### Mass-Mass Problems

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 samples below.

Here is an image of the steps involved in solving mass-mass problems. It is offered without comment.

As you can see, the bottom portion includes mass-volume problems. These type problems are not discussed in this file, but in another.

Example #1: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3 by this reaction:

2AuCl3 ---> 2Au + 3Cl2

Solution:

1) The provided equation is balanced correctly.

2) The conversion of moles of AuCl3 to grams:

$\frac{\mathrm{64.0 g}}{\mathrm{303.32 g/mol}}$ = 0.210998 mol of AuCl3

I picked AuCl3 to convert from grams to moles because a gram amount of AuCl3 was provided in the problem. The ChemTeam has heard many variations of this:

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

3) Use a molar ratio involving AuCl3 and Cl2:

$\frac{\mathrm{AuCl3}}{\mathrm{Cl2}}$

$\frac{2}{3}$ = $\frac{\mathrm{0.210998 mol}}{x}$ <--- the 2/3 comes from the coefficients of the balanced equation

x = 0.316497 mol of Cl2

This is the hardest step. Constructing the proper ratio and proportion appears to be hard to understand.

4) Convert 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 #2: 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) Let us convert grams of AlCl3 to moles:

$\frac{\mathrm{200. g}}{\mathrm{133.341 g/mol}}$ = 1.499914 mol of AlCl3 <--- look for the substance has a gram amount associated with it

2) Use a molar ratio involving AgCl and AlCl3:

$\frac{\mathrm{AgCl}}{\mathrm{AlCl3}}$

$\frac{3}{1}$ = $\frac{x}{1.499914 mol}$

x = 4.499742 mol of AgCl

The 'x' in the right-hand ratio comes from the substance we are trying to calculate an amount for. Look for phrases like "Calculate the mass of . . ." or "Determine the mass of . . . "above

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:

$\frac{1}{3}$ = $\frac{1.499914 mol}{x}$

Example #3: 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. You'll get unbalanced equations soon enough.

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

$\frac{\mathrm{30.0 g}}{\mathrm{165.998 g/mol}}$ = 0.180725 mol

3) Construct a ratio and proportion:

This ratio:
$\frac{2}{1}$

comes from the coefficients of the balanced equation. This ratio:

$\frac{0.180725 mol}{x}$

comes from a consideration of the data in the problem. The ratio and proportion to solve is this:

$\frac{2}{1}$ = $\frac{0.180725 mol}{x}$

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.

4) Convert moles to grams:

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

Example #4: 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:

$\frac{\mathrm{92.0 g}}{\mathrm{26.98 g/mol}}$ = 3.4099 mol

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

$\frac{\mathrm{Al}}{\mathrm{Al\left(NO3\right)3}}$

$\frac{2}{2}$ = $\frac{3.4099 mol}{x}$

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

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 #5: 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:

$\frac{\mathrm{100.0 g}}{\mathrm{70.906 g/mol}}$ = 1.41032 mol

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

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

$\frac{3}{2}$ = $\frac{\mathrm{1.41032 mol}}{x}$

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