DO NOW: (Refresh your memory)

Classify the following as an atom, molecule, ion, or formula unit:

1.Fe _________

2.F2 _________

3.H2O _________

4.Na _________

5.NaCl _________

6.Na+1 _________

DO NOW: (Refresh your memory)

Classify the following as an atom, molecule, ion, or formula unit:

1.Fe ATOM

2.F2 MOLECULE

3.H2O MOLECULE

4.Na ATOM

5.NaCl FORMULA UNIT

6.Na+1 ION

Chapter 10THE MOLE!!!

What is a mole?

What is a mole? It is a unit of measure. 6.02 x 1023

If we write this out (standard notation),

we get:

602,000,000,000,000,000,000,000

A mole is equal to: 6.02 x 1023 Representative Particles

Atoms, molecules, formula units, and ions

Fe F2 NaCl Na+1

Now we have a conversion unit to use: 1mole

6.02 x 1023 r.p. (any of the four choices above)

Lets do some problems

2.1 moles of Fe = how many r.p. of Fe? First what r.p. is Fe?

(atom)

2.1 moles Fe 6.02 x 1023 atoms Fe1mole Fe

2.1moles Fe = 1.3x 1024 atoms Fe

140 moles of MgCl2 = how many r.p. of Cl-? First what r.p. is MgCl2 and then Cl-?

(formula unit, ion)

140 moles MgCl2 6.02 x 1023 f.u. MgCl2 2 ions Cl-

1mole MgCl2 1 f.u. MgCl2

140 moles MgCl2 = 1.7 x 1026 ions Cl-

Gram Formula Mass & Gram Molar Mass (molar mass) Use individual atomic masses to determine

overall mass In 1 mole of NaCl, there are 58 g of NaCl

23 + 35 = 58 In 1 mole of H2O, there are 18 g of H20

2(1) + 16 = 18

A mole is equal to:

The molar mass of a substance Now we have a conversion unit to use:

1mole

molar mass of a substance

EX: what is the gfm of C6H12O6

180g/mol

So, in 1 mole of C6H12O6, there are 180 g

6(12) + 12(1) + 6(16) = 180

Lets do some problems

11.3 moles of C6H12O6 = how many grams?

11.3 moles C6H12O6 180 g C6H12O6

1mole C6H12O6

11.3 moles C6H12O6 = 2030 g C6H12O6

= 2.03 x 103 g C6H12O6

A mole is equal to:

The 22.4L of gas at STP STP= Standard Temperature and Pressure = 0o C or 273 K and (101.3 kPa or 1 atm or 760 mmHg or 760 torr)

Now we have a conversion unit to use: 1mole

22.4L of a gas @ STP

THE MOLE ISLAND!!!!

1 mole of a

substance

Molar Mass gfm or gmm

of a substance

6.02x1023

r.p.(Avogadro’s

#)

22.4 Lof gas per mole @

STP

Lets do some problems

5.3x1024 molecules of CH4 = how many grams CH4?

!!!!! You must go from the r.p. island then to the mole island, and then to the molar mass island!!!!!

5.3x1024 molecules of CH4 1 mole CH4 16g CH4

6.02x1023r.p. CH4 1 mole CH4

5.3x1024 molecules of CH4= 140 g CH4

= 1.4 x 102 g CH4

Chapter 10Percent Composition

Calculating % Composition

Chemists use this calculation when new compounds are created in the lab and they would have to determine the formula of the

cmpd.

Formula

% mass of element = grams of element x 100

grams of cmpd

Lets do some problems

An 8.20g piece of magnesium combines completely with 5.40g of oxygen to form a compound. What is the % composition of Magnesium and Oxygen the cmpd?

Step 1 : Add masses to get total

8.20g + 5.40g = 13.60g Step 2 : Find the % of each element.

Mg: (8.20g/13.60g)*100 = 60.3%

O: (5.40g/13.60g)*100 = 39.7%

Step 3 : Make sure your %s add up to 100.

Chapter 10Empirical Formulas

Calculating Empirical Formulas

Gives the lowest whole number ratio of the atoms of the elements in a cmpd.

Remember the Poem

% to mass

Mass to mole

Divide by the smaller #

And multiply until whole

Lets do some problems

What is the empirical formula of the cmpd that is 25.9% N and 74.1% O?

1. 25.9% N = 25.9g N

74.1% O = 74.1g O

2. 25.9g N 1mole N =1.85 mole N

14 g N

74.1 g O 1 mole O =4.63 mole O

16 g O

1.85 = 1.00 mole for N 4.63 = 2.50 mole for O

1.85 1.85

Make into whole #s by multiplying by 2

1 mole N x 2 = 2 moles N

2.5 mole O x 2 = 5 moles O

So, the empirical formula is N2O5

Empirical vs. Molecular Formulas

(lowest whole # ratio) vs (multiple of empirical)

Calculating Molecular Formulas1. Go through all of the steps of an empirical

formula problem. (poem) Ex: CH3

2. Add up the mass of the empirical formula.

Ex: 15g/mol (efm)

3. Divide the mass of the molecular formula (gfm), which will be given in the problem (30g/mol), by the mass of the empirical formula (efm). (gfm/efm)

Ex: 30g/mol = 2 15g/mol

4. Multiply all of the subscripts in the empirical formula by the 2.

This will be the new molecular formula.

CH3 (empirical) x 2=C2H6 (molecular formula)

* Try #38 and #39 p. 312.