from http://molarmath.info/
Ratio #4: The Mole Ratio of Two Substances in a Chemical Equation
In most mole ratio problems, we are given a balanced chemical equation (or we must balance it before we begin).
How to Find Mole Ratios from a Chemical Equation
A mole ratio is the ratio of moles of one species in a chemical reaction, to the number of moles of another.
(A “species” used in this context means “a particular chemical,” such as hydrogen or oxygen.)
In the equation below, there are three species: hydrogen, oxygen and water.
Here is the balanced equation for the formation of water.
From this equation, we know that 2 molecules of hydrogen plus 1molecule of oxygen combine to make 2 molecules of water.
The coefficients (large numbers) in front of the chemical formulas represent a ratio of components.
Thus, we have a 2:1:2 ratio between the different species in the reaction.
From the above equation, we get 3 different “mole ratios.”
We get the mole ratios from the coefficients.
Study the diagram below and see if you can figure out how to find mole ratios.
Using Mole Ratios to Solve Problems
The following problems will give you practice in solving mole ratio problems.
Introducing the Multiple Conversion Flow Chart
Problem 9: Find Grams of Water from Liters of Hydrogen.
Given 17.3 Liters of diatomic hydrogen gas, how many grams of water will be produced, if there is an excess of oxygen?
Analysis: Start units are “Liters of hydrogen.” End units are “grams of water.”
Set up the problem. This time there will be 3 conversion ratios.
The middle ratio will be one of the mole/mole ratios from the balanced chemical equation.
This kind of problem can be made EASY by charting the “conversion flow” on the “Multiple Conversion Flow Chart” as shown below.
Now we will show how to use the “Multiple Conversion Flow Chart” with the problem above.
(1) Write the starting units and ending units under the particular chemical species to which they pertain.
We start with 17.3 Liters of hydrogen gas.
Place 17.3 Liters in the bottom box directly under hydrogen in the equation.
We are looking for grams of water, so we must write “grams” with a question mark under water, or H2O.
(2) Draw an arrow directly up from the start unit to the mole/mole box and write the word “moles.”
(For this problem, it will mean “moles of H2.”)
(3) Draw an arrow directly down from the mole/mole box pointing towards the ending units,
which in this case are “grams of water.”
Write the word “moles” in the mole/mole box where the down arrow begins.
This will mean “moles of water.”
(4) Finally, draw an arrow from the moles of the starting chemical (in this case, hydrogen gas) to the moles of the ending chemical (in this case, water).
You have now graphically set up the flow of the conversion factors you will use.
(5) Once you have these units correctly placed, writing the conversion ratios is simply a matter of following the arrows.
(6) Finally, input the numbers, cancel units, do the calculation, determine significant figures, and write the final answer in correct form.
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Problem 10: Find Liters H2 at STP from Grams O2.
If you want to use up 56.7 grams of oxygen to make water, how many Liters of hydrogen gas at STP will you need?
Analysis: Start units “56.7 grams of oxygen.” (The words “use up” are the clue.) End units “Liters hydrogen gas.”
Set up the equation showing start and end units, leaving space for the conversion factors.
Next, set up the “Multiple Conversion Flow Chart” with the equation as shown below. We are starting with 56.7 grams of oxygen and ending with Liters of hydrogen gas at STP. (Notice that the direction of the arrow in the mole/mole box goes in the opposite direction from the previous problem.)
Next, set up the conversion factor ratios with the units in the correct places for cancelling.
Now input the numbers.
If you are not sure how to find the grams per mole of any substance, such as that there are 32 grams per mole of diatomic oxygen gas, click here. [Not active yet.]
Finally, cancel units to make sure you have them correctly placed, do the calculation, round to the correct number of significant figures and report the answer in final form.
Recognize that in the real world of doing chemistry, we do not go to all the trouble of actually drawing the “Multiple Conversion Flow Chart.” We just remember the structure and apply it. So my actual figuring of the above problem on paper would look more like this:
From here, I would cancel the units and do the calculation.
Hopefully this has been helpful. If you see something confusing in this presentation, or if you find errors, or if you have questions, please contact me, Lynda Jones, at sing-smart.com. My goal is to make chemistry EASY! How am I doing?