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### HW Problem E17

Posted: Wed Oct 02, 2019 12:39 pm
In part C of this problem, it asks to compare which has the higher amount of moles. It gives 7.36 x 10^27 atoms of Ru and 7.36 x 10^27 atoms of Fe. I'm confused on how to convert this to moles because the question doesn't give us the mass, only the amount of atoms. The molarity for Ru is 101.07 g/mol and Fe is 55.85 g/mol.

### Re: HW Problem E17

Posted: Wed Oct 02, 2019 12:41 pm
Nevermind, I think I just divide the atoms by Avagrados number to convert it to moles.

### Re: HW Problem E17

Posted: Wed Oct 02, 2019 12:45 pm
Yes, you can convert the atoms to moles by dividing by Avogadro's number which is 6.022x10^23. Knowing that 1 mol Ru has 6.022x10^23 atoms of Ru, you can divide 7.36x10^27 by 6.022x10^23 to get the moles of Ru which is 1.22x10^4. Doing the same fo 7.36x10^27 atoms of Fe would show you that there are also 1.22x10^4 mol Fe. Thus, both samples contain the same number of moles.

### Re: HW Problem E17

Posted: Wed Oct 02, 2019 4:54 pm
Number of moles is directly proportional to number of atoms, because Avogadro's number is a constant. Number of atoms has no relation to atomic mass.

### Re: HW Problem E17

Posted: Wed Oct 02, 2019 5:09 pm
There are the same number of atoms, 6.022 * 10^23, in a mole of any element. Therefore, seeing that Ru and Fe are present in the same number of atoms, they should be present in the same number of moles as well.

### Re: HW Problem E17

Posted: Wed Oct 02, 2019 5:15 pm
Also, by the way, in the last sentence, instead of "molarity" I think it should be "molar mass". Molarity refers to the concentration of a solution in mol/L. No big deal but just something to look out for!