Determining the Molecular Formula
1. Determine the atoms in the material using
elemental analysis
2. Use various chemical techniques like
combustion to determine the relative
number of atoms in the substance.
This is the empirical formula.
3. Determine the molecular mass of the
substance using mass spectrometry, gas
density measurements, etc.
4. Deduce the molecular formula
Empirical / Molecular Formulas
Determine the empirical formula of the
compounds with the following
compositions:
a. 0.0130 mol C, 0.0390 mol H,
and 0.0065 mol O
b. 11.66 g iron and 5.01 g oxygen
c. 40 % C, 6.7 % H
and 53.3 % O by mass
What is the molecular formula of each of the
following compounds?
a. empirical formula CH2,
molar mass = 84 g/mol
b. empirical formula NH2Cl,
molar mass = 51.5 g/mol
Molecular formula from combustion
Given that the material contains C, H & O
Burn 0.225 g of CxHyOz
C x H y Oz + O2 ( g ) → CO2 ( g ) + H 2O( g )
form 0.209 g H2O & 0.512 g CO2
1. The number of moles of CO2 is the
number of moles of C in CxHyOz
2. The number of moles of H2O is twice the
number of moles of H in CxHyOz
3. The number of moles of O in CxHyOz
must be determined by difference.
moles of CO2 =
0.512g
= 0.0116 mol
44.01 g/mol
0.209g
= 0.0116 mol
mole of H 2O =
18.00g/mol
grams of C = 0.0116(12.00) = 0.1392
grams of H = 2(0.0116)(1.00) = 0.0232
grams of O = 0.225 – 0.1392 – 0.0232 = 0.061
moles O =
0.061g
= 0.0038 mol
16.00g/mol
atoms of C relative to O = 0.0116/0.0038 = 3.04
atoms of H relative to O = 0.0232/0.0038 = 6.08
C3H6O
Formula Mass is 58.09
Molecular mass as 116 g/mol
C6H12O2
Limiting Reactants
Reaction will stop when one of the reactants
is gone
3H 2 ( g ) + N 2 ( g ) → 2 NH 3 ( g )
How many moles of product will result from
3.0 mol N2 and 6.0 mole H2
3.0 mol N2 can use 9.0 mol H2, have 6.0
6.0 mol H2 can use 2 mol N2, have 3.0
All H2 will be consumed and so it is the
limiting reactant
Given the reaction
2 Na3 PO4 ( aq ) + 3Ba( NO3 )2 ( aq ) → Ba3 ( PO4 )2 ( s ) + 6 NaNO3 ( aq )
3.50 g Na3 PO4 and 6.40g Ba(NO3 )2
How many grams of Ba3 (PO4 )2 will be formed?
Which species is the limiting reactant?
Compound
Na3 PO4
Moles
0.0213
Ba(NO3 )2
0.0245
Ba(NO3 )2 is limiting reactant
Moles of Ba3 (PO4 )2 formed
1
moles of Ba(NO3 )2 reacted
3
which is 0.00816
Concentration of Solutions
molarity =
moles of solute
liters of solution
n
M=
L
Molarity of solution prepared by mixing
23.4g of Na2SO4 with enough water to make
a solution whose volume is 125ml?
Formula weight of Na2 SO4 is 142.0 g/mol
M=
23.4 / 142.0
= 1.32
0.125
Dilution Problem
Given a solution with a known molarity one
is to dilute it to form a solution of a lesser
molarity.
Given a 1.00M solution of CuSO4, make
250ml of a 0.1M solution.
How many moles of CuSO4 do we need?
n = ( 0.250 )x( 0.1 ) = 0.025 mol
What volume of the 1.0M solution contains
this amount ?
0.025mol
V=
= 0.025 L
1.00mol / L
Add 25ml of 1.00M solution to 225ml of water
Titration
The process of reacting a solution of an
unknown concentration with one of a known
concentration
One can determine the Cl- concentration in a
water sample by titrating with AgNO3.
Ag + ( aq ) + Cl − ( aq ) → AgCl( s )
How many grams of Cl- are present in a
sample if 20.2 ml of 0.100M Ag+ is needed
to react with all of the Cl- ?
moles Ag+ = moles Clmoles Ag+ = (0.100mol/L)(0.020L) =
0.0020 mol
g of Cl- = (0.0020 mol)(35.5g/mol) = 7.17 x 10-2