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How do you do Molar Math in Chemistry?

(Site updated: November 16, 2010)

Remember, chemistry IS easy! (If you have the right tools to help you learn it.)

Introduction to the Mole

Before you can do molar math successfully, you must first know what a mole IS. If you are NOT completely clear about what a mole IS, quickly browse the page at the following link. Then come back when you think you understand enough to go on. What is a Mole?

We use the unit of a "mole" in science to talk about extremely large numbers of extremely tiny countable objects, such as molecules, atoms or electrons, objects which are obviously way too small to see.

We use the unit "mole" just like we use the unit "dozen." One dozen means "12 of something." In chemistry, one "mole" means "602,200,000,000,000,000,000,000 of something." ("Six-hundred-two-sextillion-two-hundred-quintillion of something.") The "something" can be anything which is "countable" by individual units. "Countable" as opposed to something without a definite particle nature, such as light or energy. The unit of a "mole" is usually used to count molecules, atoms or electrons, but technically it can also be used as a numerical unit even for something like apples, if you really wanted to take the time to count that long. (Actually you could not count that many apples in an entire lifetime, even with the help of many friends!)

The numerical value of the "mole" is frequently written using scientific notation as 6.022 x 10^23. ("^23" means "to the 23rd power, or something multipled by 10 and then multipled by 10 again and again and again, until 23 times.)

So basically, a "mole" is just a very, VERY large number.


Four Simple Steps to Mastering Molar Math

Molar Math is actually VERY EASY! With a little patience, you can learn the 4 simple steps listed below. We will present these concepts to you one by one so you do not get overwhelmed. Be sure you understand each step before going onto the next one, so you can build a solid foundation. Otherwise you may get confused.

1. Be able to find "starting units" and "ending units" in a problem.

2. Be able to set up "conversion ratios" to cancel the unwanted starting units and leave you with the desired ending units.

3. Be able to find any of the 4 mole ratios listed below. (There are ONLY 4!)

(1) Know that a mole is "6.022 x 10^23 of anything."
(2) Know that a mole of gas at STP (Standard Temperature and Pressure) occupies a volume of 22.4 Liters.
(3) Be able to find the molar mass of any substance.
(4) Be able to find the "mole ratio" between two reactants, between two products, or between a reactant and a product from any balanced chemical equation.

4. Know how to set up a series of ratios to cancel the units you DON'T want and end up with the units you DO want. This is the SECRET to being able to solve ALL mole problems in chemistry.

That's all there is to it!


(1) Be Able to Find Starting and Ending Units in a Problem

What is a unit and why is it important? A unit is one of whatever it is you are counting or measuring. For example, if you have 10 gallons of gasoline, "gallons" is the unit you are counting. If you drive 40 miles, "miles" is the unit of distance you are measuring. Units are extremely important in science. Without units, our numbers don't mean anything. If I said, "I have 30," what would your question be? Wouldn't it be "30 of what?" Reporting WHAT we are measuring or counting is essential to understanding. So we must always use units.

All problems in molar math are really just "unit conversion" problems.

Let's take a non-chemistry example first. Say I want to convert units of "feet" to units of "inches." My starting units are "feet" and my ending units are "inches." Using this information, I would set up a conversion structure as shown below, leaving space for a "conversion factor" in between.

Conversion factor set up with an arrow showing the starting units, feet, and the ending units, inches.  There is an empty horizontal fraction bar in between where the conversion factor will go, once we determine what that is.  feet x (?/?) = inches.

We will talk about "conversion factors" later. But first you need to be sure you can find starting and ending units from the wording of problems. Practice with the following examples. We will use nonsensical units so you can concentrate on the LANGUAGE which gives you clues to the starting and ending units.

(1) How many bars are in a plick? Ask yourself, "what are you trying to find?" The answer is "bars." Which words give you the clue that these are your ending units? "How many." What are your starting units? "plick" Which words give you the clue? "are in." So you would set up the structure this way:

Conversion set up showing starting units, plick, and ending units, bars, with a blank horizontal fraction bar in the middle.  Plick x (?/?) = bars.

(2) Convert jupos to babs. What are your ending units? "babs." Which word gives you the clue? "to." What are your starting units? "jupos." Which word gives you the clue? "Convert." Set up the structure like this:

Conversion set up with start units, jupos, ending units, babs, and a blank horizontal fraction bar in between.  jupos x (?/?) = babs.

(3) Given 15 beebos, how many gibbers do you have? What are your ending units? "gibbers." What word clue tells you that? "how many." What are your starting units? "beebos." What word clue tells you that? "Given." Set up the structure like this:

Conversion set up showing start units, beebos, ending units, gibbers, and a blank fraction bar in between.  beebos x (?/?) = gibbers.

You may find other language variations in problems, but you should always be able to find the ending units and starting units and set up a structure as we have done above.


(2) Be Able to Set Up Conversion Ratios to cancel the unwanted starting units and leave you with the desired ending units.

What is a Conversion Ratio?

A "conversion ratio" is a ratio which converts one unit into another.

Here is an example of a conversion ratio using feet and inches.
12 inches/1 foot

We can use this conversion ratio to convert feet to inches as follows:

feet x (12 inches/1 foot) = inches, but feet is cancelled with foot.  The middle ratio is labeled as the conversion ratio.

Notice that using the conversion ratio, "12 inches/1 foot," the starting units of "feet" cancel, leaving the ending units, "inches." Placing ratios correctly to cancel units is the second basic skill needed to solve ALL mole problems in chemistry.


Turning Conversion Ratios Upside Down in Order to Cancel Units


The next thing to know about using "conversion ratios" in chemistry is that you can turn ANY ratio "upside down" in order to get the answer you are seeking.

For example, the ratio "12 inches/1 foot," may be turned "upside down" as shown below.

Arrow pointing up with the words 'right side up' over the factor 12 inches/1 foot and an arrow pointing down with the words 'upside down' over 1 foot/12 inches. 

The second form of the ratio is used when we want to convert from inches to feet, instead of from feet to inches. We set up the structure to START with inches and END with feet, as shown below.

start units, inches, end units, feet.  inches x (?/?) = feet.

It is important to ALWAYS choose the correct form of the ratio to cancel the units we DON'T want and end up with the units we DO want. To convert inches to feet, which of the above forms of the ratio should we use?

comparison between the two possible set ups with conversion ratios in between.  which one will cancel out?  We have inches x (12 inches/1 foot) = feet  or  inches x (1 foot/12 inches) = feet.  Which one will cancel?  The first way is incorrect.  The second way is correct.

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Use ONLY Horizontal Fraction Bars

You also have to know is that it is FAR better to write all ratios in science with horizontal fraction bars, rather than diagonal bars. This keeps everything easy to see when you have a string of ratios in a problem.

30 miles/hour written with a horizontal fraction bar.  30 miles/hour written with a diagonal fraction bar, but crossed out to emphasize that we should only use horizontal fraction bars in chemistry from now on.

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showing a series of fractions, a/b, b/c, c/d, d/e, written with once with horizontal fraction bars and once with diagonal fraction bars.  It is obvious that the second writing leads to errors with long strings of factors as we have in chemistry.


(3) Be able to find any of the 4 ratios related to the mole listed below. (There are ONLY 4!)

Ratio #1: The Number of Particles in a Mole
(This value is always the same, regardless of the type of particle.)
6.022 x 10exp23 / mole
Ratio #2: The Volume of a Mole of Gas at STP
(This value is always the same, no matter which gas you have.)
22.4 Liters/mole

Ratio #3: The Molar Mass of a Specific Substance
(This value changes depending on which substance you have.)

? grams/mole of substance
Ratio #4: The Mole Ratio of Two Substances in a Chemical Equation
(This value changes depending on the coefficients in your chemical equation.)
? mole of substance A/? mole of substance B

 

 

 

 

 

 

 

 


Ratio # 1: The Number of Particles in a Mole

602,200,000,000,000,000,000,000, or 6.022 x 10^23 in scientific notation, is the number of particles in a mole. While anything "countable" may be considered a particle, in chemistry, particles usually means molecules, ions or electrons. Without going into explicit details here, the numerical value of this ratio is given BY DEFINITION based on experiments in the real world. This relationship may be written in one of two ways:

6.022 x 10exp23 particles/mole or 1 mole/6.022 x 10exp23 particles

And remember, it doesn't matter WHAT the particles are. The number of particles in a mole is ALWAYS the same.

Now let's use this ratio in solving two example problems. In the first few examples, I go into excruciating detail, but after you get a feel for this kind of problem solving, it goes very quickly and is really very EASY. (My assumption is that you already understand scientific notation and significant figures.)

Problem 1: Find the Number of Particles from Moles

You have 3.0 moles of substance. How many particles of substance do you have?

Analysis: You are given moles and are asked to find particles. So you start with units of moles and end up with units of particles in your answer. Before you even begin to worry about the numbers, be sure to place the starting and ending units where they need to go.

conversion set up showing starting units, moles, ending units, particles and a blank horizontal fraction bar in between.  moles x (?/?) = particles.

Notice that we placed the units where they go first, without even worrying about the numbers. We also left a blank space to write in the conversion ratio, once we determine what that is.

Next, we need to choose the correct version of the ratio between moles and particles to make the conversion work. Again, not worrying about numbers, concentrate on just finding the UNITS which will cancel out moles and leave us with particles. Below we see the two possible ways that units in the Conversion Ratio may be placed. Which one do you think is correct?

two alternate set ups.  A is moles x (particles/mole) = particles.  B is moles x (mole/particles) = particles.

If you chose possibility "A," you are correct. That placement allows us to cancel the unit "moles" and keep the unit "particles."

conversion set up.  moles x (particles/mole) = particles.  But the units of moles are cancelled, leaving us with particles in the end.

Now that our units are placed correctly, we can insert the numbers. According to the problem, we start with 3.0 moles of substance. The number of particles in a mole is constant, so our conversion factor is constant. (I have also placed grey numeral "1's" into the problem to help us remember that even when there are no numbers or units written, the number "1" is always there.)

Putting the numbers in, 3.0 moles x (6.022 x 10exp23 particles/1 mole) = ? particles.  The middle factor, 6.022 x 10exp23 particles/mole is labeled the conversion factor.

Our final steps are to check the unit cancellation, do the calculation, determine significant figures and put the answer in final form.

3.0 moles x (6.022 x 10esp23 particles/mole) = ? particles.  Units of moles cancel, leaving particles.  Now put in numbers.  Get 18.066 x 10exp23 particles =1.8066 x 10exp24 particles (rounded to 2 significant figures, because of 3.0 moles) = 1.8 x 10exp24 particles -- Final Answer

 

Problem 2: Find the Number of Moles from Particles

Instead of being given moles as in Problem 1, you are now given a number of particles and must convert back to the units of moles.

You have 2.56 x 10^25 particles of substance. Find out how many moles you have.

Analysis: You are given particles and are asked to find how many moles you have. So this time you start with particles and end up with units of moles in your answer. Follow the same pattern as before. Begin by placing your units first.

Conversion set up:  particles x (?/?) = moles.

Next, choose which version of the conversion ratio you need to cancel out the units you DON'T want and end up with the units you DO want.

Choose A or B:  A -- particles x (particles/mole) = moles    or    B -- particles x (mole/particles) = moles

In this case, if you chose possibility "B," you are correct. Particles will cancel and you will end up with moles, which is what we want.

 B -- particles x (mole/particles) = moles.  Units of particles are cancelled.

Now put the numbers in, cancel units, do the calculations, determine significant figures and report the answer in final form.

2.56 x 10exp25 particles x (1 mole/6.022 x 10exp23 particles) =  ? moles.  Units of particles are cancelled.   = 0.4251 x 10exp2 moles = 42.51 moles  (round to 3 significant figures, because of 2.56 x 10exp25 particles.)  = 42.5 moles - FINAL ANSWER.

 

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Review of the Four Steps to Solve Mole Problems

The number of particles in a mole is a constant, 6.022 x 10^23. This is ALWAYS true. In doing simple conversions back and forth between the number of particles and the mole, we can use the conversion ratio either "right side up" or "upside down," depending on what we're converting FROM and what we're converting TO. The steps to solving these kinds of problems are:

1. Place the starting and ending units in their correct locations.

2. Choose the correct conversion ratio to make unwanted units cancel.

3. Plug in the numbers and do the calculation.

4. Put the answer in final form. EASY!


Ratio # 2: The Volume of a Mole of Gas at STP

This quantity is used specifically ONLY for particles in the gaseous state. Its value is FIXED at 22.4 Liters per mole at STP (Standard Temperature and Pressure). STP was arbitrarily set and agreed upon by the scientific commmunity to mean 25 degrees C and 1 atmosphere pressure, the pressure at sea level. Note: At DIFFERENT temperatures and pressures, the volume occupied by one mole of gas is different and can be calculated using the laws of relationship known as the "gas laws." Nevertheless, for many calculations in chemistry, volume is measured and compared at STP, so this is a useful ratio. The ratio representing the volume of a mole of gas at STP may be written in either of two ways:

22.4 Liters/mole  or  1 mole/22.4 Liters

Problem 3: Find the Number of Liters from Moles

You have 4.0 moles of pure nitrogen gas at STP. How many Liters of gas do you have?

Analysis: You are given moles and are asked to find Liters. You start with units of moles and end up with units of Liters. Place your units as shown below.

moles x (?/?) = Liters

Choose the correct conversion ratio to make the unwanted units cancel.

Choose A or B:  A -- moles x (Liters/mole) = Liters  or  B -- moles x (mole/Liters) = Liters.

If you chose "A," you are correct. Next, insert the numbers, cancel units, do the calculation, round to the correct number of significant figures and report the answer in final form.

4.0 moles x (22.4 Liters/1 mole) = ? Liters.  Units of mole cancel.  = 89.6 Liters.  (Round to 2 significant figures, because of 4.0 moles.)  = 90 Liters.  (Must write the answer in scientific notation to preserve the fact that 0 (zero) in this case counts as a significant figure.)  = 9.0 x 10exp1 Liters --- FINAL ANSWER.

 

Problem 4: Find the Number of Moles from Liters

You have 49.7 Liters of oxygen gas at STP. How many moles do you have?

Analysis: You are given Liters and are asked to find moles. You start with units of Liters and end up with units of moles. You should place your beginning and ending units as shown below.

Liters x (?/?) = moles

From here, which conversion factor do you choose? The one which has Liters on top or moles on top? Choose the correct conversion factor, plug in the numbers and proceed through to the final answer.

49.7 Liters x (1 mole/22.4 Liters) = ? moles.  = 2.21875 moles.  (Round to 3 significant figures, because of both 49.7 and 22.4.)  = 2.22 moles  (Third digit is rounded up to 2 because of 8 being the next digit.)  = 2.22 moles - FINAL ANSWER.


The Secret to Solving ALL Mole Chemistry Problems

Hopefully by now you find it EASY to choose the correct conversion ratio to cancel the unwanted units and leave us with the units we want.

Here is the Secret! ALL mole chemistry problems you have from here on out are simply extensions of the skill of being able to find the correct conversion ratio which cancels unwanted units and leaves wanted units. EASY!

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Using Two Ratios in One Problem

Problem 5: Find the Number of Molecules from Liters.

In this problem, you will be using TWO conversion factors to get the answer.

How many molecules of hydrogen gas are in 54.6 Liters of pure hydrogen at STP?

Analysis: You are asked to start with Liters and end with molecules. (The clue that you end with molecules is given from the words "how many.") While we have no direct one-shot conversion between Liters and molecules, we DO have "Liters to moles" and "moles to molecules." So this time we can use TWO conversion factors to get our answer.

Set it up like this:

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

Next, find the conversion factors needed to do the conversion.

Conversion set up:  Liters x (mole/Liters) x (molecules/mole) = molecules.  Middle two ratios are labeled conversion factors.  Units of Liters cancel.  Units of mole cancel.

Finally, plug in the numbers and do the calculation.

54.6 Liters x (mole/22.4 Liters) x (6.022 x 10exp23 molecules/mole) =  ? molecules.  Units of Liters cancel.  Units of mole cancel.  = 14.678 x 10exp23 molecules.  = 1.4678 x 10exp24 molecules.  (Round to 3 significant figures, because of 54.6 Liters and 22.4 Liters.)  = 1.47 x 10exp24 molecules - FINAL ANSWER.

 

Notice that the entire strategy for solving problems like this, even with more than one conversion factor, is simply setting up the starting and ending units and then filling in the gaps with ratios either "right side up" or "upside down" to make unwanted units cancel and keep the units you want in the answer. EASY!


Ratio #3: The Molar Mass of a Specific Substance

Molar mass is different from the first two ratios, because it is NOT a single constant value which never changes. Although any given substance has the same unchanging, constant molar mass, each different substance has its own molar mass. Look at the three examples below.

A mole of hydrogen molecules (which are made up of two hydrogen atoms bonded together) has a mass of 2 grams.

2.0 grams/mole H2  or  1 mole H2/2.0 grams

A mole of helium gas has a mass of 4.0 grams.

4.0 grams/mole He  or  1 mole He/4.0 grams

A mole of water has a mass of 18.0 grams.

18.0 grams/mole H2O  or  1 mole H2O/18.0 grams

If you do not know how to find the molar mass of a substance, click here. [Not active yet.]

Now let's use molar mass to solve problems.

Problem 6: Find Moles from Grams.

How many moles of water are in 45.0 grams of water?

Analysis: The words "how many" point to the needed "ending units." The words "are in" indicate the "starting units."

Start units --> grams.  End units --> moles H2O.  Conversion set up:  grams x (?/?) = moles H2O

For water, the two possible ratios are:

18.0 g/mole or 1 mole/18.0 g

Now choose which of the above conversion ratios can be used to solve this problem.

Starting units are grams.  Ending units are moles H2O.

Hopefully you chose the conversion factor shown below.

Conversion factor units are mole H2O over grams.

Finally, write in the numbers, do the calculation, determine significant figures and report the answer in final form.

45.0 grams x (1 mole H2O/18.0 grams) = ? moles H2O.  Units of grams cancel.  = 2.50 moles H2O -- FINAL ANSWER.

 

Problem 7: Find Grams from Moles.

How many grams of diatomic hydrogen gas are in 5.7 moles of hydrogen?

Analysis: The words "how many" point to the ending units. The words "are in" point to the starting units.

Start by setting up the units

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

Choose the correct conversion ratio.

2.0 grams/mole H2  or  1 mole H2/2.0 grams

Write in the numbers, cancel units, do the calculation, determine significant figures and report the answer in final form.

5.7 moles H2 x (2.0 grams/1mole H2) = ? grams.  Units of mole cancel.  = 11.4 grams.  (2 significant figures, because of 5.7 and 2.0)   = 11 grams - FINAL ANSWER.


Using Two Ratios in One Problem
(Part 2)

Now let's use molar mass ratios along with the first two molar ratios to solve problems.

Problem 8: Find Liters at STP from Grams.

How many Liters of space will 25.3 grams of methane gas occupy at STP?

Analysis: The words "how many" indicate the ending units. What are the starting units? And what are the conditions? (STP) The condition of STP is important, because the conversion ratio of gas volume at STP is valid ONLY at standard temperature and pressure. That's why is must be stated explicitly in the problem.

Set up the problem.

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

Find and choose the correct conversion ratios.

Two sets of possible conversion factors:  First set = 16.0 grams/mole CH4  or  1 mole CH4/16.0 grams.  Second set = 22.4 Liters/mole  or  1 mole/22.4 Liters.

Put in the numbers, cancel units, do the calculation, determine correct number of significant figures and report the final answer.

25.3 grams x (1 mole CH4/16.0 grams) x (22.4 Liters/mole) = ? Liters.  Units of grams cancel.  Units of mole CH4 cancel.  = 35.42 Liters CH4.  (Round to 3 significant figures, because of 25.3, 16.0 and 22.4.)  = 35.4 Liters CH4 - FINAL ANSWER.

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Ratio #4: The Mole Ratio of Two Substances in a Chemical Equation

This last relationship using moles often confuses chemistry students, because it changes all the time depending on the specific problem we are considering. We find the mole ratio, or mole-mole ratio, from a balanced 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 1 molecule 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  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.

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  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, so we must 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) Next, 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  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  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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Page updated November 16, 2010 You may email Lynda directly at chembyrd@yahoo.com