Hon Chemistry 10-19-09 Moles, Molar Mass & Avogadro’s Number
HON CHEMISTRY: Hey guys, here’s the lecture from Monday on moles, molar mass, and Avogadro’s number. Remember, it’s all in terms of one mole. The mass of one mole is the average atomic mass of the element in grams (periodic table). And the number of atoms in one mole is …. well you know that. Speaking of moles, isn’t he cute? Well, beauty being in the eye of the beholder and all. Hmmmm………
Image source blog.ibts.eu/wp-content/uploads/2009/02/mole2.jpg
October 20th, 2009 at 5:04 PM
isnt molar mass the same as the average atomic mass?
[Reply]
Ms. Skinner Reply:
October 20th, 2009 at 9:13 PM
RP – Yep, the only diff is the unit. AAM is is amu or u and molar mass is in grams.
[Reply]
October 20th, 2009 at 7:21 PM
im confused on what an atomic mass unit is… does anyone know?
And is there a help session in the moring tomorrow?
[Reply]
Ms. Skinner Reply:
October 20th, 2009 at 9:13 PM
Drew – help session tomorrow morning at 7:15 A.M.
[Reply]
Ms. Skinner Reply:
October 20th, 2009 at 9:16 PM
Drew – AAM is just the average mass of all the isotopes of a particular element. It’s the mass that you find on the periodic table. The unit is amu or u. You can look it up or you can calculate it yourself with the formula you have in your notes. Good luck!
[Reply]
October 20th, 2009 at 8:18 PM
Drew, I’m pretty sure there is a help session. I will be there.
[Reply]
October 20th, 2009 at 8:48 PM
What is the formula to find mass when given the number of atoms?
[Reply]
Ms. Skinner Reply:
October 20th, 2009 at 9:12 PM
CF – Go from atoms to moles to mass. Remember you have just two conversion factors: the number of atoms in one mole and the mass of one mole. Good luck!
[Reply]
October 21st, 2009 at 9:12 AM
i’m a bit perplexed… how can avogadro’s number always be true and exact if it’s based on the average atomic mass? how do we know the average is exactly accurate? it’s impossible to measure EXACTLY how much of each isotope is in the ENTIRE earth. so, how can we know that number is accurate?
[Reply]