Calculate the mass in grams of 8.35 *10^22 molecules of CBr4
2007-05-04 09:09:52 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
8.35*10^22 molecules* mole / 6.023*10^23 molecules
= 0.1386 moles
CBr4 Mw = 331.6 g/mole
0.1386 mole * 331.6 g/mole = 45.9 g
2007-05-04 09:17:13 · answer #1 · answered by Dr Dave P 7 · 1⤊ 0⤋
Divide by Avogadro's number and multiply by the Mr of CBr4.
2007-05-04 09:14:58 · answer #2 · answered by Gervald F 7 · 1⤊ 0⤋
The molar mass of carbon tetrabromide is 331.63 g/mole
One mole is 6.023x10^23 molecules.
You have 8.35 x 10^22 molecules, thus you have:
(8.35x10^22)/(6.023x10^23) = 0.1386 moles
0.1386 moles * 331.63 g/mole = 45.98 grams
2007-05-04 09:16:52 · answer #3 · answered by Dave_Stark 7 · 0⤊ 0⤋
(28.0g / mole) x (1 mole / 6.022x10^23 molecules) = 4.65x10^-23g and if want that in AMU... 28.0g / mole of molecules = 28.0 AMU / molecule
2016-05-20 07:46:59 · answer #4 · answered by maranda 3 · 0⤊ 0⤋
1 mole = 6.02 x 10^23 molecules
8.35 x 10^22 / 6.02 x 10^23 = 0.139 moles
Molecular mass = 331.6 g/mol
0.139 mol x 331.6 = 46.1 g
2007-05-04 09:16:11 · answer #5 · answered by Anonymous · 1⤊ 0⤋
Add the atomic weights of carbon and 4xbromine. Divide by Avogadro's number. Take the inverse.
2007-05-04 09:17:15 · answer #6 · answered by Chad H 3 · 0⤊ 1⤋