# 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.

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: ${\text{2Al" + "3I"_2 → "2AlI}}_{3}$

What is the experimental molar ratio of $\text{Al}$ to ${\text{I}}_{2}$ if 1.20 g $\text{Al}$ reacts with 2.40 g ${\text{I}}_{2}$?

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 $\text{Al}$ to ${\text{I}}_{2}$ of $\frac{0.04448}{0.009456}$

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 $\text{Al}$ to ${\text{I}}_{2}$ is then $\frac{0.04448}{0.009456} = \frac{4.70}{1}$ (3 significant figures)

The experimental molar ratio of ${\text{I}}_{2}$ to $\text{Al}$ is $\frac{1}{4.70}$

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 $\text{AgCl}$ to ${\text{BaCl}}_{2}$?

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 $\text{AgCl}$ to ${\text{BaCl}}_{2}$ is $\frac{0.1012}{0.04899} = \frac{2.07}{1}$

Here is a video example:

video from: Noel Pauller