2.8-2.9 Atomic Mass & Molar Mass
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2.8-2.9 Atomic Mass & Molar Mass

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2.8-2.9 Atomic Mass & Molar Mass




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Presentation on theme: "2.8-2.9 Atomic Mass & Molar Mass"— Presentation transcript:

Slide1

2.8-2.9 Atomic Mass & Molar Mass

Slide2

Atomic Mass

What is an elements atomic mass?

The weighted average of its naturally occurring isotopes

.

When looking at an element on the periodic table of elements, where is atomic mass found?

Usually the top left hand corner of each elements box. However, on the periodic table that you will use for your AP Exam(s), this number will be located below each elements chemical symbol.

How is it calculated? Show all work for the following problem:

Magnesium has three naturally occurring isotopes with masses 23.99

amu

, 24.99

amu

and 25.98

amu

and natural abundances of 78.99%, 10.00% and 11.01% respectively. Calculate the atomic mass of magnesium.

Mg

iso1

+ Mg

iso2

+ Mg

iso3

= (23.99 X .7899) + (24.99 X .1000) + (25.98 X .1101) =

Atomic mass of Mg = 24.31

amu

Slide3

Mass Spectrometry

A technique that separates particles according to their mass.

Can be used to identify atoms that make up a molecule, and the percent abundances of isotopes of an element.

After a sample is ionized, and the ions travel through the mass spectrometer, they strike a detector that puts out signals, and ultimately produces a mass spectrum of the sample to be analyzed.

Slide4

Mass Spectrum of Boron

Boron has an atomic mass of around 10.8

amu

If you analyze this spectrum you can see that there are two lines, the larger at 11

amu

. (m/z = mass charge ratio)

The two lines indicate that boron, which is monatomic, has two isotopes boron-10 and boron-11.

On the following page, we will use the spectrum to calculate boron’s atomic mass:

Slide5

Mass Spectrum of Boron (Continued)

First, we need to calculate relative abundance:

Abundance of B-10 = 25.0% / (25.0% + 100.0%) = 20.0%

Abundance of B-11 = 100.0% / (25.0% + 100.0%) = 80.0%

Next, let’s calculate atomic mass:

B atomic mass = (.2000 X 10) + (.8000 X 11) = 10.8

amu

Slide6

Avogadro’s Number

What is

A

vogadro’s number, and how does it get its specific value?

6.022x10

23

mol

-

(You don’t need to memorize this number) is equal to the amount of atoms in 12.0 g of pure carbon 12.

Slide7

Mole Conversions (atoms to moles)

A pure silver ring contains 2.80X10

22

silver atoms. How many moles of silver atoms does it contain?

2.80X10

22

atoms Ag 6.022x10

23

atoms Ag

------------------------------ = ------------------------------

X 1.00

mol

Ag

X = 4.65x10

-2

mol

Ag

Slide8

Molar Mass and mass to mole conversions

Molar mass g/

mol

= Atomic mass in

amu

Example: carbons molar mass = 12.01 g/

mol

carbon’s atomic mass = 12.01

amu

12.01 g carbon = 1 mol carbon = 6.022X10

23

atoms carbon

Calculate the number of moles of copper in 35.8 g pure copper.

35.8 g Cu 63.55 g Cu

--------------- = --------------------

X = 5.63X10-1 mol Cu X 1.00 mol Cu

Slide9

Mole Conversions (atoms to mass)

Calculate the mass of 2.25X10

22

tungsten atoms.

This is a two step problem:

First, convert atoms to moles:

2.25X10

22

atoms W 6.022X10

23

atoms W

---------------------------------------- = ------------------------------------------------

X = 3.74X10

-2

mol

W

X 1.00 mol WNext, convert moles to atoms

3.74X10

-2

mol

W 1.00

mol

W

----------------------------------------

= ------------------------------------------------

X

= 6.87 g W

X

183.8 g W

Slide10

Pgs.

81-82 #’s 76, 80, 82, and 91.

Study for Chapter 2 Quiz