What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen?
-I found this to be C3H3O and WebAssign says it's right. I need help with part b.
If the molar mass of the above compound is 110 grams/mole, what is the molecular formula?
Take a 100 g sample to give you
65.5 g C atoms
5.5 g H atoms
29.0 g O atoms
Convert each to mols. mols = grams/molar mass
C = 65.5/12 = estimated 5.5
H = 5.5/1 = 5.5
O = 29.0/16 = about 1.8
Now find the ratio of the elements to one another with the smallest whole number being 1.00 The easy way to do that is to divide all the numbers by the smallest number. The smallest number is 1.8
So C is 5.5/1.8 = 3.05; round to 3.0
H is 5.5/1.8 = 3.05; round to 3.0
O is 1.8/1.8 = 1.00 = 1.0
empirical formula is C3H3O and the empirical mass is 3*12 + 3*1 + 16 = about 55
mols mass is 110.
empirical formula x ?whole number = 110
?whole number = 110/55 = 2 so tahe molecular formula is (C3H3O)2 or C6H6O2 with a molar mass of 72 + 6 + 32 = 110
To find the empirical formula, we need to determine the ratio of the elements present in the compound. The percentages given represent the mass percentage.
Step 1: Convert the percentages to grams:
Assume we have a 100g sample. In that case,
- Carbon = 65.5g
- Hydrogen = 5.5g
- Oxygen = 29.0g
Step 2: Calculate the number of moles for each element:
- Carbon: Moles of Carbon = (Mass of Carbon / Atomic Mass of Carbon)
= 65.5g / 12.01g/mol = 5.46 moles
- Hydrogen: Moles of Hydrogen = (Mass of Hydrogen / Atomic Mass of Hydrogen)
= 5.5g / 1.01g/mol = 5.44 moles
- Oxygen: Moles of Oxygen = (Mass of Oxygen / Atomic Mass of Oxygen)
= 29.0g / 16.00g/mol = 1.81 moles
Step 3: Divide the number of moles by the smallest number of moles:
- Divide all moles by 1.81 (since it is the smallest)
- Carbon: 5.46 moles / 1.81 moles = 3.02 (rounded to 2 decimal places)
- Hydrogen: 5.44 moles / 1.81 moles = 3.01 (rounded to 2 decimal places)
- Oxygen: 1.81 moles / 1.81 moles = 1.00
Step 4: Write the empirical formula:
The empirical formula is obtained by simplifying the ratio to the nearest whole number, giving us the ratio:
C3H3O.
This matches the empirical formula you found, which is C3H3O.
Now, we can move on to part b to determine the molecular formula.
To find the molecular formula, we need the molar mass of the compound.
Step 5: Find the molar mass of the empirical formula:
The molar mass of C3H3O can be calculated as follows:
C3H3O: (3 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 36.03 g/mol + 3.03 g/mol + 16.00 g/mol
= 55.06 g/mol
Step 6: Calculate the empirical formula mass:
The empirical formula mass is the sum of the atomic masses in the empirical formula. In this case, it is 55.06 g/mol.
Step 7: Calculate the ratio of the molar mass to the empirical formula mass:
Ratio = Molar mass / Empirical formula mass
= 110 g/mol / 55.06 g/mol
= 2 (rounded to the nearest whole number)
This ratio indicates that the molecular formula is two times the empirical formula, which means that:
Molecular formula = 2 * empirical formula
= 2 * C3H3O
= C6H6O2
Therefore, the molecular formula of the compound is C6H6O2.
To determine the empirical formula of a molecule, we need to find the simplest whole number ratio of atoms present in the compound.
To calculate the empirical formula, we first need to assume we have a 100 gram sample of the compound. This allows us to directly convert the given percentages into grams.
Given:
Percentage of Carbon (C) = 65.5%
Percentage of Hydrogen (H) = 5.5%
Percentage of Oxygen (O) = 29.0%
Assuming 100 grams of the compound, the mass of each element present would be as follows:
Mass of Carbon (C) = 65.5 grams
Mass of Hydrogen (H) = 5.5 grams
Mass of Oxygen (O) = 29.0 grams
The next step is to convert the masses of each element into moles. We need to divide the mass of each element by its respective molar mass:
Molar Mass of Carbon (C) = 12.01 g/mol
Molar Mass of Hydrogen (H) = 1.01 g/mol
Molar Mass of Oxygen (O) = 16.00 g/mol
Number of moles of Carbon (C) = Mass of Carbon (C) / Molar Mass of Carbon (C) = 65.5 g / 12.01 g/mol ≈ 5.46 mol
Number of moles of Hydrogen (H) = Mass of Hydrogen (H) / Molar Mass of Hydrogen (H) = 5.5 g / 1.01 g/mol ≈ 5.45 mol
Number of moles of Oxygen (O) = Mass of Oxygen (O) / Molar Mass of Oxygen (O) = 29.0 g / 16.00 g/mol ≈ 1.81 mol
Next, we need to find the ratio of atoms using the moles:
Carbon (C) : Hydrogen (H) : Oxygen (O) ≈ 5.46 mol : 5.45 mol : 1.81 mol
Now, we need to simplify this ratio by dividing all of the numbers by the smallest number of moles, which in this case is 1.81:
Carbon (C) : Hydrogen (H) : Oxygen (O) ≈ 5.46 mol / 1.81 mol : 5.45 mol / 1.81 mol : 1.81 mol / 1.81 mol
Simplified Ratio: C3H3O
Therefore, the empirical formula of the compound is C3H3O.
Now, let's move on to part b:
To determine the molecular formula, we need to know the molar mass of the compound. The molar mass given in the question is 110 grams/mol.
To find the molecular formula, we need to compare the molar mass of the empirical formula (C3H3O) to the molar mass of the compound.
Firstly, we calculate the molar mass of the empirical formula:
Molar Mass of C3H3O = (3 * molar mass of Carbon) + (3 * molar mass of Hydrogen) + (1 * molar mass of Oxygen)
Molar Mass of C3H3O = (3 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 16.00 g/mol) ≈ 42.06 g/mol
Now, we compare the molar mass of the empirical formula (C3H3O) to the molar mass given in the question (110 g/mol):
Molar Mass of the compound (Molecular Formula) / Molar Mass of C3H3O = n (integer)
110 g/mol / 42.06 g/mol = 2.61
Since n should be an integer, we round 2.61 to the nearest whole number, which is 3. Therefore, the molecular formula of the compound is 3 times the empirical formula:
Molecular Formula = (Empirical Formula)3
Molecular Formula = C3H3O3
So, the molecular formula of the compound is C3H3O3.