1. Stoichiometry and Quantitative Analysis Using Mole Ratios
is the study of the relative quantities
of reactants and products in a
chemical reaction
you can use the number of moles for a given reactant or product to find the moles for any other reactant or product

2. 2:7:4:6 Example Consider the following chemical reaction:
2 C2H6(g) O2(g)  4 CO2(g) H2O(g)
a) Write the ratio for all components of the reaction.
2:7:4:6
b) What amount, in moles, of CO2(g) is formed if 2.50 mol of C2H6(g) reacts?
2 C2H6(g) O2(g)  4 CO2(g) H2O(g) 4 2
n = 2.50 mol
n = 2.50 mol 
= 5.00 mol

3. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
c) What amount, in moles, of O2(g) is required to react with 10.2 mol of C2H6(g)?
2 C2H6(g) O2(g)  4 CO2(g) H2O(g) 7 2
n = 10.2 mol
n = 10.2 mol 
= 35.7 mol
d) What amount, in moles, of H2O(g) is formed when 100 mmol of CO2(g) is formed?
2 C2H6(g) + 7 O2(g)  4 CO2(g) H2O(g) 6 4
n = 100 mmol
n = 100 mmol 
= 150 mmol
= mol

4. B. Gravimetric Stoichiometry
mass measurements
Steps
1. Write a including the states. Write the information
balanced equation
given.
2. Find the of the species using
moles
given
n=m M
3. Find the of the species using
moles
wanted
mole ratio
wanted
given
4. Calculate of the wanted species using
mass
m=nM

5. 1 Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) Example 1
Iron is produced by the reaction of iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. What mass of iron(III) oxide is required to produce 1000 g of iron? g w 1
Fe2O3(s) + 3
CO(g)  2
Fe(s) + 3
CO2(g)  
m=?
M = g/mol
m = 1000 g
M = g/mol 1 2
n = 17.90… x
n = m M
= g
55.85g/mol
= 17.90… mol
= 8.95… mol
m = nM
= (8.95… mol)( g/mol)
= g  
= 1430 g

6.  1 Cu2(CO3)(OH)2(s) 2 CuO(s) + 1 CO2(g) + 1 H2O(g) Example 2
The decomposition of the mineral malachite, Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon dioxide and water vapour. What mass of copper(II) oxide is produced from 1.00 g of malachite? g w  1
Cu2(CO3)(OH)2(s) 2
CuO(s) + 1
CO2(g) + 1
H2O(g)
m = 1.00 g
M = g/mol
m=?
M = g/mol
n = … x
n = m M
= g
g/mol
= … mol
= … mol
m = nM
= ( … mol)(79.55 g/mol)
= g
= g