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I need help with Mole


Fastguy

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Ok, I understand that mole is 6.02x 10^23. My teacher gave me this question

 

Moles of Sodium Barcarbonate in a teaspoon

 

Mass of Teaspoon: 10.02g

 

Mass of Teaspoon of Sodium Bicarbonate: 20.19g

 

Sodium Bicarbonate: NaHCO3

 

what is the mole, molecules, and oxygen atoms. Could someone please explain to me how to do this.I really don't understand :(

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well 1 mole is the number you stated, molecules of the substance in this case NaHCO3, so you need to work out the mass of one molecule, the mass of all the molecules of that in the teaspoon and then from that you have too masses, total mass of molecules and mass of one mole. So if we use a bit of an annolgy of apples in a box, where 1 apple is 1 mole, some apples in a box, the mass of the apples is 10Kg, 1 apple is 1Kg, how many apples are there...

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There are 10 apples in that box. Did I get it right?

 

Ok from what you said I need to find the mass of one molecule, the total mass, and then the mass of one mole. Well I if I subtract the two numbers given I get the total mass of just the Sodium Bicarbonate alone

 

20.19g-10.02g= 10.17g

 

I'm guessing that from that number, I need to divide something from it to get the mass of one molecule, right? How do I do that tho? Also, how do you find the mass of mole. I thought mole (itself) was just a number or unit. Could you be possibly be talking about moles of SB in a teaspoon. I'm sorry if I'm being confusing, I'm just trying to understand :-(

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  • 2 weeks later...

hello

you r too fast in order to understand mole

u pick the wrong concept

mole is actually a molecular mass or atomic mass of a molecule or an atom expressed in grams

and 6.02x10^23 is the no. of molecules or no. of atoms per mole

 

for mole u can find out from the equation of disociation of NaHCO3 it would be three moles of oxygen now no of moles=mass/mol mass

 

no of molecules = Avogadros no.(6.02x10^3)x mass of substance/molecular mass

 

no of atoms = Avogadro's n0x mass of sub/atomic mass

 

simple now i think your problem is solved u can ask more freely:-)

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