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Chemistry Questiom! How do you convert atoms into moles/grams?

 Feb 20, 2015

Best Answer 

 #1
avatar+109 
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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:

 

$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$

 

Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:

 

$$\left({\mathtt{4.87}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{22}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right)}}\right) = {\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}$$

 

To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:

 

$${\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{gC}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}}\right) = {\mathtt{9.077}}{\mathtt{\,\times\,}}{\mathtt{gC}}$$

 Feb 20, 2015
 #1
avatar+109 
+5
Best Answer

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:

 

$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$

 

Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:

 

$$\left({\mathtt{4.87}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{22}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right)}}\right) = {\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}$$

 

To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:

 

$${\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{gC}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}}\right) = {\mathtt{9.077}}{\mathtt{\,\times\,}}{\mathtt{gC}}$$

Sorasyn Feb 20, 2015

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