1. Molecular Formula
number and type of atoms
covalent compounds
non-metals
glucose
C6H12O6
mix
C(s)
H2(g)
O2(g)
1 atom of C
 2 x g
1 atom of 12C
(6 protons + 6 neutrons)
 12 amu C
= amu
different isotopes
12C 98.9%
13C 1.10%
ave. =
(0.989 x 12)
+ (0.001 x ) =

2. C
relative masses
= amu O
exact masses
(g)
= amu H
= amu
12 g 12C
= x 1023 atoms 12C
Avogadro’s number
(NA)
= mole
12 g 12C
6.022 x 1023 atoms
= 12 g 12C
6.022 x 1023 atoms
mole
mol C
= g/mol O
= g/mol H
= g/mol

3. Start with 15.43 g carbon How much H2(g) and O2(g) ? C (s) H2 (g)
= g/mol O
= g/mol H
= g/mol
Start with g carbon
How much H2(g) and O2(g) ?
15.43 g C
1 mol =
mol C
= mol C g
2.569 mol H
g H
= g H
= g H
1 mol H
1.285 mol O
g O
= g O
= g H
1 mol O
C (s)
H2 (g)
O2 (g)
C6H12O6
15.43 g
2.59 g
20.56 g
1 atom 2
atoms 1
atom
1 mol C
2 mol H
1 mol O
1.285 mol C
2.569 mol H
1.285 mol O

4. How many grams of glucose will be produced? 15.43 + 2.59 + 20.56
= g glucose
How many moles of glucose is this?
1 mol glucose =
6 mol C
+ 12 mol H
+ 6 mol O
= g
6 x
12 x
6 x
38.58 g glucose
1 mol
= mol glucose g
C (s)
+ H2 (g)
+ O2 (g)
C6H12O6
15.43 g
2.59 g
20.56 g

5. Percent composition C6H12O6 % (by mass) of each element molar mass
= g/mol
6 mol C
= (6 x g/mol) =
= 40.00% C
mol C6H12O6
(1 x g/mol)
12 mol H
= (12 x g/mol) =
= 6.72% H
mol C6H12O6
(1 x g/mol)
6 mol O
= (6 x g/mol) =
= 53.28% O
mol C6H12O6
(1 x g/mol)

6. Used to determine formula of unknown
40.92% carbon
4.58% hydrogen
54.50% oxygen 1.
Assume you have 100 grams
40.92 g C
4.58 g H
54.50 g O 2.
Calculate moles of elements 3.
Determine empirical formula
3.41 mol C
4.54 mol H
C H O
3.41
4.54
3.41
3.41 mol O
C H O 1
1.33 1
C H O 3 4 3

7. C H O 3 4
C H O 3 4
88.06 g/mol
C H O 6 8
g/mol
C H O 9 12
g/mol
need molecular weight
Combustion reaction
CnHnOn
+ O2
CO2
+ H2O
11.2 g
of an unknown compound is burned,
producing
22.0 g CO2
and 13.5 g H2O. 1.
Change grams to moles

8. 11.2 g
of an unknown compound is burned,
producing
22.0 g CO2
and 13.5 g H2O. 1.
Change grams to moles
0.500 moles CO2
0.500
mol C
0.75 moles H2O
1.50
mol H 2.
How much O came from unknown?
mass unknown
= 11.2 g
mass unknown
= mass C
+ mass H
+ mass O
mass unknown
= 11.2 g
= 6.00 g g
+ mass O
mass O =
4.0 g
4.0 g O
= mol O
0.25

9. mol H
1.50
mol O
0.25
mol C
0.500
C H O .5
1.5
.25
C H O 2 6
Empirical formula