A 11.0g sample of a compound containing only oxygen and carbon produce 3.0g carbon and 8.0g oxygen. Determine the empirical formula of the compound.
mols C = 3.0/12 = ?
mols O = 8.0/16 = ?
Find the ratio of the two to each other with the smallest number being 1.00. The easy way to do that is to divide both numbers by the smaller of the two. That will give you whole numbers with the smallest being 1.0
A 11.0 g sample of a compound containing only oxygen and carbon produce 3.0 g carbon and 8.0 g oxygen. Determine the empirical formula for the compound.
the empirical fomula of carbon and oxygen.
1 : 2
there are one carbon and two oxygen
Simple and easy
Well, this compound seems to have experienced quite a weight loss program! Let's see if we can figure out its empirical formula.
To determine the empirical formula, we need to find the ratio of atoms in the compound. We have 3.0g of carbon and 8.0g of oxygen.
Step 1: Convert grams to moles
Carbon has a molar mass of approximately 12.01 g/mol, so we have 3.0g ÷ 12.01 g/mol = 0.25 mol of carbon.
Oxygen has a molar mass of approximately 16.00 g/mol, so we have 8.0g ÷ 16.00 g/mol = 0.5 mol of oxygen.
Step 2: Find the simplified mole ratio
Divide the number of moles of each element by the smallest number of moles (0.25 mol in this case):
Carbon: 0.25 mol ÷ 0.25 mol = 1
Oxygen: 0.5 mol ÷ 0.25 mol = 2
The empirical formula is therefore C1O2. So, it looks like the compound is CO2, which happens to be carbon dioxide!
To determine the empirical formula of a compound, we need to know the mass percent composition of each element.
In this case, we are given the masses of carbon and oxygen produced, but we need to determine the mass of carbon and oxygen in the original compound.
First, let's find the mass of carbon in the original compound.
Mass of carbon = 3.0g
Next, let's find the mass of oxygen in the original compound.
Mass of oxygen = 8.0g
Now, let's calculate the mass percent composition of carbon and oxygen in the compound.
Mass percent of carbon = (mass of carbon / mass of compound) x 100%
= (3.0g / 11.0g) x 100%
≈ 27.3%
Mass percent of oxygen = (mass of oxygen / mass of compound) x 100%
= (8.0g / 11.0g) x 100%
≈ 72.7%
The next step is to convert the mass percent composition to moles. Divide the mass percent of each element by the atomic mass of that element.
The atomic mass of carbon (C) is 12.01 g/mol, and the atomic mass of oxygen (O) is 16.00 g/mol.
Moles of carbon = (Mass percent of carbon / Atomic mass of carbon)
= 27.3% / 12.01 g/mol
≈ 2.27 mol
Moles of oxygen = (Mass percent of oxygen / Atomic mass of oxygen)
= 72.7% / 16.00 g/mol
≈ 4.54 mol
Now, we need to find the simplest whole-number ratio of carbon to oxygen.
Divide the number of moles of each element by the smaller number of moles. Round to the nearest whole number.
Carbon (C): 2.27 mol / 2.27 mol = 1
Oxygen (O): 4.54 mol / 2.27 mol = 2
The empirical formula of the compound is CO2, indicating that there is one carbon atom and two oxygen atoms in the compound.