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# 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