I posted this response to someone else who was converting 4.87*10^22 atoms of Carbon to Grams. It should serve as a good example of how to convert between the three. Note that all units being converted are cancelled out.
Here's the link to the question: http://web2.0calc.com/questions/how-much-carbon-is-in-4-87-10-22
Assuming that you're speaking in units of atoms, you can convert them like so:
Grams⇔Mols⇔Atoms
Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:
(4.87×1022×atomsC)×((1×molC)(6.022×1023×atomsC))=0.8087×molsC
To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:
0.8087×molsC×((12×gC)(1×molC))=9.077×gC
I posted this response to someone else who was converting 4.87*10^22 atoms of Carbon to Grams. It should serve as a good example of how to convert between the three. Note that all units being converted are cancelled out.
Here's the link to the question: http://web2.0calc.com/questions/how-much-carbon-is-in-4-87-10-22
Assuming that you're speaking in units of atoms, you can convert them like so:
Grams⇔Mols⇔Atoms
Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:
(4.87×1022×atomsC)×((1×molC)(6.022×1023×atomsC))=0.8087×molsC
To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:
0.8087×molsC×((12×gC)(1×molC))=9.077×gC