molecular weight is 284.5 g/mol what is the molecular formula.76.0% C, 12.8% H, 11,2%O.
Take a 100 g sample. That gives you
76.0g C
12.8 g H
11.2 g O
Convert to mols. mol = grams/atomic mass.
Then find the ratio; the easy way to do that is to divide the smallest number by itself then divide all of the other numbers by the same small number. That will give you the empirical formula. Now you need an empirical formula mass.
(284.5/empirical formula) = n. Round n to a whole number and the molar formula is (empirical formula)n
Post your work if you get stuck.
To determine the molecular formula, we need to calculate the empirical formula first. The empirical formula gives the simplest whole number ratio of the elements present in a compound.
First, we assume we have 100g of the compound. Therefore, in 100g of the compound, we have:
- Carbon (C): 76.0g
- Hydrogen (H): 12.8g
- Oxygen (O): 11.2g
Next, we need to convert the masses of each element to moles. We can do this by dividing the mass of each element by its molar mass.
The molar masses are as follows:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
Converting the masses to moles:
- Moles of Carbon (C) = 76.0 g / 12.01 g/mol ≈ 6.33 mol
- Moles of Hydrogen (H) = 12.8 g / 1.01 g/mol ≈ 12.67 mol
- Moles of Oxygen (O) = 11.2 g / 16.00 g/mol ≈ 0.70 mol
Next, we need to determine the empirical formula. To do this, we divide the number of moles of each element by the smallest number of moles.
- Divide the number of moles of Carbon by 0.70: 6.33 mol / 0.70 mol ≈ 9.04
- Divide the number of moles of Hydrogen by 0.70: 12.67 mol / 0.70 mol ≈ 18.10
- Divide the number of moles of Oxygen by 0.70: 0.70 mol / 0.70 mol = 1.00
Now, we have the molar ratios of the elements: C9.04H18.10O1.00
Since we want a whole number ratio, we should round the subscripts to the nearest whole number: C9H18O.
Therefore, the empirical formula is C9H18O.
To determine the molecular formula, we need to calculate the ratio of the molar mass of the molecular formula to the molar mass of the empirical formula.
Molar mass of the empirical formula (C9H18O):
= (9 * atomic mass of C) + (18 * atomic mass of H) + (1 * atomic mass of O)
= (9 * 12.01 g/mol) + (18 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 108.09 g/mol + 18.18 g/mol + 16.00 g/mol
= 142.27 g/mol
Next, divide the given molecular weight by the molar mass of the empirical formula:
Molecular weight / Molar mass of empirical formula = 284.5 g/mol / 142.27 g/mol
≈ 2
This means that the molecular formula is twice the empirical formula: C9H18O * 2 = C18H36O2.
Therefore, the molecular formula for the compound is C18H36O2.
To find the molecular formula, we need to determine the empirical formula first, which represents the simplest whole-number ratio of atoms in a compound. To do this, we need to find the subscripts of each element in the empirical formula using the given percentages.
Given:
- Carbon (C) = 76.0%
- Hydrogen (H) = 12.8%
- Oxygen (O) = 11.2%
1. Assume you have 100 grams of the compound, so you can convert the percentages into grams:
- Carbon (C) = 76.0 grams
- Hydrogen (H) = 12.8 grams
- Oxygen (O) = 11.2 grams
2. Convert the grams to moles for each element using their respective atomic weights:
- Carbon (C) atomic weight = 12.01 g/mol
- Hydrogen (H) atomic weight = 1.008 g/mol
- Oxygen (O) atomic weight = 16.00 g/mol
- Moles of Carbon (C) = 76.0 grams / 12.01 g/mol = 6.33 moles
- Moles of Hydrogen (H) = 12.8 grams / 1.008 g/mol = 12.7 moles
- Moles of Oxygen (O) = 11.2 grams / 16.00 g/mol = 0.7 moles
3. Divide each number of moles by the smallest number of moles to find the simplest whole-number ratio:
- Simplest ratio of moles: C:6.33, H:12.7, O:0.7
(dividing all by 0.7 gives)
- Simplest ratio of moles: C:9, H:18, O:1
So, the empirical formula is C9H18O1.
Now, to find the molecular formula, we need the molecular weight of the compound.
Given:
- Molecular weight = 284.5 g/mol
4. Calculate the empirical formula's molar mass:
- Molar mass of C9H18O1 = (9 * 12.01 g/mol) + (18 * 1.008 g/mol) + (1 * 16.00 g/mol) = 144.23 g/mol
5. Divide the molecular weight by the empirical formula's molar mass to find the whole-number multiple:
- Molecular weight / Empirical formula molar mass = 284.5 g/mol / 144.23 g/mol ≈ 1.97
6. Round this number to the nearest whole number multiple:
- 1.97 ≈ 2
7. Multiply the subscripts in the empirical formula by the whole-number multiple obtained:
- Molecular formula = (C9H18O1) * 2 = C18H36O2
Therefore, the molecular formula for this compound is C18H36O2.