Ten Examples

All Examples & Problems (no solutions)

**Example #1:** A mixture of Mg and Zn with a combined mass of 1.0875 g was burned in oxygen producing MgO and ZnO with a combined mass of 1.4090 g. How many grams of zinc was in original mixture?

**Solution #1:**

1) Write the chemical equations:

2Zn + O_{2}---> 2ZnO

2Mg + O_{2}---> 2MgO

2) Assign a variable:

Let z be the mass of zinc to be found.

Then 1.0875 − z is the mass of Mg in the original mixture.

3) Zinc:

z 2 mol 81.3794 g –––––––––––– x ––––– x –––––––– = 1.244676z 65.3820 g/mol 2 mol 1 mol In other words, the amount of zinc oxide produced is 1.244676 times the mass of zinc in the original mixture.

4) Magnesium:

1.0875 − z 2 mol 40.3044 g ––––––––––––– x ––––– x –––––––– = 1.80337 − 1.65827z 24.30506 g/mol 2 mol 1 mol In other words, the amount of magnesium oxide produced is 1.80337 minus 1.65827 times the mass of magnesium in the original mixture.

5) Add the two masses of oxides and set the sum equal to the given total mass:

(1.244676z) + (1.80337 − 1.65827z) = 1.4090 gz = 0.9535 g (of zinc)

**Solution #2:**

1) Let us assign the mass of Mg as x and the mass of Zn as y. We can write two simultaneous equations in two unknowns. The first is this:

x + y = 1.0875

2) The second is a bit more complex. Here it is:

(40.3044 / 24.30506) (x) + (81.3794 / 65.3820) (y) = 1.4090

3) Some explanation:

Write the equation for Mg becoming MgO:Mg +

^{1}⁄_{2}O_{2}---> MgOWe see that 1 mole of Mg will produce 1 mole of MgO. Converting to grams, we see that 24.30506 g of Mg will produce 40.3044 g of MgO. The producing of MgO from Mg will always be like this:

For any mass of Mg present, the amount of MgO produced will always be 40.304406/24.305 the amount of Mg originally present.

4) Solve:

1.658272x + 1.244676y = 1.4090Eliminate the y:

1.658272x + (1.244676) (1.0875 − x) = 1.4090

1.658272x + 1.353585 − 1.244676x = 1.4090

0.413596x = 0.055415

x = 0.134 g

**Example #2:** A metal container of a fixed volume contains water vapor at 200. °C at a pressure of 4.50 atm. The water is then electrolytically split to generate hydrogen (H_{2}) and oxygen (O_{2}) gases. What will be the pressure (in atm) of the container at 200. °C after the water has been completely split?

**Solution:**

1) The balanced chemical equation is this:

2H_{2}O(g) ---> 2H_{2}(g) + O_{2}(g)

2) Gay-Lussac's Law:

since the volume and temperature are constant, the amount (moles) of the gas and the pressure are directly proportional.

3) Therefore:

2 4.50 atm ––– = ––––––– 3 x x = 6.75 dm

^{3}The three comes from the fact that every two volumes of water vapor produce three volumes (two of hydrogen gas and one of oxygen gas) of total gas.

**Example #3:** A metal container of fixed volume is filled with 2.50 atm of H_{2} gas and 2.50 atm of O_{2} at 200. °C. The gases then react to generate water vapor (H_{2}O gas) through a combustion reaction. What is the total pressure (in atm) in the container at 200. °C when the reaction has progressed to the point where it has produced the maximum amount of H_{2}O gas?

**Solution:**

1) Two preliminary points:

(a) this is a limiting reagent problem.

(b) because the temperature and volume are constant, the pressure and amount of gas (in moles) are in direct proportion.

2) The balanced chemical reaction:

2H_{2}(g) + O_{2}(g) ---> 2H_{2}O(g)

3) Determine the limiting reagent:

H_{2}---> 2.50 / 2 = 1.25

O_{2}---> 2.50 / 1 = 2.50Hydrogen gas is the limiting regent.

4) Determine the amount of water that is produced:

2 2.50 atm ––– = ––––––– 2 x x = 2.50 atm

5) Determine the amount of O_{2} that reacts:

2 2.50 atm ––– = ––––––– 1 x x = 1.25 atm

6) Determine the amount of O_{2} that is unreacted:

2.50 atm − 1.25 atm = 1.25 atm

7) Determine the total pressure in the container after reaction is complete:

2.50 atm + 1.25 atm = 3.75 atm

**Example #4:** 1 mole of hydrocarbon of formula C_{n}H_{2n} was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192 grams of oxygen. Determine the formula of the hydrocarbon.

**Solution:**

1) Determine moles of O_{2} that combusted:

192 g / 32.0 g/mol = 6 mol2) Balance the chemical equation of the reaction:

C_{n}H_{2n}+ O_{2}---> CO_{2}+ H_{2}OC

_{n}H_{2n}+ O_{2}---> nCO_{2}+ nH_{2}OC

_{n}H_{2n}+ (3/2)nO_{2}---> nCO_{2}+ nH_{2}O2C

_{n}H_{2n}+ 3nO_{2}---> 2nCO_{2}+ 2nH_{2}O

3) The mole ratio of C_{n}H_{2n} to O_{2} is 2 : 3n. Therefore:

2 1 <--- the one is from one mole of C _{n}H_{2n}given in the problem––– = ––– 3n 6 <--- six moles of O _{2}required to combust one mole of C_{n}H_{2n}3n = 12

n = 4

4) Formula of the hydrocarbon:

C_{4}H_{8}

**Example #5:** A 13.43 g mixture of CH_{4} and C_{2}H_{6} is completely burned in excess oxygen. The mixture of CO_{2} and H_{2}O that results weighs 64.84 g. Determine the fraction of CH_{4} in the original mixture.

**Solution:**

1) This problem requires two simultaneous equations in two unknowns. Here is the first equation:

x + y = 13.43Where 'x' is the mass of CH

_{4}in the mixture and 'y' is the mass of C_{2}H_{6}in the mixture.

2) The second equation is this:

(80/16) (x) + (142/30) (y) = 64.84 gThe (80/16) and (142/30) will be explained!

3) The (80/16) factor:

Look at the balanced equation for the complete combustion of CH_{4}:

CH _{4}+ 2O _{2}--->CO _{2}+ 2H _{2}O16 44 36 <--- that last is 2 moles of water Look at the gram (not molar!) amounts of reactants and products:

16 grams of CH_{4}produce 44 g of CO_{2}plus 36 g of H_{2}OTherefore, 'x' grams of CH

_{4}will produce (80/16) times 'x' grams of CO_{2}+ H_{2}O

4) The argument for the (142/30) factor is the same:

C _{2}H_{6}+ 3.5O _{2}--->2CO _{2}+ 3H _{2}O30 88 54 <--- 2 moles of carbon dioxide and 3 moles of water 'y' grams of C

_{2}H_{6}will produce (142/30) times 'y' grams of CO_{2}+ H_{2}O

5) Use y = 13.43 − x to eliminate y in the second equation:

(80/16) (x) + (142/30) (13.43 − x) = 64.84Algebra!

5x + (4.733) (13.43 − x) = 64.84

5x + 63.57 − 4.733x = 64.84

0.267x = 1.27

x = 4.75655 g

6) The problem asked for the fraction of CH_{4}. Let us assume that means mass percent:

(4.75655 g / 13.43 g) * 100 = 35.4%

**Example #6:** You are given 1.446 g of a mixture of KClO_{3} and KCl. When heated, the KClO_{3} decomposes to KCl and O_{2}:

2KClO_{3}(s) ---> 2KCl(s) + 3O_{2}(g)

and 355 mL of O_{2} is collected over water at 26.0 °C. The total pressure of the gases in the collection flask is 745.0 torr. What is the weight percentage of KClO_{3} in the sample?

**Solution:**

1) Use Dalton's Law to determine the pressure of the dry O_{2}:

P_{total}= P_{O2}+ P_{H2O}745.0 torr = P

_{O2}+ 25.2 torrP

_{O2}= 719.8 torr

2) Determine moles of O_{2} produced:

PV = nRT(719.8 torr / 760.0 torr/atm) (0.355 L) = (n) (0.08206 L atm / mol K) (299 K)

n = 0.01370326 mol

3) Determine moles of KClO_{3} that decomposed. The KClO_{3} to O_{2} molar ratio is 2 to 3. Therefore:

2 x ––– = –––––––––––––– 3 0.01370326 mol x = 0.0091355 mol

4) Determine mass of KClO_{3} and its percentage in the mixture:

(0.0091355 mol) (122.55 g/mol) = 1.120 g(1.120 g / 1.446 g) (100) = 77.46%

**Example #7:** A mixture of potassium chloride and potassium bromide weighing 3.595 g is heated with chlorine gas, which converts the mixture completely to potassium chloride. The total mass of the KCl after the reaction is 3.129 g. What percent of the original mixture is KBr?

**Solution:**

1) We need two equations in two unknowns. The first is this:

x + y = 3.595 gwhere 'x' is the mass of KCl and 'y' is the mass of KBr.

2) Some explanation:

(a) total moles present at end:3.129 g / 74.55 g / mol = 0.04197 mol(b) this is true:

Mole of KCl + Mole of KBr = 0.04197 molbecause the moles of KBr are replaced by moles of KCl in a 1:1 molar ratio.

3) Therefore, this is the second equation:

[x / 74.55 g/mol] + [y / 119.0 g/mol] = 0.04197 mol

4) Rearrange first equation, then substitute it into second equation (I dropped all units.):

y = 3.595 − x[x / 74.55] − [(3.595 − x) / 119.0] = 0.04197

5) Solve:

[(x) (119.0)] + [(3.595 − x) (74.55)] = (0.04197) (119.0) (74.55)119.0x − 74.55x + [(3.595) (74.55)] = 372.33

44.45x + 268.01 = 372.33

44.45x = 104.32

x = 2.347 g (of KCl)

y = 3.595 g − 2.347 g = 1.248 g (of KBr)

%KBr = (1.248 g / 3.595 g) (100) = 34.71%

**Example #8:** An alloy contain aluminum, copper, and zinc. You take a 12.657 g sample of the alloy and react it with an HCl solution. The reaction converts all of the aluminum and zinc in the alloy to aluminum chloride and zinc chloride in addition to producing hydrogen gas. The copper does not react with the HCl. Upon completion of the reaction, a total of 10.2 L of hydrogen gas was collected at a pressure of 726 torr and a temperature of 27.0 °C. Additionally, 2.642 g of unreacted copper is recovered. What is the percent composition of the alloy?

**Solution:**

1) Use PV = nRT:

(726 torr / 760 torr/atm) (10.2 L) = (n) (0.08206 liter-atm/mole K) (300.0 K)n = 0.395795 mol

2) Determine combined grams of Al + Zn:

12.657 g minus 2.642 g = 10.015 g

3) The following balanced equations are required:

2Al + 6HCl ---> 2AlCl_{3}+ 3H_{2}Zn + 2HCl ---> ZnCl

_{2}+ H_{2}

4) Write a molar relationship between Al & Zn consumed and H_{2} produced:

the molar relationship between H_{2}produced and Al consumed is 3:2. That means this:(3/2) (moles Al) = the moles of H_{2}produced by AlIn like manner:

(1/1) (moles Zn) = the moles of H_{2}produced by Zn

5) We know the total moles of H_{2} to be 0.395795 mol, which gives us this relationship:

[(3/2) (moles Al)] + [(1/1) (moles Zn)] = 0.395795 mol

6) Let y = grams Al. Then (10.015 − y) = grams Cu. Using the molar masses for Al and Zn gives us:

[(1.5) (y / 26.98)] + [(1/1) ((10.015 − y) / 65.409)] = 0.3957950.055597y + 0.1531135 − 0.0152884y = 0.395795

0.0403086y = 0.2426815

y = 6.020 g <--- grams Al

grams Zn ---> 10.015 g − 6.020 g = 3.995 g

7) Let's make sure everything adds up:

2.642 g + 6.020 g + 3.995 g = 12.657 gYay!

**Example #9:** A sample of sodium bicarbonate (NaHCO_{3}) is known to contain some impurities. It is found that Na makes up 18.00% of the entire mass of the sample. All of the Na comes from the NaHCO_{3}. Find the mass percent of NaHCO_{3} in the sample.

**Solution:**

1) Let us determine the mass percent of Na in pure NaHCO_{3}:

22.99 g / 84.0059 g = 0.27367

2) By how much of a factor would we have to increase the impurities to make the sodium percentage become 18%?

0.27367 / 0.18 = 1.52039

3) What is that increased amount in grams?

(84.0059 g) (1.52039) = 127.7 g

4) What is the mass percentage of the sodium bicarbonate in the 127.7 g sample?

84.0059 g / 127.7 g = 65.78%

Here's another way

Assume 84.0059 g of NaHCO_{3}is present in the sample.This means 22.99 g of Na is present.

That 22.99 g represents 18% of the entire mixture.

What is the mass of the entire mixture?

22.99 is to 18% as x is to 100%

x = 127.7 g

The mass percent of NaHCO

_{3}is:84.0059 g / 127.7 g = 65.78%

**Example #10:** An ore of barium contains BaCO_{3}. A 1.495 g sample of the ore was treated with HCl to dissolve the BaCO_{3}. The resulting solution was filtered to remove insoluble material and then treated with H_{2}SO_{4} to precipitate BaSO_{4}. The precipitate was filtered, dried, and found to weigh 1.152 g. What is the percentage by mass of barium in the original sample?

**Solution:**

1) Molar mass of BaSO_{4}:

233.391 g/mol

2) Moles of BaSO_{4} in the precipitate:

1.152 g / 233.391 g/mol = 0.0049359 moles

3) Moles of barium in the precipitate:

0.0049359 molesBy the 1:1 molar ratio between Ba and BaSO

_{4}

4) Mass of barium in the precipitate:

(0.0049359 mol) (137.327 g/mol) = 0.67783 g

5) Barium percent by mass in original sample:

(0.67783 g / 1.495 g) (100) = 45.34%

**Bonus Example:** A solution contains Cr^{3+} and Mg^{2+} ions. The addition of 1.00 L of 1.55 M NaF solution is required to cause the complete precipitation of these ions as CrF_{3}(s) and MgF_{2}(s). The total mass of the precipitate is 50.2 g .

Find the mass of Cr^{3+} in the original solution.

**Solution:**

1) Write chemical equations:

Cr^{3+}+ 3NaF ---> CrF_{3}+ 3Na^{+}

Mg^{2+}+ 2NaF ---> MgF_{2}+ 2Na^{+}

2) Assign the following:

Let Z be the mass (in grams) of the CrF_{3}precipitate.

Let 50.2 − Z be the mass of the MgF_{2}precipitate.

3) Determine mol of NaF required to precipitate all the Cr^{3+}.

Z / (108.99131 g CrF_{3}/mol) x (3 mol NaF / 1 mol CrF_{3}) = (0.0275251 Z) mol NaF

4) Determine mol of NaF required to precipitate all the Mg^{2+}.

(50.2 − Z) / (62.30181 g MgF_{2}/mol) x (2 mol NaF / 1 mol MgF_{2}) = (1.61151 − 0.0321018 Z) mol NaF for Mg

5) Determine total moles of NaF:

(1.00 L) x (1.55 mol/L NaF) = 1.55 mol NaF

6) Set the sum of the two expressions for partial moles of NaF equal to the total moles of NaF:

(0.0275251 Z) + (1.61151 − 0.0321018 Z) = 1.55

7) Let there be algebra:

Z = 13.4398 g CrF_{3}

8) Determine mass of the Cr^{3+} ion:

(13.4398 g CrF_{3}) / (108.99131 g CrF_{3}/mol) x (1 mol Cr^{3+}/ 1 mol CrF_{3}) x (51.9961 g Cr/mol) = 6.41 g Cr^{3+}