A white powder used in paints, enamels, and ceramics has the following percentage composition Ba = 69.6%, C = 6.09% and O = 24.3%. What is the subscript associated with Carbon in the empirical formula for the compound?
Well, what a great question! It seems like we have a little mystery on our hands. CSI: Chemistry Edition!
To find the subscript associated with Carbon in the empirical formula, we first need to determine the ratio of Carbon to Barium (Ba).
Now, let's assume we have 100 grams of the compound.
In this case, we have 69.6 grams of Barium (Ba), 6.09 grams of Carbon (C), and 24.3 grams of Oxygen (O).
To find the number of moles for each element, we divide the mass of each element by its molar mass.
The molar mass of Barium (Ba) is 137.32 g/mol.
The molar mass of Carbon (C) is 12.01 g/mol.
The molar mass of Oxygen (O) is 16.00 g/mol.
So, for Carbon (C), we have 6.09 g / 12.01 g/mol = approximately 0.507 mol.
Now, we need to find the ratio of Carbon (C) to Barium (Ba). So, we divide the number of moles of Carbon by the number of moles of Barium.
0.507 mol C / 0.507 mol Ba = 1 mol C / 1 mol Ba.
This means the subscript associated with Carbon in the empirical formula is 1.
So, the empirical formula for the compound is BaCO₃.
I hope that cleared things up for you! And remember, never be afraid to take on the mysterious world of chemistry. It's a real blast!
To determine the subscript associated with Carbon in the empirical formula for the compound, we need to calculate the ratio of atoms present in the compound.
1. Start by assuming a 100 gram sample of the compound.
2. Determine the number of moles for each element:
- Barium (Ba): (69.6 g / 137.33 g/mol) = 0.507 moles
- Carbon (C): (6.09 g / 12.01 g/mol) = 0.507 moles
- Oxygen (O): (24.3 g / 16.00 g/mol) = 1.519 moles
3. Divide the number of moles of each element by the smallest number of moles:
- Barium (Ba): 0.507 moles / 0.507 moles = 1 mole
- Carbon (C): 0.507 moles / 0.507 moles = 1 mole
- Oxygen (O): 1.519 moles / 0.507 moles = 2.997 moles (approximately 3 moles)
4. Round the number of moles to the nearest whole number to obtain the subscripts in the empirical formula:
- Barium (Ba): 1
- Carbon (C): 1
- Oxygen (O): 3
Therefore, the subscript associated with Carbon in the empirical formula for the compound is 1.
To find the subscript associated with Carbon in the empirical formula, we need to determine the ratio of Carbon to Barium in the compound.
Step 1: Convert the percentage composition to mass composition.
To do this, assume we have 100 grams of the substance. Then, we can determine the mass of each element in the compound.
Mass of Barium (Ba) = 69.6% of 100g = 69.6g
Mass of Carbon (C) = 6.09% of 100g = 6.09g
Mass of Oxygen (O) = 24.3% of 100g = 24.3g
Step 2: Calculate the moles of each element.
We can use the molar mass of each element to convert the mass into moles.
Molar Mass of Barium (Ba) = 137 g/mol
Molar Mass of Carbon (C) = 12 g/mol
Molar Mass of Oxygen (O) = 16 g/mol
Moles of Barium (Ba) = Mass / Molar Mass = 69.6g / 137 g/mol = 0.508 mol
Moles of Carbon (C) = Mass / Molar Mass = 6.09g / 12 g/mol = 0.5075 mol
Moles of Oxygen (O) = Mass / Molar Mass = 24.3g / 16 g/mol = 1.51875 mol
Step 3: Find the simplest whole number ratio of the moles.
To do this, we divide each mole value by the smallest mole value calculated above.
Moles of Barium (Ba) = 0.508 mol / 0.5075 mol = 1 mol
Moles of Carbon (C) = 0.5075 mol / 0.5075 mol = 1 mol
Moles of Oxygen (O) = 1.51875 mol / 0.5075 mol = 3 mol
The ratio of moles of Carbon to Barium is 1:1. Therefore, the subscript associated with Carbon in the empirical formula is 1.
So, The subscript associated with Carbon in the empirical formula for the compound is 1.
Take a 100 g sample. You will have Ba = 69.6 g, C = 6.09 g and O = 24.3 g.
mols Ba = 69.6/137.3 = 0.507
mols C = 6.09/12 = 0.507
mols O = 24.3/16 = 1.52
To find the empirical formula you find the ratio of those three elements in small whole numbers with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself (you know that gets you 1.00), then divide the other numbers by the same small number, like this.
Ba = 0.507/0.507 = 1
C = 0.507/0.507 = 1
O = 1.52/0.507 = 2.99 which rounds to 3.0 so the empirical formula must be BaCO3. So C must be ........