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Chem Question

ManBearPig

Diamond Member
If 1.14 moles of a make beleive element weighs 354.98 g, then what would be the mass of a single atom in grams? Its atomic mass is 311.3852 amu.

thanks
 
multiply Avogadro's Number by 1.14 (to find out the total amount of atoms in 1.14 moles) and then divide 354.98g by that number.

edit: or, yes, you could find out how many grams are in 1 amu and go from there.
 
1.14 moles of stuff * (6.023 x 10^23 atoms/mole) = about 6.86 x 10^23 atoms

354.98grams / 6.86 x 10^23 atoms = 5.16 x 10^-22 g/atom
 
Originally posted by: Elderly Newt
multiply Avogadro's Number by 1.14 (to find out the total amount of atoms in 1.14 moles) and then divide 354.98g by that number.

edit: or, yes, you could find out how many grams are in 1 amu and go from there.


i tried that and got 1.8751x10 to the 26 and it didnt work 🙁
 
Originally posted by: Heen05
Originally posted by: Elderly Newt
multiply Avogadro's Number by 1.14 (to find out the total amount of atoms in 1.14 moles) and then divide 354.98g by that number.

edit: or, yes, you could find out how many grams are in 1 amu and go from there.


i tried that and got 1.8751x10 to the 26 and it didnt work 🙁

1 atom weighs 1.8751 *10^26 grams????? WOW, that's one HEAVY atom!!
 
Originally posted by: Heen05
If 1.14 moles of a make beleive element weighs 354.98 g, then what would be the mass of a single atom in grams? Its atomic mass is 311.3852 amu.

thanks

Rather than screwing around with Avogadro's number, why not simply convert from amu's to grams?? Especially since:
A) 3 significant digits in 1.14 moles
B) 5 significant digits in 354.98 grams
C) None of you posted Avogadro's number beyond 4 significant digits

D) 311.3854amu is 7 significant digits. Use that number in the calculation along with 1 amu = 1.66053886*10^-24 grams. That way, you have an answer with the most significant digits.

Also, here's a cool Google trick (Everyone click and learn)
try this (note what the search term is)
 
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