### Stoichiometry - AP levelProblems #1-10

Problem #1: A salt contains only barium and one of the halide ions. A 0.1480 g sample of the salt was dissolved in water and an excess of sulfuric acid was added to form barium sulfate, which was filtered, dried and weighed. Its mass was found to be 0.1660 g. What is the formula for the barium halide?

Solution:

1) Calculate mass of barium ion in precipitate:

0.166 g x (137.33 / 233.395) = 0.097674672 g

2) Determine mass of halide ion in dissolved sample:

0.1480 g − 0.097674672 g = 0.050325328 g

3) How many moles of barium ion were present in dissolved sample?

0.097674672 g / 137.33 g mol¯1 = 0.000711241 mol

4) Barium halide compounds are known to take the formula BaX2. How many moles of halide were present in the dissolved sample?

0.000711241 mol x 2 = 0.001422481 mol

5) Which halide has 0.001422 mole of it weigh 0.05032 grams?

F: 0.001422481 mol times 18.9984 g mol¯1

Cl: 0.001422481 mol times 35.453 g mol¯1

Br: 0.001422481 mol times 79.904 g mol¯1

I: 0.001422481 mol times 126.90447 g mol¯1

6) BaCl2

Problem #2: A 4.000 g sample of M2S3 is converted to MO2 and loses 0.277 g. What is the atomic weight of M?

Solution:

1) Write the chemical equation:

M2S3 + 5O2 ---> 2MO2 + 3SO2

2) Some facts that I can't think of a good title for:

a) The grams of M in M2S3 equals the grams of M in 2MO2. (Notice the inclusion of the coefficient.)

b) Let x = the atomic weight of M.

3) Construct gravimetric factors for M:

M2S3 ⇒ 2x / (2x + 96)

2MO2 ⇒ 2x / (2x + 64)

Comment: there are 2 M and 4 O in 2MO2.

4) Calculate grams of M2O3 and 2MO2:

M2S3 ⇒ (4.000 g) times [2x / (2x + 96)]

2MO2 ⇒ (3.723 g) times [2x / (2x + 64)]

5) Set them equal to each other and solve:

(4.000 g) times [2x / (2x + 96)] = (3.723 g) times [2x / (2x + 64)]

You do the math.

x = 183

In addition to the alternate solution, there are three additional forms of this problem, two which have solutions appended.

Problem #3a: A 5.000 gram sample of a dry mixture of potassium hydroxide, potassium carbonate and potassium chloride is reacted with 0.100 L of 2.00-molar HCl solution. A 249.0 mL sample of dry carbon dioxide gas, measured at 22.0 °C and 740.0 torr, is obtained from this reaction. What was the percentage of potassium carbonate in the mixture?

Solution to 3a:

1) Calculate moles of CO2 using PV = nRT: 2) Recall the reaction between K2CO3 and HCl:

K2CO3 + 2HCl ---> 2KCl + CO2 + H2O

3) Note the 1:1 molar ratio between K2CO3 and CO2. From this we conclude:

0.0100153 mol K2CO3 present in the dry mixture

4) Calculate grams K2CO3, then its mass percentage:

0.0100153 mol x 138.2057 g/mol = 1.384 g

(1.384 g / 5.000 g) x 100 = 27.68%

Problem #3b: The excess HCl in problem 3a was found by titration to be chemically equivalent to 86.60 mL of 1.50-molar sodium hydroxide. What was the percentage of KOH and of KCl in the original mixture?

Solution to 3b:

1) Calculate excess HCl:

We know this: HCl + NaOH ---> NaCl + H2O

There is a 1:1 molar ratio between NaOH and HCl

Use NaOH data to determine moles of excess HCl:

(0.08660 L) (1.50 mol L¯1) = 0.1299 mol

2) Calculate total moles of HCl in the 0.100 L of 2.00-molar HCl solution:

(0.100 L) (2.00 mol L¯1) = 0.200 mol

3) Calculate amount of HCl used:

0.200 mol − 0.1299 mol = 0.0701 mol

However, this is the combined amount of HCl used to titrate K2CO3 AND KOH

4) Calculate HCl used to titrate KOH by subtracting the HCl used to titrate K2CO3:

0.0701 mol − 0.0200 mol = 0.0501 mol

The 0.02 comes from the fact that 2 HCl were required to neutralize one K2CO3. See step 2 in problem 3a.

5) Calculate grams of KOH

KOH + HCl ---> KCl + H2O

From the 1:1 ratio between the reactants, we know there are 0.0501 mol of KOH in the original sample.

0.0501 mol x 56.1057 g mol¯1 = 2.811 g

6) Determine KCl by subtraction:

1.384 g + 2.811 g = 4.195 g

5.000 g − 4.756 g = 0.805 g

7) You may do the weight percentages for KOH and KCl on your own.

Problem #4a: For the reaction below, when 0.5000 g of XI3 reacts completely, 0.2360 g of XCl3 is obtained. Calculate the atomic weight of element X and identify it.

2XI3 + 3Cl2 ---> 2XCl3 + 3I2

Solution:

1) From the balanced equation, we know that the moles XI3 used equals moles XCl3 produced. Therefore:

0.5000 g / (x + 381 g/mol) = 0.2360 g / (x + 106.5 g/mol)

The 381 is the weight of three iodines and the 106.5 is the weight of three chlorines.

2) Solving for x, we find it equal to 138.9. This is the atomic weight of lanthanum.

Problem #4b: If 0.520 grams of XCl3 are treated with iodine, 0.979 g of XI3 are produced. What is the chemical symbol for this element?

2XCl3 + 3I2 ---> 2XI3 + 3Cl2

Solution:

1) From the balanced equation, we know that the moles XCl3 used equals moles XI3 produced. Therefore:

0.520 g / (x + 106.5 g/mol) = 0.979 g / (x + 381 g/mol)

The 106.5 is the weight of three chlorines and the 381 is the weight of three iodines.

2) Solving for x, we find it equal to 204.5. This is the atomic weight of thallium and its symbol is Tl.

Problem #4c: Consider the reaction involving unknown element X:

F2 + 2XBr ---> Br2 + XF

When 5.500g of XBr reacts, 3.693g of Br2 is produced. Identify element X.

Solution:

1) Moles of Br2:

3.693 g / 159.808 g/mol = 0.023109 mol

2) Moles of XBr:

0.023109 mol times 2 = 0.046218 mol

3) Molar mass of XBr:

5.500 g / 0.046218 mol = 119.0 g/mol

4) Identity of X

119.0 − 79.904 = 39.0

X is potassium

Problem #5: A 2.077 g sample of an element, which has an atomic mass between 40 and 55, reacts with oxygen to form 3.708 g of an oxide. Determine the formula mass of the oxide (and identify the element).

Solution:

1) Determine moles of oxygen involved:

3.708 g − 2.077 g = 1.631 g

1.631 g / 15.9994 g/mol = 0.10194 mol of oxygen in MxOy.

2) Determine the atomic weight of M, if the formula is MO:

2.077 g / 0.10194 mol = 20.37 g/mol

The formula of MxOy is not MO since the atomic weight of M is known to be between 40 and 55.

3) Determine the atomic weight of M, if the formula is MO2:

2.077 g / 0.05097 mol = 40.75 g/mol

This is within the acceptable range.

Before going on, I would like to point out that a M2O3 formula leads to an atomic weight of approximately 34 and a M2O formula leads to approxmately 10.2. You may do the math on those two possibilities.

4) The closest value to our desired range is potassium and yes, it does form the compound KO2, known as potassium superoxide.

If we were to look for a +4 forming ion (in other words, something to satisfy the MO2 requirement) in the 40-55 range, we find titanium. However, its atomic weight is about 48, which is too high a value predicted by the MO2 formula.

Only KO2 provides an atomic weight within the parameters specified by the problem.

Problem #6: A 12.5843 g sample of ZrBr4 was dissolved and, after several steps, all of the combined bromine was precipitated as AgBr. The silver content of AgBr was found to be 13.2160 g. Assume the atomic masses of silver and bromine to be 107.868 and 79.904. What value was obtained for the atomic mass of Zr from this experiment?

Solution:

1) calculate the moles Ag in the AgBr:

13.2160 g / 107.868 g/mol = 0.12252 mole of Ag in the AgBr

2) Since AgBr has a 1:1 molar ratio of silver to bromine, Br in sample is 0.12252 mole. Calculate the grams Br in the sample:

0.12252 g times 79.904 g/mol = 9.78985 g of Br

3) calculate Zr in sample:

12.5843 − 9.78985 = 2.79445 g of Zr

4) determine moles of Zr present:

The 0.12252 mol of Br represents the 4 moles of Br in the formula ZrBr4.

Therefore, 0.12252 / 4 equals the moles of Zr present.

moles Zr = 0.030635 mol

5) Determine molecular weight of Zr:

2.79445 g / 0.030635 mol = 91.2 g/mole

Problem #7: Two different chloride compounds of platinum are known, compound X and Y. When 3.45 g of compound X is heated, 2.72 g of compound Y is formed along with some chlorine gas. Upon further heating, the 2.72 g of compound Y is decomposed to 1.99 g of platinum metal and some more chlorine gas. Determine the formulas of compounds X and Y.

Solution:

1) Determine how much Cl2 is produced in each reaction:

Heat 3.45 g X to get 2.72 g Y and Cl2 gas
Mass of Cl2 gas is 3.45 − 2.72 = 0.73 g

Heat 2.72 g Y to get 1.99 g Pt and Cl2 gas
Mass of Cl2 gas in compound Y is 2.72 − 1.99 = 0.73 g

2) Determine formula for compound Y:

moles of Pt atom = 1.99 g / 195.1 g/mol = 0.01 mol
moles of Cl atom (not chlorine molecules!) = 0.73 g / 35.45 g/mol = 0.02 mol

simplest mole ratio:
Pt = 0.01 / 0.01 = 1
Cl = 0.02 / 0.01 = 2

formula of Y = PtCl2

3) Determine formula for compound X:

mass of Pt = 1.99 g
mass of Cl2 = 0.73 g from 1st decomposition + 0.73 g from 2nd decomposition = 1.46 g total in X

mole of Pt atom = 1.99 g / 195.1 g/mol = 0.01 mol
mole of Cl atom (not chlorine molecules!) = 1.46 g / 35.45 g/mol = 0.04 mol

simplest mole ratio:
Pt = 0.01 / 0.01 = 1
Cl = 0.04 / 0.01 = 4

Formula of X = PtCl4

Problem #8: The active ingredients of an antacid tablet contained only magnesium hydroxide and aluminum hydroxide. Complete neutralization of a sample of the active ingredients required 48.5 mL of 0.187 M hydrochloric acid. The chloride salts from this neutralization were obtained by evaporation of the filtrate from the titration; they weighed 0.4200 g. What was the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet?

Solution:

1) Determine moles of chloride ion used:

(0.187 mol L¯1) (0.0485 L) = 9.0685 x 10¯3 mol

2) Detemine moles of chloride ion used to make the MgCl2:

9.0685 x 10¯3 mol x 0.40 = 3.6274 x 10¯3 mol

3) Explanation of the 0.40:

Out of every 5 Cl¯ used, three go to make one AlCl3 and two go to make one MgCl2

2 out of 5 is 40%.

4) Determine moles of MgCl2 that are present:

3.6274 x 10¯3 mol / 2 = 1.8137 x 10¯3 mol

Remember, one MgCl2 is present for every two Cl¯

5) Repeating (2), (3) and (4) with appropriate modifications yields 1.8137 x 10¯3 mol for AlCl3. (Use 0.60, not 0.40 and divide by 3, not 2.)

6) Determine moles of Mg(OH)2 and Al(OH)3 in original mixture:

First, write the equations which produce MgCl2 and AlCl3
Mg(OH)2 + 2HCl ---> MgCl2 + 2H2O

Al(OH)3 + 3HCl ---> AlCl3 + 3H2O

Second, note the following molar ratios:

Mg(OH)2 : MgCl2 is 1 : 1

Al(OH)3 : AlCl3 is 1 : 1

This means we have the following molar amounts in the original sample:

Mg(OH)2 = 1.8137 x 10¯3 mol

Al(OH)3 = 1.8137 x 10¯3 mol

7) Determine grams of Mg(OH)2 and Al(OH)3:

This is left to the reader.

8) Determine percent by mass of Mg(OH)2 in the original sample.

This is left to the reader.

Problem #9: When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of FeO and Fe2O3. In a certain experiment, 20.00 g of iron metal was reacted with 11.20 g of oxygen gas. After the experiment the iron was totally consumed and 3.56 g oxygen gas remained. Calculate the amounts of FeO and Fe2O3 formed in this experiment.

Solution:

1) Determine the amount of oxygen gas (in grams, then moles) consumed:

11.20 g − 3.56 g = 7.64 g of O2

7.64 g / 31.9988 g mol¯1 = 0.238759 mol (I kept several guard digits.)

2) Determine how many moles of O2 goes to form FeO and how many moles goes to form Fe2O3:

2Fe + O2 ---> 2FeO
4Fe + 3O2 ---> 2Fe2O3

25% goes to form FeO
75% goes to form Fe2O3

0.238759 mol x 0.25 = 0.05968974 mol of O2 used to form FeO
0.238759 x 0.75 = 0.17906922 mol of O2 used to form Fe2O3

3) Determine moles of FeO and moles of Fe2O3 formed:

FeO ⇒ the O2 to FeO ratio is 1:2, therefore double the amount of O2 used to get FeO produced:
0.1193795 mol of FeO produced.

Fe2O3 ⇒ the O2 to Fe2O3 ratio is 3:2, therefore double the amount of O2 used to get Fe2O3 produced and then divide that value by three:

0.1193795 mol of Fe2O3 produced.

4) Determine grams of FeO and grams of Fe2O3 formed:

FeO ⇒ 0.1193795 mol x 71.844 g mol-1 = 8.5767 g
Fe2O3 ⇒ 0.1193795 mol x 159.687 g mol-1 = 19.06335 g

The above values (you may round them off on your own) are the answer to the problem, but I thought one more step would be fun.

5) Determine grams of Fe in FeO and grams of Fe in Fe2O3:

FeO ⇒ 8.5767 g x (55.845 / 71.844) = 6.667 g of Fe in FeO
Fe2O3 ⇒ 19.06335 g x (111.69 / 159.687) = 13.33

This, within rounding errors, totals to the 20.00 g of Fe mentioned in the problem.

Problem #10: 0.197 g of magnesium is burned in air:

2Mg + O2 ---> 2MgO

However, some of the magnesium reacts with nitrogen in the air to form magnesium nitride instead:

3Mg + N2 ---> Mg3N2

So you have a mixture of MgO and Mg3N2 weighing 0.315 g. Determine what percentage of the Mg formed the nitride in the initial reaction.

Solution:

1) This problem is solved with two simultaneous equations in two unknowns:

First equation: x + y = 0.315 g

Second equation: (24.305 / 40.304) x + (72.915 / 100.929) y = 0.197 g

Explanation:

x = the mass of MgO in the mixture of MgO and Mg3N2
y = the mass of Mg3N2 in the mixture of MgO and Mg3N2

(24.305 / 40.304) is the percentage of Mg in MgO
(72.915 / 100.929) is the percentage of Mg in Mg3N2

2) Substitute x = 0.315 − y into the second equation:

(24.305 / 40.304) (0.315 − y) + (72.915 / 100.929) y = 0.197

3) Solve:

0.189958 − 0.60304y + 0.72244y = 0.197

0.1194y = 0.007042

y = 0.058978 g

4) Percent of Mg3N2 in original sample:

(0.058978 g / 0.315 g) x 100 = 18.7% (to three sf)