Download - DO NOW: (Refresh your memory)
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.