Chapter 7

Representative particle- refers to
whether a substance commonly exists
as atoms, ions, or molecules
 Ex.
Elements- Representative
Particle is the atom
7
elements exist as diatomic molecules
H 2
N2 O2 F2 Cl2 Br2 I2
Representative
Particle of a molecular
compound is a molecule
Example
CO SO3
Formula unit (FU)
-Representative Particle of ionic
compounds
Example
NaCl AgNO3 BaS
Ca(C2H3O2)2

Mole- Chemists quantity of a substance
that represents 6.02x1023
representative particles of that
substance- called Avagadro’s number
 ex. 1 dozen eggs
 How many moles of Mg atoms in
3.01x1022 atoms of Mg?
#
of moles in 1.20x1025 atoms of P?
#
of atoms in .750 mol of Zn?
#
of molecules in 4 mol of glucose,
C6H12O6?
#
of moles in 1.20x1024 molecules
of CO2 ?

To find the # of atoms in 1 mole of a
compound, you must determine the # of
atoms in a representative formula of that
compound
#
of fluoride ions in 1.46 mol of
Aluminum fluoride?
#
of C atoms in a mixture of 3 mol
C2H2 and .7 mol carbon monoxide?

Atomic Mass – amu
 Mass of single atom
 Ex) C = 12 amu
 Ex) H = 1 amu (Periodic Table)

Gram Atomic Mass – gam
 # of grams of an element that is
numerically equal to the atomic
mass in amu.
 Ex) Carbon – gam is 12 g
 Ex) Oxygen – gam is 16 g

GAM – mass of 1 mol of atoms of
a mono-atomic element
 Ex) Carbon – gam is 12 g/mol
 Ex) Oxygen – gam is 16 g/mol
 GMM
– mass of 1 mol of that
compound.
 Ex) GMM of H2O2
2 mol H
x 1gH
1 mol H
2 mol O x 16.0 g O
1 mol O
= 2.0 g H
= 32.0 g O
34 g
Examples

Find the GMM of C6H4Cl2



C
H
Cl
6 x 12.01 = 72.06
4 x
1 = 4
2 x 35.45 = 70.9
146.96 g

GFM – Mass of 1 mol of an ionic
compound.
 Ex) GFM of Ammonium Carbonate
2 mol N
8 mol H
1 mol C
3 mol O
x
14gN
1 mol N
x
1gH
1 mol H
x
12 g C
1 mol C
x 16 g O
1 mol O
= 28 g
=
8g
= 12 g
= 48 g
96 g
Molar Mass

Mass of a mole of any
element or compound.
Ex) O2 = 32 g/mol,
O = 16 g/mol
Mole
Mass Conversion
Ex) # of grams in 7.20 mole of
dinitrogen trioxide

2 mol N
x
3 mol O
x
14 g N
1 mol N
16 g O
1 mol O
= 28 g N
= 48 g O
76 g
Example Cont’d…
7.20 mol N2O3 x 76 g N2O3
1mol N2O3
= 5.47 x 102 g N2O3
Grams

Moles
Ex) find # of moles 922g of
iron(III) oxide.
922 g Fe2O3 x 1 mol Fe2O3
159.6 g Fe2O3
= 5.78 mol Fe2O3

The Volume of a gas at Standard
Temperature and Pressure is 22.4 L
(STP).
 Std Temp = 0°C
 Std Press = 1 atmosphere (atm)
 22.4
= molar volume of a gas = 22.4 L
1 mol

Ex) Determine the Volume in L of
0.600 mol of Sulfur Dioxide gas @
STP.
 Mol
L
 Known : 1 mol SO2 = 22.4 L
.600 mol SO2
x
22.4 L
1 mol SO2
= 13.4 L SO2




Gas Density and Gram Molecular
Mass
Density of gas – units g/L
Ex. Density of carbon and oxygen is 1.969 g/L at
STP. Determine gfm. Is compound CO or CO2 ?
Densities of A, B, and C are 1.25, 2.86, and .714
g/L at STP. Calculate gfm of each. Identify each
substance as ammonia, sulfur dioxide, chlorine,
nitrogen or methane.
Volume of gas
at STP
Representative
1 mol/ 22.4 L
1 mol/6.02x1023 part
22.4 L/ 1 mol
particles
6.02x1023 part/ 1 mol
Mole
1 mol/gfm
gfm/ 1 mol
Mass

Ex. How many Carbon atoms are in a 50
carat diamond that is pure carbon?



50 carats= 10 g
Mass in grams of an atom of nickel?
How many molecules are in a 6 L balloon
filled with carbon dioxide (@ STP)?
Percent Composition

the percent by mass of each
element in a compound.
Examples
 Find
the percent composition
of K2CrO4
% mass = grams of element x 100
grams of compound
 40.3
%K
 26.8 % Cr
 32.9 % O
 They must add up to equal
100%
Example

An 8.20g piece of Mg combines
completely with 5.40g of oxygen
to form a compound. Calculate
the % composition of the
compound.
8.20g + 5.40g = 13.60g
% Mg =
mass of Mg
x
mass of compound
8.2 x
13.6
100 = 60.3%
100
Cont’d
%O = mass of O
x 100
mass of compound
= 5.40
13.6
x 100 = 39.7%
Example
 29g
of silver combines with 4.3g
of sulfur. Calculate %
composition.
29
x
33.3
100
= 87.1% Ag
4.3 x
33.3
100
=
12.9%S
Example
 222.6g
of Sodium combines
with 77.4g of Oxygen.
Calculate % composition.
222.6
300
x 100 = 74.2 % Na
77.4
300
x 100
= 25.8% O
% Composition of a known
compound
% mass =
grams of element in 1 mol of cmpd x 100
gfm of compound
Examples
 Calculate
the % composition of
ethane, C2H6.
Cont’d
C
H -
2 x 12 = 24g
6 x 1 = 6g
30 g
Cont’d
24
30
x 100 = 80 % C
6
30
x 100 = 20 % H
Examples
 Calculate
the % composition
of:
 a) C3H8
 b) Calcium Acetate
 c) Hydrogen Cyanide
 a)
81.8%C, 18.2% H
 b) 25.4% Ca, 30.4% C, 3.8% H,
40.5% O
 c) 3.7% H, 44.4% C, 51.9% N
Examples
 Calculate
the mass of
Carbon in 82 g of C2H6.
82g C2H6
x
80g C = 66g C
100g C2H6
*Based on previous example,
Composition of C in C2H6 is 80% C
& 20% H
Examples
 Calculate
the amount of
Hydrogen in the following
compounds.
350 g C3H8
350 g C3H8 x 18.2g H
100g C3H8
63.7g H
=
124g Calcium Acetate
124 g Ca(OAc)2
4.71g H
x 3.8g H
=
100 g Ca(OAc)2
378g Hydrogen Cyanide
378 g HCN
x 3.7 g H
=
100 g HCN
14 g H
Empirical Formulas
Empirical Formula  The
lowest whole number ratio
of the element in a compound.


An empirical formula may not
be the same as the molecular
formula
Examples
 CO2 is both empirical and
molecular
 N2H4 is molecular formula
 NH2 is empirical formula
because it is the simplest
ratio of N : H
Examples
 Find
the empirical formula
of a compound that is
25.9% nitrogen and 74.1%
oxygen.
25.9 g N
x 1 mol N
14 g N
= 1.85 mol N
74.1g O
x
= 4.63mol O
1 mol O
16 g O
1.85 mol
1.85 mol
= 1 mol N
4.63 mol
1.85 mol
=
2.5 mol O
N) 1 x 2 = 2
O) 2.5 x 2 = 5
empirical formula - N2O5
Examples
 Find
the empirical formulas
for the following
compounds.
79.8 %C , 20.2 %H
Empirical
CH3
Formula
67.6% Mercury , 10.8%
Sulfur, 21.6% Oxygen
 Empirical
Formula
HgSO4
17.6% Sodium, 39.7%
Chromium, 42.7% Oxygen
 Empirical
Formula
Na2Cr2O7
Calculating Molecular
Formulas
 You
must know:
Empirical Formula
GFM
Examples
-gfm = 60g
-Empirical formula = CH4N
-efm = 30
gfm = 60 = 2
efm
30
2(CH4N) = C2H8N2
 gfm
= 181.5 g
 Empirical formula = C2HCl
 efm = 60.5
gfm = 181.5 = 3
efm
60.5
3(C2HCl) = C6H3Cl3
Methyl butanoate smells
like apples. The Percent
Composition is 58.8% C,
9.8%H, 31.4% O. The gfm
is 102 g/mol - what is the
molecular formula?
58.8 gC
x 1 mol C = 4.9 mol C
12g C
9.8 gH
x
1 mol H = 9.8 mol H
1g H
31.4 gO
x 1 mol O = 1.96 mol O
16g O
4.9
1.96
= 2.5 mol C
9.8
1.96
= 5 mol H
1.96
1.96
= 1 mol O
C) 2.5 x 2 = 5
H) 5 x 2 = 10
0) 1 x 2 = 2
C5H1002