# Mole Ratios

How to Write Mole Ratios

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

The mole ratio is the ratio of moles of one substance to the moles of another substance in a balanced equation.

#### Explanation:

2 H2(g) + O2(g) â†’ 2 H2O(g)

The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are formed.

The mole ratio between H2 and H2O is 1:1.

For every two moles of H2 used, 2 moles of H2O is formed.

• Mole ratios are used as conversion factors between products and reactants in stoichiometry calculations.

For example, in the reaction

2Hâ‚‚(g) + Oâ‚‚(g) â†’ 2Hâ‚‚O(g)

The mole ratio between Oâ‚‚ and Hâ‚‚O is (1 mol Oâ‚‚)/(2 mol Hâ‚‚O).

The mole ratio between Hâ‚‚ and Hâ‚‚O is (2 mol Hâ‚‚)/(2 mol Hâ‚‚O).

Example:

How many moles of Oâ‚‚ are required to form 5.00 moles of Hâ‚‚O?

Solution:

5.00 mol Hâ‚‚O Ã— (1 mol Oâ‚‚)/(2 mol Hâ‚‚O) = 2.50 mol Oâ‚‚.

If the question had been stated in terms of grams, you would have had to convert grams of Hâ‚‚O to moles of Hâ‚‚O, then moles of Hâ‚‚O to moles of Oâ‚‚ (as above), and finally moles of Oâ‚‚ to grams of Oâ‚‚.

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: ${\text{2Al" + "3I"_2 â†’ "2AlI}}_{3}$

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") Ã— "1 mol Al"/(26.98 cancel("g Al")) = "0.044 48 mol Al"

2.40 cancel("g Iâ‚‚") Ã— ("1 mol I"_2)/(253.8 cancel("g Iâ‚‚")) = "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") â†’ "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 BaClâ‚‚") Ã— ("1 mol BaCl"_2)/(208.2 cancel("g BaClâ‚‚")) = "0.048 99 mol BaCl"_2

14.5 cancel("g AgCl") Ã— "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

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