Find the molecular formula of a compound that is 48.38% carbon, 8.12% hydrogen, and 53.5% oxygen by mass and has a molecular weight of 328 g/mol.
To find the molecular formula of a compound, you need to know the percent composition of each element by mass and the molecular weight of the compound. Here's how you can calculate the molecular formula:
1. Convert the percent composition of each element to grams:
- Carbon (C): 48.38% of 328g = 158.9g
- Hydrogen (H): 8.12% of 328g = 26.7g
- Oxygen (O): 53.5% of 328g = 175.5g
2. Convert the mass of each element to moles by dividing by their respective atomic masses:
- Carbon (C): 158.9g / atomic mass of carbon = moles of carbon
- Hydrogen (H): 26.7g / atomic mass of hydrogen = moles of hydrogen
- Oxygen (O): 175.5g / atomic mass of oxygen = moles of oxygen
3. Find the simplest mole ratio of the elements by dividing each element's moles by the smallest number of moles calculated in step 2.
4. Round the mole ratios to the nearest whole number. These whole number ratios represent the subscripts in the molecular formula.
5. Write the molecular formula using the subscripts found in step 4.
Given that we have calculated the moles of each element, we can assume the atomic masses as follows:
- Carbon (C) has an atomic mass of 12.01 g/mol
- Hydrogen (H) has an atomic mass of 1.01 g/mol
- Oxygen (O) has an atomic mass of 16.00 g/mol
Let's calculate the moles of each element:
Moles of Carbon (C) = 158.9g / 12.01 g/mol = 13.23 mol
Moles of Hydrogen (H) = 26.7g / 1.01 g/mol = 26.4 mol
Moles of Oxygen (O) = 175.5g / 16.00 g/mol = 10.97 mol
Now, let's find the simplest mole ratio:
Dividing each element's moles by the smallest number of moles (10.97 mol) gives us:
Carbon: 13.23 mol / 10.97 mol = 1.203 (rounded to 1)
Hydrogen: 26.4 mol / 10.97 mol = 2.404 (rounded to 2)
Oxygen: 10.97 mol / 10.97 mol = 1
Therefore, the empirical formula is CH2O (where the subscripts represent the mole ratios).
To determine the molecular formula, you need to compare the empirical formula's formula mass to the given molecular weight (328 g/mol).
Calculate the formula mass (molar mass) of the empirical formula:
- Carbon (C): 1 * 12.01 = 12.01 g/mol
- Hydrogen (H): 2 * 1.01 = 2.02 g/mol
- Oxygen (O): 1 * 16.00 = 16.00 g/mol
Formula mass = 12.01 g/mol + 2.02 g/mol + 16.00 g/mol = 30.03 g/mol
Divide the given molecular weight (328 g/mol) by the formula mass (30.03 g/mol):
328 g/mol / 30.03 g/mol = 10.93
Since the molecular weight is approximately 10.93 times the empirical formula mass, we can multiply the subscripts in the empirical formula by this factor to find the molecular formula:
Molecular formula = (C1H2O1) * 10.93 = C10.93H21.86O10.93
Rounding the subscripts to the nearest whole number, the molecular formula is C11H22O11.
To find the molecular formula of the compound, we need to determine the empirical formula first. The empirical formula represents the simplest whole-number ratio of atoms present in a compound.
Step 1: Calculate the mass of each element in the compound.
- Carbon (C) = 48.38% of 328 g/mol = 158.69 g/mol
- Hydrogen (H) = 8.12% of 328 g/mol = 26.70 g/mol
- Oxygen (O) = 53.5% of 328 g/mol = 175.4 g/mol
Step 2: Determine the number of moles for each element.
- Moles of C = Mass of C / Molar mass of C = 158.69 g / 12.01 g/mol = 13.21 mol
- Moles of H = Mass of H / Molar mass of H = 26.70 g / 1.01 g/mol = 26.44 mol
- Moles of O = Mass of O / Molar mass of O = 175.4 g / 16.00 g/mol = 10.96 mol
Step 3: Divide the number of moles of each element by the smallest number of moles obtained.
- Divide all moles by 10.96 since it is the smallest value.
- Moles of C = 13.21 mol / 10.96 mol = 1.203 ≈ 1
- Moles of H = 26.44 mol / 10.96 mol = 2.408 ≈ 2
- Moles of O = 10.96 mol / 10.96 mol = 1
The empirical formula is therefore CH2O.
Now, to find the molecular formula, we need to know the molecular weight of the empirical formula. The molecular weight of CH2O is:
- (C: 12.01 g/mol × 1) + (H: 1.01 g/mol × 2) + (O: 16.00 g/mol × 1) = 30.03 g/mol
Finally, divide the molecular weight of the compound by the molecular weight of the empirical formula:
- Molecular weight of compound / Molecular weight of empirical formula
- 328 g/mol ÷ 30.03 g/mol = 10.92
Since we obtained a value close to 11, the molecular formula is approximately 11 times the empirical formula.
Therefore, the molecular formula is C11H22O11.
Take a 100 g sample to give you
48.38 g C
8.12 g H
53.5 g O.
Convert grams to mols. mols = grams/molar mass
uh oh! Percent adds up to more than 100% (110% to be exact about it). You must have made a typo or the problem is printed wrong. My best educated guess is that 53.5% O should be 43.5% O.