Oxygen: Atomic mass of oxygen (O) = 16 g/mol
Moles of oxygen = Mass of oxygen / Atomic mass of oxygen = 53.3g / 16 g/mol = 3.33 mol
3. Divide each element's moles by the smallest number of moles found in step 2. This step ensures we have the simplest ratio of atoms.
- Carbon: 3.33 mol / 3.33 mol = 1
- Hydrogen: 6.7 mol / 3.33 mol = 2
- Oxygen: 3.33 mol / 3.33 mol = 1
Therefore, the empirical formula of compound X is CH2O.
(ii) To find the molecular formula of compound X, we need additional information about the relative molecular mass (Mr).
1. Calculate the empirical formula mass:
- Carbon: Atomic mass of carbon (C) = 12 g/mol
Moles of carbon (from empirical formula) = 1
Mass of carbon in empirical formula = Atomic mass of carbon * Moles of carbon = 12 g/mol * 1 = 12 g/mol
- Hydrogen: Atomic mass of hydrogen (H) = 1 g/mol
Moles of hydrogen (from empirical formula) = 2
Mass of hydrogen in empirical formula = Atomic mass of hydrogen * Moles of hydrogen = 1 g/mol * 2 = 2 g/mol
- Oxygen: Atomic mass of oxygen (O) = 16 g/mol
Moles of oxygen (from empirical formula) = 1
Mass of oxygen in empirical formula = Atomic mass of oxygen * Moles of oxygen = 16 g/mol * 1 = 16 g/mol
Empirical formula mass = Mass of carbon in empirical formula + Mass of hydrogen in empirical formula + Mass of oxygen in empirical formula
= 12 g/mol + 2 g/mol + 16 g/mol
= 30 g/mol
2. Divide the relative molecular mass (Mr) by the empirical formula mass to find the whole number ratio, "n."
- Relative molecular mass (Mr) = 60 g/mol
- n = Mr / Empirical formula mass = 60 g/mol / 30 g/mol = 2
3. Multiply the subscripts in the empirical formula by "n" to get the molecular formula.
Molecular formula = Empirical formula * n
CH2O * 2 = C2H4O2
Therefore, the molecular formula of compound X is C2H4O2.