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Molar Math 3

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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.

Chemical equation:  2 H2 + O2 <=> 2 H2O.  Underneath this equation is a graphic showing colored circles representing atoms.  Small blue circles represent hydrogen atoms.  Large red circles represent oxygen atoms.  To represent molecules, the circles are drawn slightly overlapping.  Drawn under 2 H2 are 2 pairs of overlapping small blue circles, representing the 2 H2 molecules.  Under O2 is one pair of large red overlapping circles representing 1 molecule of O2. Under 2 H2O are two groups of overlapping circles representing the 2 molecules of H2O.  Each molecule is represented by three overlapping circles, the central large red circle representing 1 oxygen atom and the 2 smaller blue circles representing 2 hydrogen atoms.  It can be seen from the graphic that the total number of small blue circles on each side of the equation is the same.  Also, the number of large red circles on each side of the equation is the same.

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.

This chart contains several parts.  The first part shows the equation for the synthesis of water:  2 H2 + O2 Ô 2 H2O.  There are three two-headed arrows drawn around the equation, each pointing to coefficients in the equation that define mole ratios.  Mole Ratio 1 points to the 2 in front of H2 and the 1 in front of O2.  Mole Ratio 2 points to the 1 in front of O2 and the 2 in front of H2O.  Mole Ratio 3 points to the 2 in front of H2 and the 2 in front of H2O.</p>
<p>The next part of the chart shows the mole ratios derived from the above equation written in the form of a conversion factor.  Mole Ratio 1 gives  (2 H2/1 O2)  or  (1 O2/2 H2).  The mole ratio between hydrogen and oxygen is 2:1.  Mole Ratio 2 gives (1 O2/2 H2O)  or  (2 H2O/1 O2).  The mole ratio between oxygen and water is 1:2.  Mole Ratio 3 gives (2 H2/2 H2O).  The 2's cancel, so we then get  (H2/H2O) or (H2O/H2).  The mole ratio between H2 and H2O is 1:1.


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.

Start units --> Liters H2.  End units --> grams H2O.  The conversion set up now requires 3 conversion factors:  Liters H2 x (?/?) x (?/?) x (?/?) = grams H2O.

This kind of problem can be made EASY by charting the “conversion flow” on the “Multiple Conversion Flow Chart” as shown below.

Multiple Conversion Flow Chart.  This chart consists of 4 boxes, placed one under the other in a single column.  The first box contains the title, 'Multiple Conversion Flow Chart.'  The second box contains instructions:  Place the balanced chemical equation here.  The example provided is the equation for the synthesis of water.  2 H2 + O2 Ô 2 H2O.  The third box is colored pink and has the instructions 'Place mole/mole conversions here.'  The 4th box is green and is separated from the pink box by a dotted line.  It contains the instructions 'Place Start Units and End Units here.'

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.

This is a Multiple Conversion Flow Chart as described above, but with added information to show how it is used.  Again, we have the equation for the synthesis of H2O.  2 H2 + O2 Ô 2 H2O.  Since we are converting from Liters of H2 to grams of H2O, we place 17.3 Liters in the green box of the chart, where Start and End units are placed, and we place the '17.3 Liters' directly below H2 in the equation.  From there, we draw an arrow directly up into the pink box, crossing the dotted line.  The arrow should be pointing up in the direction of H2 in the equation.  Then we place '? grams' in the green box, where the Start and End units go, directly under H2O in the equation.  We then draw an arrow DOWN from the pink box to the '? grams.'  Then, in the pink box, where mole/mole conversions are displayed, we write the word moles under both H2 and H2O, so that the up arrow points directly to moles under H2 and the down arrow goes directly from moles under H2O.  Finally, we draw an arrow horizontally in the pink box going from moles H2 to moles of H2O.  Therefore, the arrow should point to the right.  Now that we have filled in this chart, we should be able to merely read off our conversion factors to fill in our conversion set up for this problem.  We go from 17.3 Liters H2 to moles of H2 (step 1).  Then we go from moles of H2 to moles of H2O (step 2).  Finally, we go from moles of H2O to grams of H2O.  (step 3).

(5) Once you have these units correctly placed, writing the conversion ratios is simply a matter of following the arrows.

Conversion set up showing the three conversion factors from steps 1, 2 and 3.  Conversion set up:  Liters H2 x (moles H2/Liters H2) x (moles H2O/moles H2) x (grams H2O/moles H2O) = grams H2O.

(6) Finally, input the numbers, cancel units, do the calculation, determine significant figures, and write the final answer in correct form.

Conversion set up:  17.3 Liters H2 x (1 mole H2/22.4 Liters H2) x (2 moles H2O/2 moles H2) x (18 grams H2O/1 mole H2O) = ? grams H2O.  Liters H2 cancel.  Mole H2 cancel.  Moles H2O cancel.  = 13.90178 grams H2O.  (Round to 3 significant figures, because of 17.3 and 22.4.  By definition the mole ratio 2:2 is infinitely accurate, and the molar mass of water is 18.0 grams/mole.)  = 13.9 grams H2O - FINAL ANSWER.

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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.

Start units --> grams O2.  End units -> Liters H2.  Conversion set up:  grams O2 x (?/?) x (?/?) x (?/?) = Liters H2.

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.)

Multiple Conversion Flow Chart for Problem 10.  The balanced equation is  2 H2 + O2 Ô 2 H2O.  Start units, 56.7 grams, is placed in the green box directly under O2 with an arrow pointing up from it into the pink box.  End units, ? Liters, are placed in the green box directly  under H2 with an arrow pointing down to it from the pink box.  In the pink box, under both H2 and O2, write moles.  Finally, draw an arrow from moles of O2 to moles of H2.  Your arrow should be pointing to the left this time.  So step 1 is conversion from 56.7 grams O2 to moles O2.  Step 2 is conversion from moles of O2 to moles of H2.  Step 3 is conversion from moles of H2 to Liters of H2.

Next, set up the conversion factor ratios with the units in the correct places for cancelling.

Conversion set up:  grams O2 x (moles O2/grams O2) x (moles H2/moles O2) x (Liters H2/moles H2) = Liters H2.  Start units are grams O2.  End units are Liters H2.  The three conversion factor in the middle are labeled with steps 1, 2 and 3.

Now input the numbers.

56.7 grams O2 x (1 mole O2/32 grams O2) x (2 moles H2/1 mole O2) x 22.4 Liters H2/1 moles H2) = ? Liters H2.

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.

56.7 grams O2 x (1 mole O2/32 grams O2) x (2 moles H2/1 mole O2) x (22.4 Liters H2/1 mole H2) = ? Liters H2.  Units of grams O2 cancel.  Units of moles O2 cancel.  Units of moles H2 cancel.  We are left with units of Liters H2.  = 79.38 Liters H2.  (Round to 3 significant figures, because of 56.7 grams.  The molar mass of O2 is really 32.0 grams per mole, even though we often write just 32, because we use it so often.  The mole ratio of 2:1 moles H2/moles O2 is an infinitely exact ratio defined by the equation, so it does not affect significant figures.  Therefore, the number of significant figures here is 56.7 grams.)  = 79.4 Liters H2 - FINAL ANSWER.

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:

This is a hand written version of problem showing how I really solve this problem in practice.  I do not draw the chart, but I keep the spatial orientation in my mind, and I use abbreviations.  I wrote:  2 H2 + O2 ' 2 H2O.  Underneath O2, two lines down, I placed 56.7 g, standing for 56.7 grams with an arrow pointing up into the mole/mole space.  I wrote mol at the end of the arrow.  Under H2 I wrote ?L, standing for ? Liters, with an arrow down from the mole/mole space.  I wrote mol at the other end of the arrow.  Then I drew an arrow from mol to mol, pointing to the left in the mole/mole space, pointing from moles O2 to moles H2.  Then I set up my conversion string.  56.7 g O2  x 1 mol O2/32 g O2 x   2 mol H2/1 mol O2  x  22.4 L/mol H2 = ? L.  From there I would do the calculation, find significant figures and report the final answer.

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?

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