Question

Question #dc373

Answer 1

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