Atomic Atomic StructureStructure

Subatomic ParticlesSubatomic Particles(Particles that make up an atom)(Particles that make up an atom)

● ● Proton (p+)Proton (p+)- Positively charged - Positively charged - Found in the nucleus- Found in the nucleus- Large mass- Large mass

●● Neutron (nNeutron (n00))- A neutral particle - A neutral particle - Found in the nucleus- Found in the nucleus- Large mass- Large mass

●● Electron (e-)Electron (e-)- Negatively charged particle - Negatively charged particle - Found outside of the nucleus in the electron - Found outside of the nucleus in the electron

cloudcloud- Very small mass- Very small mass

SummarySummary

●● The nucleus has almost all of the mass & it The nucleus has almost all of the mass & it has a + chargehas a + charge●● The electron cloud has a – charge & The electron cloud has a – charge & creates the atom’s creates the atom’s volumevolume

How to read a box on the How to read a box on the periodic tableperiodic table

11

Na 22.98

● Atomic Number - # above symbol

- Always determines # of protons (can never change!)

- Determines # of electrons if atom is neutral (0 charge)

- We assume the periodic table is neutral (same # of p+ & e-)

● Summary:

- Sodium’s atomic number is 11

- Sodium has 11 protons & 11 electrons

Atomic #Symbol

11

Na

22.98● Average atomic mass - # below the symbol in decimal form

- The average mass of an atom

- Not all sodiums have the same mass due to different number of neutrons (isotope)

Average atomic mass

● Mass Number – rounding the a.a.m. to a whole number

- Mass # = # of protons + number of neutrons

- Therefore, use to find number of neutrons mass # - # of p = # of n

● Summary- Na’s ave. atomic mass = 22.98 amu

(atomic mass units)- Na’s mass # = 23- Number of neutrons in Na:

23 – 11 = 12 neutrons

11

Na

22.98

Mass #

(23)

You just have to try one!You just have to try one!

47

Ag 107.87

Determine:

1. Atomic # =

2. # of protons =

3. # of electrons =

4. Ave. atomic mass =

5. Mass number =

6. # of neutrons =

IsotopesIsotopes● Atoms of the same element can have different numbers of neutrons, therefore, different masses

- Remember, neutrons have mass!

- Changing the number of neutrons, changes the mass

● Let’s look at 2 isotopes of carbon as an example:

- Carbon ALWAYS has 6 protons

- But it can have a mass of 12 amu

(6p + 6n) C or C-12

- and it can have a mass of 14 amu

(6p + 8n) C or C-14

12

14

6

6

Perfect practice makes Perfect practice makes perfect!perfect!

● Here is an isotope of oxygen: O- How many protons are present? __________

- What is the mass number? __________

- How many neutrons are present? __________

- How many electrons are present? __________

● Write the shorthand form of a nitrogen isotope that has 13 neutrons.

_ N or N - __

188

_

Mole ConversionsMole Conversions● Moles (mol) are a unit of measurement

● 1 mole = 6.02 x 1023 units (atoms, molecules, formula units,

ions, etc)

● 6.02 x 1023 is Avogadro’s number

● Mole Conversions

1 mole = 6.02 x 1023 units = formula weight (grams)

What is formula weight?What is formula weight?

● Formula weight is the weight of an element or compound in grams

● How is formula weight determined?

- Use your periodic table and find the ave. atomic mass

- Formula weight of H2O

- H’s ave. atomic mass = 1.01 g (x 2) = 2.02 g

- O’s ave. atomic mass = 16.00 g

2.02 g + 16.00 g = 18.02 g H18.02 g H22OO

What is the formula What is the formula weight of…weight of…

AlAl BrBr22

MgFMgF22

CHCH44

CaCa33(AsO(AsO44))22

ConversionsConversions1. Moles to grams

# of moles x formula weight (g) = _____ grams 1 1 mole

● Example: How many grams are in 5.00 moles of CaCl2?

Formula weight of CaCl2:

● Ca = 40.08 g Cl = 35.45 g (x2) = 70.90 g

● 40.08 g + 70.90 g = 110.98 g CaCl2

5.00 moles x 110.98 g CaCl2 = 554.9 = 1 1 mole

555 g CaCl2

2. Grams to moles

# of grams x ___1 mole _ = _______ moles 1 formula wt (g)

● Example: How many moles are in 25.00 g of NaCl?

25.00 g of NaCl x _ 1 mole___ = 1 58.44 g NaCl

0.4278 moles of NaCl

3. Moles to units (atoms, molecules, formula units, ions, etc.)

# of moles x 6.02 x 1023 units = ____ units1 1 mole

● Example: How many atoms are in 0.250 moles of neon?

0.250 moles of Ne x 6.02 x 1023 atoms = 1 1 mole1.51 x 1023 atoms

of Ne

4. Units to moles

# of units x ___1 mole____ = ____ moles 1 6.02 x 1023 units

● Example: How many moles are in 4.23 x 1024 molecules of H2O?

4.23 x 1024 molecules x ______1 mole______ = 1 6.02 x 1023 molecules

7.03 moles of H2O

5. Grams to units

# of grams x 6.02 x 1023 units = ____ units 1 formula wt (g)

● Example: How many formula units are in 35.0 g of K2O?

35.0 g K2O x 6.02 x 1023 formula units = 1 94.20 g

K2O 2.24 x 1023 formula units of K2O

6. Units to grams

# of units x _formula wt (g)_ = ____ grams 1 6.02 x 1023 units

● Example: How many grams are in 9.75 x 1025 atoms of Ag?

9.75 x 1025 atoms x __107.87 g Ag__ = 1 6.02 x 1023

atoms

17500 g Ag

Mass Percent CompositionMass Percent Composition● Determining what percentage of each element is in a specific formula

● Example: Find the mass % of each element in NaHCO3.

- Step 1: Find their individual ave. atomic masses from the PT & multiply by the number of atoms of each (subscript)

Na = 22.99 g (1) = 22.99 g H = 1.01 g (1) = 1.01 gC = 12.01 g (1) = 12.01 gO = 16.00 g (3) = 48.00 g

84.01 g NaHCO3

- Step 2: Add them to get the total weight of the formula.

- Step 3: Find the mass % of each!

-Remember: Na = 22.99 g (1) = 22.99 g H = 1.01 g (1) = 1.01 gC = 12.01 g (1) = 12.01 gO = 16.00 g (3) = 48.00 g

84.01 g of NaHCO3

- Take the elements individual total weight and divide by the total weight of the formula. Then Multiply by 100.

- Mass % of Na = 22.99g /84.01 (100) = 27.36 %

- Mass % of H = 1.01g /84.01 (100) = 1.20 %

- Mass % of C = 12.01g /84.01 (100) = 14.30 %

- Mass % of O = 48.00g /84.01 (100) = 57.14 %

- Add %’s to make sure they add up to 100%

Getting the formula from Getting the formula from mass %mass %

● Do the opposite of finding the mass %

● Example: What is the formula of a substance that is made of 27.29% C & 72.71% O. The total weight of the substance is 44.01 g.

- Step 1: Divide each % by 100 then multiply by the total weight

C : 27.29/100 = 0.2729 (44.01 g) = 12.01 g C

O: 72.71/100 = 0.7271 (44.01 g) = 32.00 g O

- Step 2: Divide the totals by their average atomic mass (from PT)

12.01 g C/12.01 g C = 1

32.00 g O/16.00 g O = 2

- Step 3: Put the formula together

CO2

Finding the relative Finding the relative atomic massatomic mass

● Where does the periodic table get its average atomic masses from?

● Here’s an example:

There are two isotopes of chlorine which consists of atoms of relative isotopic masses 35.0 (75.0 %) and 37.0 (25.0 %).

% abundance Isotope mass

Cl-35 75.0 35.0 amu

Cl-37 25.0 37.0 amu

(75.0/100) x 35.0 amu + (25.0/100) x 37.0 amu =

35.5 amu

The answer matches Cl on the periodic table!