### StoichiometryMass-Mass Examples

This is the most common type of stoichiometric problem in high school.

There are four steps involved in solving these problems:

1. Make sure you are working with a properly balanced chemical equation.
2. Convert grams of the substance given in the problem to moles.
3. Construct two ratios - one from the problem and one from the chemical equation and set them equal. The ratio from the problem will have an unknown, 'x.' Solve for "x."
4. Convert moles of the substance just solved for into grams.

1. Double check the equation. The ChemTeam has seen lots of students go right ahead and solve using the unbalanced equation supplied in the problem (or test question for that matter).
2. DON'T use the same molar mass in steps two and four. Your teacher is aware of this and, on a multiple choice test, will provide the answer arrived at by making this mistake. You have been warned!
3. Don't multiply the molar mass of a substance by the coefficient in the problem BEFORE using it in one of the steps above. For example, if the formula says 2H2O in the chemical equation, DON'T use 36.0 g/mol, use 18.0 g/mol.
4. Don't round off until the very last answer. In other words, don't clear your calculator after step two and write down a value of 3 or 4 significant figures to use in the next step. Round off only once after all calculations are done.

## STOP!!!

Go back to the start of this file and re-read it. Notice that I give four steps (and some advice) in how to solve the example problems just below. My advice is to keep going back to those steps as you examine the examples below.

Example #1: How many grams of hydrogen gas are needed to react completely with 54.0 g of oxygen gas, given the following unbalanced chemical reaction:

H2 + O2 ---> H2O

Solution:

1) Balance the chemical equation:

2H2 + O2 ---> 2H2O

2) Convert grams of the substance given:

54.0 g / 32.0 g/mol = 1.6875 mol of O2

Note the use of 32.0 and not 16.0. The chemical substance is O2. Students have been known to sometimes forget to write the subscript of 2 on a diatomic element (H2, N2, O2, F2, Cl2, Br2, I2)

3) Construct two molar ratios and set them equal to each other. The first molar ratio is from the coefficients of the balanced chemical equation. The two substances are:

 H2 ––– O2

and the numerical ratio is this:

 2 ––– 1

4) The second ratio is found within the problem statement. The H2 is our unknown because the problem says "how many grams of hydrogen" and the O2 mole amount is the other value. Like this:

 x ––––– 1.6875

I left the mol unit off for convenience. Note also that I did not round off. I'll do that at the end.

5) We need to set the two ratios equal to each other and solve:

 2 x ––– = ––––– 1 1.6875

x = 3.375 mol of H2 required

6) Convert the calculated moles from step #3 into grams:

(3.375 mol) (2.016 g/mol) = 6.80 g (to three sig figs)

Note: if you did not balance the equation, you'd wind up using an incorrect 1:1 molar ratio rather than the correct 2:1 ratio.

Example #2: How many grams of hydrogen gas are needed to produce 105.0 grams of water, given the following unbalanced chemical reaction:

H2 + O2 ---> H2O

Solution:

1) Balance the chemical equation:

2H2 + O2 ---> 2H2O

2) Convert grams of the substance given:

105.0 g / 18.015 g/mol = 5.82848 mol of H2O

I rounded off some, but I made sure to keep more digits than what I will round off to at the end.

3) Construct two molar ratios and set them equal to each other. The two substances in in the first ratio are these:

 H2 –––– H2O

and the numerical ratio from the coefficients of the chemical equation is this:

 2 ––– 2

4) The second ratio comes from information in the problem:

 x ––––– 5.82848

5) Setting equal and solving:

 2 x ––– = ––––––– 2 5.82848

x = 5.82848 mol of H2 required

Note: this can be an area of confusion. Since the ratio is a 1:1 ratio, the answer of 5.82848 mol is arrived at easily. However, many students will forget that the 5.82848 mol answer is now that of the OTHER substance, the hydrogen.

It seems that, because the number (the 5.82848) didn't change, the substance didn't change. Consequently, the student will enter the next (and last) step thinking the 5.82848 still refers to water.

6) Convert moles to grams:

(5.82848 mol) (2.016 g/mol) = 11.75 g of H2 (to four sig figs)

Example #3: How many grams of hydrogen gas are needed to produce 85.2 grams of ammonia, given the following unbalanced chemical reaction:

N2 + H2 ---> NH3

Solution:

1) Balance the chemical equation:

N2 + 3H2 ---> 2NH3

2) Convert the given grams to moles:

85.2 g / 17.0307 g/mol = 5.00273 mol

3) Construct two molar ratios and set them equal to each other. The two substances in our ratios are these:

 H2 –––– NH3

4) The two ratios set equal to each other are:

 3 x ––– = ––––––– 2 5.00273

x = 7.504095 mol of H2

5) Convert the calculated moles to grams:

(7.504095 mol) (2.016 g/mol) = 15.8 g (to three sig figs)

Example #4: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3 by this reaction:

AuCl3 ---> Au + Cl2

Solution:

1) The provided equation must be balanced correctly:

2AuCl3 ---> 2Au + 3Cl2

2) Convert grams of AuCl3 to moles:

Let x = the moles of AuCl3

 64.0 g x = –––––––––––– 303.32 g/mol

x = 0.210998 mol of AuCl3

The ChemTeam has heard many variations of this:

"But how did you know to convert grams of AuCl3 to moles?"

I picked AuCl3 to convert from grams to moles because a gram amount of AuCl3 was provided in the problem.

3) Use two molar ratio involving AuCl3 and Cl2:

 AuCl3 ––––– Cl2

4) The two molar ratios set equal to each other:

 2 0.210998 ––– = ––––––– 3 x

x = 0.316497 mol of Cl2

This is the hardest step. Constructing the proper ratio and proportion can cause a great deal of confusion.

5) Convert the calculated moles to grams:

(0.316497 mol) (70.906 g/mol) = 22.4 g (to three sig figs)

One question I often get is "Where did the value of 303.32 come from?" Answer - it's the molar mass of AuCl3. Keep this answer in mind as you wonder about where other numbers come from in a given solution.

You might also want to consider looking at the solution to the problem and try to fit it to the list of steps given above. I know what I am suggesting is horrible and very mean, but then, I'm a teacher. What the heck do I know?

Example #5: Calculate the mass of AgCl that can be prepared from 200. g of AlCl3 and sufficient AgNO3, using this equation:

3AgNO3 + AlCl3 ---> 3AgCl + Al(NO3)3

Solution:

1) Since the chemical equation is already balanced, let us convert grams of AlCl3 to moles:

 200. g –––––––––––– = 1.499914 mol of AlCl3 133.341 g/mol

I picked AlCl3 because it was the substance has a gram amount associated with it in the problem.

2) Use a proportion with molar ratios involving AgCl and AlCl3:

 AgCl ––––– AlCl3

 3 x ––– = ––––––– 1 1.499914

x = 4.499742 mol of AgCl

The 'x' in the right-hand ratio is associated with the substance we are trying to calculate an amount for (the AgCl). Look for phrases like "Calculate the mass of . . ." or "Determine the mass of . . . " in the problem statement.

3) Convert moles to grams:

(4.499742 mol) (143.323 g/mol) = 645 g (to three sig figs)

By the way, what if you had used the ratio of 1 over 3, with the AlCl3 value in the numerator? Then, the other ratio would have been reversed and the answer would have been the same. The ratio and proportion would have looked like this:

 1 1.499914 ––– = ––––––– 3 x

Example #6: Given this equation:

2KI + Pb(NO3)2 ---> PbI2 + 2KNO3

calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2

Solution:

1) The equation is balanced. Sometimes you're given an unbalanced equation on the test when all the classroom examples used already-balanced equations. Make sure you do these problems with a balanced chemical equation.

2) We are given 30.0 g of KI. Change it to moles:

 30.0 g –––––––––––– = 0.180725 mol of KI 165.998 g/mol

3) Construct a ratio and proportion:

This ratio:
 2 ––– 1

comes from the coefficients of the balanced equation.

This ratio:

 0.180725 –––––––– x

comes from a consideration of the data in the problem.

Setting the two ratios equal to each other gives us the proportion to solve:

 2 0.180725 –– = –––––––– 1 x

x = 0.0903625 mol <--- this is moles of PbI2

The substance associated with the 'x' is not the one for which the grams are given in the problem statement. The 'x' is associated with the substance for which a phrase like "Determine how much . . ." is used.

Notice that a third substance (the Pb(NO3)2) is mentioned, but the word excess is used to describe it. As you learn more about stoichiometry, the excess substance will be brought into the calculations. Not yet, however. Look for it in a section called 'limiting reagent.'

4) Convert moles to grams:

(0.0903625 mol) (461.01 g/mol) = 41.6 g (to three sig figs)

Example #7: If 92.0 g of aluminum is produced, how many grams of aluminum nitrate reacted?

Al(NO3)3 + Mg ---> Mg(NO3)2 + Al

Solution:

1) An unbalanced equation was given in the problem. It needs to be balanced:

2Al(NO3)3 + 3Mg ---> 3Mg(NO3)2 + 2Al

2) Grams of aluminum is given. Convert it to moles:

 92.0 g –––––––––– = 3.4099 mol of Al 26.98 g/mol

3) Use a ratio and proportion involving aluminum and aluminum nitrate:

 Al –––––––– Al(NO3)3
 2 3.4099 –– = ––––––– 2 x

x = 3.4099 mol <--- this is moles of Al(NO3)3, NOT moles of Al

Warning: there will be a real temptation in the next step to use the wrong molar mass

4) Determine grams of the unknown, the aluminum nitrate:

(3.4099 mol) (212.994 g/mol) = 726 g (to three sig figs)

It is quite common in a problem like this for the student to use the molar mass of Al in this step. I think it is because they see the same value (the 3.4099 mol) in this step as in the second step. The conclusion is that it must be the same substance. And that is in error.

In the second step, we had 3.4099 mol of aluminum, but after solving the ratio and proportion, we now have 3.4099 mol of aluminum nitrate.

Be careful on the point, especially if the amount you got at the end equals the amount you had at the beginning (the 92 grams).

Example #8: How many grams of AuCl3 can be made from 100.0 grams of chlorine by this reaction:

2Au + 3Cl2 ---> 2AuCl3

Solution:

1) The equation is balanced. Yay!

2) 100.0 g of chlorine is given in the problem. Convert it to moles:

 100.0 g –––––––––– = 1.41032 mol of Cl2 70.906 g/mol

Notice that the element chlorine is diatomic. Students sometimes forget to write the seven diatomics with the subscripted two. The seven diatomics are: H2, N2, O2, F2, Cl2, Br2, I2

3) The ratio and proportion will involve Cl2 and AuCl3:

 3 1.41032 mol –– = ––––––––––– 2 x

x = 0.940213 mol

Notice that the values associated with chlorine (3 and 1.41032) are in the numerator and the values associated with gold(III) chloride (2 and x) are in the denominator. If you were to flip one ratio, you'd have to flip the other.

4) Convert moles of AuCl3 to grams:

(0.940213 mol) (303.329 g/mol) = 285 g

Example #9: Aluminum foil 1.00 cm square and 0.540 mm thick react with bromine to form aluminum bromide. (a) How many grams of bromine were consumed? (b) How many grams of aluminum bromide were produced?

Solution:

1) Let us determine the mass, then moles, of Al present:

volume of Al foil ---> (1.00 cm) (1.00 cm) (0.0540 cm) = 0.0540 cm3
Note the change of mm to cm.

mass of Al ---> (2.70 g/cm3) (0.0540 cm3) = 0.1458 g

Note the use of the density of aluminum.

moles of Al ---> 0.1458 g / 26.98154 g/mol = 0.0054037 mol

2) The equation for the reaction is this:

2Al + 3Br2 ---> 2AlBr3

The Al to Br2 molar ratio of 2:3 will be used to answer (a). The Al to AlBr3 molar ratio of 2:2 will be used to answer (b).

3) Use the Al to Br2 molar ratio to determine moles of Br2 consumed:

 2 0.0054037 mol –––– = ––––––––––––– 3 x

x = 0.00810555 mol (of Br2)

4) Determine grams of Br2:

(0.00810555 mol) (159.808 g/mol) = 1.30 g (to three sig figs)

5) Use the Al to AlBr3 molar ratio to determine moles of AlBr3 produced:

 2 0.0054037 mol –––– = ––––––––––––– 2 x

x = 0.0054037 mol (of AlBr3)

6) Determine grams of AlBr3:

(0.0054037 mol) (266.694 g/mol) = 1.44 g (to three sig figs)

Example #10: How many grams of oxygen are in a sample of Ca3(PO4)2 that contains 66.0 g of calcium?

Comment: stoichiometric problems are usually of the "I have one chemical substance, how much of another chemical substance"? variety. But, they don't have to be. Here is an example of a mass-mass stoichiometric problem based on the relationships within one chemical substance.

Solution:

1) Determine moles of calcium:

66.0 g / 40.078 g/mol = 1.6468 mol

2) Determine moles of oxygen in the sample, based on a 3:8 ratio between Ca and O:

 3 8 ––––––––– = ––– 1.6368 mol x

x = 4.3648 mol

3) Determine mass of oxygen:

(4.3648 mol) (16.00 g/mol) = 69.8 g

Bonus Example: Solid lithium hydroxide is used in space vehicles to removed exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. (a) What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium? (b) At STP, what is the volume of CO2 produced?

Solution:

1) Write the balanced chemical equation for the described reaction:

2LiOH + CO2 ---> Li2CO3 + H2O

2) However, there is a possible problem. The question asks for 1.00 kg of lithium, not lithium hydroxide. We need to know the molar relationship between Li and CO2. So, let's make LiOH from Li:

2Li + 2H2O ---> 2LiOH + H2

3) If I add the two reactions, I obtain this:

2Li + CO2 + H2O ---> Li2CO3 + H2

Note that two LiOH and one H2O cancel out. This third reaction gives me the Li to CO2 as 2 to 1, so I am now ready to continue on.

4) Determine moles of Li that react:

1000 g / 6.941 g/mol = 144.07 mol

5) Using the 2:1 molar ratio, I can determine the moles of CO2 consumed:

 2 144.07 mol ––– = ––––––––– 1 x

x = 72.035 mol (of CO2)

6) Convert moles to grams to get the answer for (a):

(72.035 mol) (44.009 g/mol) = 3170 g

7) To determine the volume at STP, we can use either PV = nRT or molar volume:

PV = nRT
(1.00 atm) (V) = (72.035 mol) (0.08206 L atm / mol K) (273.15 K)

V = 1614.6 L (to three sig figs, this would be 1610 L)

molar volume

(22.414 L/mol) (72.035 mol) = 1614.6 L (1610 L to three sig figs)