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QUESTION

# How do you calculate experimental mole ratio?

To get the experimental molar ratio, you divide the moles of each reactant that you actually used in the experiment by each other.

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

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