Calculating Mass of O in CaCO3: 250g = 83g O

[Sace Ver]

Homework Statement

What would be the mass of oxygen in 2.5 x 10²g of CaCO₃?

Given:
mCaCO₃=250g
M=100.09g/mol
n=2.498mol

Homework Equations

?

3. The Attempt at a Solution

mass of O=250g/3 atoms of O
mass of O=83g

Is that correct?
Bc if it is can someone please explain why it's correct.

 

[SteamKing]

Homework Statement

What would be the mass of oxygen in 2.5 x 10²g of CaCO₃?

Given:
mCaCO₃=250g
M=100.09g/mol
n=2.498mol

Homework Equations

?

3. The Attempt at a Solution

mass of O=250g/3 atoms of O
mass of O=83g

Is that correct?
Bc if it is can someone please explain why it's correct.

Remember, 250 g of CaCO₃ contains more than just Oxygen. What about the Calcium and the Carbon in the substance? What happened to them?

 

[Sace Ver]

Remember, 250 g of CaCO₃ contains more than just Oxygen. What about the Calcium and the Carbon in the substance? What happened to them?

So oxygen wouldn't be 83g?

 

[SteamKing]

So oxygen wouldn't be 83g?

Honestly, what do you think?

You can't divide the mass of a substance by the number of atoms of one constituent to obtain the mass of that constituent if other kinds of atoms are present. That should be understood, whether you're taking chemistry or not.

 

[Sace Ver]

Honestly, what do you think?

You can't divide the mass of a substance by the number of atoms of one constituent to obtain the mass of that constituent if other kinds of atoms are present. That should be understood, whether you're taking chemistry or not.

Not to put the blame on others someone showed me how to do it that way which made me confused.

So I'm assuming the final answer is 119.9g of Oxygen.

 

[SteamKing]

Not to put the blame on others someone showed me how to do it that way which made me confused.

So I'm assuming the final answer is 119.9g of Oxygen.

And how did you arrive at that figure?

Please show your work.

 

[Sace Ver]

And how did you arrive at that figure?

Please show your work.

Given:
mCaCO₃=250g
MCaCO₃=100.09g/mol
nCaCO₃=2.498mol

Required:
mO=?

Solution (I tried two methods):

Method #1:

%decimal= 16(3)/100.09g/mol= 0.4796

(0.4796)(250g of CaCO₃)= 119.9g of O

Method #2:

250gCaCO₃/100.09g/mol=2.498mol

2.498mol x 3 mol_O = 7.494mol_O

7.494mol_O x 16g_O = 119.9 g of o

 

[SteamKing]

Given:
mCaCO₃=250g
MCaCO₃=100.09g/mol
nCaCO₃=2.498mol

Required:
mO=?

Solution (I tried two methods):

Method #1:

%decimal= 16(3)/100.09g/mol= 0.4796

(0.4796)(250g of CaCO₃)= 119.9g of O

Method #2:

250gCaCO₃/100.09g/mol=2.498mol

2.498mol x 3 mol_O = 7.494mol_O

7.494mol_O x 16g_O = 119.9 g of o

Method 1:
0.4796 is the fraction of the mass of CaCO₃ which is composed of Oxygen. This fraction remains the same regardless of the how much CaCO₃ you are given.

0.4796 is also called the ratio of the mass of Oxygen to the mass of CaCO₃.

Now, do you see why your original method of calculation in the OP was faulty?

Method 2:
There are 2.498 mol of CaCO₃ in 250 g of the substance. There are 3 atoms (not moles) in each molecule of CaCO₃ .

That's why there are 2.498 mol × 3 atoms of O / molecule = 7.494 moles of Oxygen in 250 g of CaCO₃

 

[Sace Ver]

Method 1:
0.4796 is the fraction of the mass of CaCO₃ which is composed of Oxygen. This fraction remains the same regardless of the how much CaCO₃ you are given.

0.4796 is also called the ratio of the mass of Oxygen to the mass of CaCO₃.

Now, do you see why your original method of calculation in the OP was faulty?

Method 2:
There are 2.498 mol of CaCO₃ in 250 g of the substance. There are 3 atoms (not moles) in each molecule of CaCO₃ .

That's why there are 2.498 mol × 3 atoms of O / molecule = 7.494 moles of Oxygen in 250 g of CaCO₃

I now see why the original calculation was incorrect.