Problems #1-10

All Examples & Problems (no solutions)

**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 BaX_{2}. 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) BaCl_{2}

**Problem #2:** A 4.000 g sample of M_{2}S_{3} is converted to MO_{2} and loses 0.277 g. What is the atomic weight of M?

**Solution:**

1) Write the chemical equation:

M_{2}S_{3}+ 5O_{2}---> 2MO_{2}+ 3SO_{2}

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

a) The grams of M in M_{2}S_{3}equals the grams of M in 2MO_{2}. (Notice the inclusion of the coefficient.)b) Let x = the atomic weight of M.

3) Construct gravimetric factors for M:

M_{2}S_{3}⇒ 2x / (2x + 96)2MO

_{2}⇒ 2x / (2x + 64)Comment: there are 2 M and 4 O in 2MO

_{2}.

4) Calculate grams of M_{2}O_{3} and 2MO_{2}:

M_{2}S_{3}⇒ (4.000 g) times [2x / (2x + 96)]2MO

_{2}⇒ (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 #3:** 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:**

1) Calculate moles of CO_{2} using PV = nRT:

2) Recall the reaction between K_{2}CO_{3} and HCl:

K_{2}CO_{3}+ 2HCl ---> 2KCl + CO_{2}+ H_{2}O

3) Note the 1:1 molar ratio between K_{2}CO_{3} and CO_{2}. From this we conclude:

0.0100153 mol K_{2}CO_{3}present in the dry mixture

4) Calculate grams K_{2}CO_{3}, 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 #4:** 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:**

1) Calculate excess HCl:

We know this: HCl + NaOH ---> NaCl + H_{2}OThere 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 molHowever, this is the combined amount of HCl used to titrate K

_{2}CO_{3}ANDKOH

4) Calculate HCl used to titrate KOH by subtracting the HCl used to titrate K_{2}CO_{3}:

0.0701 mol − 0.0200 mol = 0.0501 molThe 0.02 comes from the fact that 2 HCl were required to neutralize one K

_{2}CO_{3}. See step 2 in problem 3a.

5) Calculate grams of KOH

KOH + HCl ---> KCl + H_{2}OFrom 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 g5.000 g − 4.756 g = 0.805 g

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

**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 g1.631 g / 15.9994 g/mol = 0.10194 mol of oxygen in M

_{x}O_{y}.

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 M_{x}O_{y} 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 MO_{2}:

2.077 g / 0.05097 mol = 40.75 g/molThis is within the acceptable range.

Before going on, I would like to point out that a M_{2}O_{3} formula leads to an atomic weight of approximately 34 and a M_{2}O 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 KO_{2}, known as potassium superoxide.

If we were to look for a +4 forming ion (in other words, something to satisfy the MO_{2} 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 MO_{2} formula.

Only KO_{2} provides an atomic weight within the parameters specified by the problem.

**Problem #6:** A 12.5843 g sample of ZrBr_{4} 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 ZrBr_{4}.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 Cl_{2} is produced in each reaction:

Heat 3.45 g X to get 2.72 g Y and Cl_{2}gas

Mass of Cl_{2}gas is 3.45 − 2.72 = 0.73 gHeat 2.72 g Y to get 1.99 g Pt and Cl

_{2}gas

Mass of Cl_{2}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 molsimplest mole ratio:

Pt = 0.01 / 0.01 = 1

Cl = 0.02 / 0.01 = 2formula of Y = PtCl

_{2}

3) Determine formula for compound X:

mass of Pt = 1.99 g

mass of Cl_{2}= 0.73 g from 1st decomposition + 0.73 g from 2nd decomposition = 1.46 g total in Xmole 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 molsimplest mole ratio:

Pt = 0.01 / 0.01 = 1

Cl = 0.04 / 0.01 = 4Formula of X = PtCl

_{4}

**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 MgCl_{2}:

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 AlCl_{3}and two go to make one MgCl_{2}2 out of 5 is 40%.

4) Determine moles of MgCl_{2} that are present:

3.6274 x 10¯^{3}mol / 2 = 1.8137 x 10¯^{3}molRemember, one MgCl

_{2}is present for every two Cl¯

5) Repeating (2), (3) and (4) with appropriate modifications yields 1.8137 x 10¯^{3} mol for AlCl_{3}. (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 MgCl_{2}and AlCl_{3}Mg(OH)_{2}+ 2HCl ---> MgCl_{2}+ 2H_{2}OAl(OH)

_{3}+ 3HCl ---> AlCl_{3}+ 3H_{2}OSecond, note the following molar ratios:

Mg(OH)_{2}: MgCl_{2}is 1 : 1Al(OH)

_{3}: AlCl_{3}is 1 : 1This means we have the following molar amounts in the original sample:

Mg(OH)_{2}= 1.8137 x 10¯^{3}molAl(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 Fe_{2}O_{3}. 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 Fe_{2}O_{3} 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 O_{2}7.64 g / 31.9988 g mol¯

^{1}= 0.238759 mol (I kept several guard digits.)

2) Determine how many moles of O_{2} goes to form FeO and how many moles goes to form Fe_{2}O_{3}:

2Fe + O_{2}---> 2FeO

4Fe + 3O_{2}---> 2Fe_{2}O_{3}25% goes to form FeO

75% goes to form Fe_{2}O_{3}0.238759 mol x 0.25 = 0.05968974 mol of O

_{2}used to form FeO

0.238759 x 0.75 = 0.17906922 mol of O_{2}used to form Fe_{2}O_{3}

3) Determine moles of FeO and moles of Fe_{2}O_{3} formed:

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

_{2}O_{3}⇒ the O_{2}to Fe_{2}O_{3}ratio is 3:2, therefore double the amount of O_{2}used to get Fe_{2}O_{3}produced and then divide that value by three:0.1193795 mol of Fe_{2}O_{3}produced.

4) Determine grams of FeO and grams of Fe_{2}O_{3} formed:

FeO ⇒ 0.1193795 mol x 71.844 g mol^{-1}= 8.5767 g

Fe_{2}O_{3}⇒ 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 Fe_{2}O_{3}:

FeO ⇒ 8.5767 g x (55.845 / 71.844) = 6.667 g of Fe in FeO

Fe_{2}O_{3}⇒ 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 + O_{2}---> 2MgO

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

3Mg + N_{2}---> Mg_{3}N_{2}

So you have a mixture of MgO and Mg_{3}N_{2} 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 gSecond 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 Mg_{3}N_{2}

y = the mass of Mg_{3}N_{2}in the mixture of MgO and Mg_{3}N_{2}(24.305 / 40.304) is the percentage of Mg in MgO

(72.915 / 100.929) is the percentage of Mg in Mg_{3}N_{2}

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 = 0.1970.1194y = 0.007042

y = 0.058978 g

4) Percent of Mg_{3}N_{2} in original sample:

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