Mole Ratios
Key Questions

Mole ratios are used as conversion factors between products and reactants in stoichiometry calculations.
For example, in the reaction
2H₂(g) + O₂(g) → 2H₂O(g)
The mole ratio between O₂ and H₂O is
#(1 mol O₂)/(2 mol H₂O)# .The mole ratio between H₂ and H₂O is
#(2 mol H₂)/(2 mol H₂O)# .Example:
How many moles of O₂ are required to form 5.00 moles of H₂O?
Solution:
5.00 mol H₂O ×
#(1 mol O₂)/(2 mol H₂O)# = 2.50 mol O₂.If the question had been stated in terms of grams, you would have had to convert grams of H₂O to moles of H₂O, then moles of H₂O to moles of O₂ (as above), and finally moles of O₂ to grams of O₂.

Answer:
To get the experimental molar ratio, you divide the moles of each reactant that you actually used in the experiment by each other.
Explanation:
EXAMPLE 1
Consider the reaction:
#"2Al" + "3I"_2 → "2AlI"_3# What is the experimental molar ratio of
#"Al"# to#"I"_2# if 1.20 g#"Al"# reacts with 2.40 g#"I"_2# ?Solution
Step 1: Convert all masses into moles.
#1.20 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "0.044 48 mol Al"# #2.40 cancel("g I₂") × ("1 mol I"_2)/(253.8 cancel("g I₂")) = "0.009 456 mol I"_2# Step 2: Calculate the molar ratios
To calculate the molar ratios, you put the moles of one reactant over the moles of the other reactant.
This gives you a molar ratio of
#"Al"# to#"I"_2# of#0.04448/0.009456# Usually, you divide each number in the fraction by the smaller number of moles. This gives a ratio in which no number is less than 1.
The experimental molar ratio of
#"Al"# to#"I"_2# is then#0.04448/0.009456 = 4.70/1# (3 significant figures)The experimental molar ratio of
#"I"_2# to#"Al"# is#1/4.70# Note: It is not incorrect to divide by the larger number and express the above ratios as 1:0.213 and 0.213:1, respectively. It is just a matter of preference.
EXAMPLE 2
A student reacted 10.2 g of barium chloride with excess silver nitrate, according to the equation
#"BaCl"_2("aq") + "2AgNO"_3("aq") → "2AgCl(s)" + "Ba(NO"_3)_2("aq")# She isolated 14.5 g of silver chloride. What was her experimental molar ratio of
#"AgCl"# to#"BaCl"_2# ?Solution
Step 1: Convert all masses into moles
#10.2 cancel("g BaCl₂") × ("1 mol BaCl"_2)/(208.2 cancel("g BaCl₂")) = "0.048 99 mol BaCl"_2# #14.5 cancel("g AgCl") × "1 mol AgCl"/(143.3 cancel("g AgCl")) = "0.1012 mol AgCl"# Step 2: Calculate the molar ratios
The experimental molar ratio of
#"AgCl"# to#"BaCl"_2# is#0.1012/0.04899 = 2.07/1# Here is a video example:
video from: Noel Pauller