2007-04-09 15:31:39 · 4 answers · asked by kkearney12iu 2 in Science & Mathematics ➔ Chemistry
1 mol CO <=> 28.0101 g CO
7.326 mol CO <=> 205.20 g CO
=>7.326 mol C present
1 mol C <=> 6.022*10^23 atoms of C
=>4.412*10^24 atoms of C present.
Good luck!
2007-04-09 15:39:19 · answer #1 · answered by Sư Ngố 4 · 0⤊ 0⤋
First we need to know how many moles. The molecular mass of CO is 12 + 16 = 28 g /mol.
Divide you mass by 28 and you have moles of CO. This also happens to be the moles of C.
Now multiply by Avogadro's number, 6.02 x10E23 atoms per mole, and you have your answer
2007-04-09 22:35:49 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
use stoichiometry.
C = 12.0111
O= 15.999
add together to get the molar mass (28 g/mol)
then divide/multiply everything out.
use avagadros number to convert to atoms
205.2 g x (1 mol/28g) x (6.02 times 10^23 atoms/1 mol) = 4.41 times 10^24 atoms
2007-04-09 22:49:35 · answer #3 · answered by suggargurl302 2 · 0⤊ 1⤋
the molar weight of CO is 28. Carbon comprises 12/28 or 42.9%, of this mass.
205.20 * 12/28 = 87.943 grams
2007-04-09 22:37:07 · answer #4 · answered by Matthew P 4 · 0⤊ 1⤋