Problems #11 - 25

**Problem #11:** 6.9832 g of FeSO_{4} **⋅** xH_{2}O is dissolved in water acidified with sulfuric acid. The solution is made up to 250. cm^{3}. 25.00 cm^{3} of this solution required 25.01 cm^{3} of 0.0200 M KMnO_{4} to titrate completely. Calculate x.

**Solution:**

1) Fe^{2+} gets oxidized by the MnO_{4}^{-}. The products are Fe^{3+} and Mn^{2+}

Fe^{2+}---> Fe^{3+}+ e^{-}

5e^{-}+ 8H^{+}+ MnO_{4}^{-}---> Mn^{2+}+ 4H_{2}Owhich leads to:

8H

^{+}+ 5Fe^{2+}+ MnO_{4}^{-}---> 5Fe^{3+}+ Mn^{2+}+ 4H_{2}O

2) The key is the 5 to 1 molar ratio between ferrous ion and permanganate. What I want to get is the number of moles of ferrous ion in the 25.00 cm^{3}. That will get me to grams of FeSO_{4} in the solution. A subtraction will give me the water in the hydrate.

moles KMnO_{4}---> (0.0200 mol/L) (0.02501 L) = 0.000500 molfor every one mole of permanganate, five moles of ferrous are oxidized.

0.000500 mol of MnO

_{4}¯ oxidizes 0.0025 of ferrous ion

3) That's the moles in 25.00 cm^{3}. We originally had 250. cm^{3}, so 0.0250 mol total of dissolved FeSO_{4} **⋅** xH_{2}O

0.0250 moles of anhydrous FeSO_{4}weighs ---> 0.025 mol times 151.906 g/mol = 3.79765 gwater in the hydrate ---> 6.9832 g − 3.79765 g = 3.18555 g

moles of water ---> 3.18555 g / 18.015 g/mol = 0.17683 mol

4) I want this molar ratio:

0.0250 to 0.17683in smallest whole number terms where the FeSO

_{4}part is 1:1 to 7.0732

5) Close enough for this:

FeSO_{4}⋅7H_{2}O

**Problem #12:** A 81.4 gram sample of BaI_{2} **⋅** 2H_{2}O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

**Solution:**

1) Dehydration of the hydrated barium iodide salt is shown by this reaction:

BaI_{2}⋅2H_{2}O(s) ---> BaI_{2}(s) + 2H_{2}O(g)

2) The following "dimensional analysis" set-up uses the concepts of mass-mass stoichiometry:

1 mol BaI _{2}⋅2H_{2}O1 mol BaI _{2}391.1 g BaI _{2}81.4 g BaI _{2}⋅2H_{2}O x––––––––––––––––– x ––––––––––––––– x –––––––––– = 74.5 g BaI _{2}427.1 g BaI _{2}⋅2H_{2}O1 mol BaI _{2}⋅2H_{2}O1 mol BaI _{2}

3) A brief explanation of the steps:

(a) 81.4 g BaI_{2}⋅2H_{2}O ---> the starting mass(b) (1 mol BaI

_{2}⋅2H_{2}O / 427.1 g BaI_{2}⋅2H_{2}O) ---> divide by the molar mass of BaI_{2}⋅2H_{2}O(c) (1 mol BaI

_{2}/ 1 mol BaI_{2}⋅2H_{2}O) ---> the 1:1 molar ratio(d) (391.1 g BaI

_{2}/ 1 mol BaI_{2}) ---> multiply by the molar mass of BaI_{2}

**Problem #13:** If the hydrated compound UO_{2}(NO_{3})_{2} **⋅** 9H_{2}O is heated gently, the water of hydration is lost. If you heat 4.05 g of the hydrated compound to dryness, what mass of UO_{2}(NO_{3})_{2} will remain?

**Solution:**

Calculate how many moles of UO_{2}(NO_{3})_{2}⋅9H_{2}O you have in 4.05 gWhen you heat it, you have only UO

_{2}(NO_{3})_{2}remaining. 1 mole of UO_{2}(NO_{3})_{2}⋅9H_{2}O leaves 1 mole of UO_{2}(NO_{3})_{2}.Multiply moles of UO

_{2}(NO_{3})_{2}by the molar mass of UO_{2}(NO_{3})_{2}to get the mass.

**Problem #14:** Solid copper(II) chloride forms a hydrate of formula CuCl_{2} **⋅** xH_{2}O. A student heated a sample of hydrated copper(II) chloride, in order to determine the value of x. The following results were obtained:

mass of crucible = 16.221 g

mass of crucible and hydrated copper(II) chloride = 18.360 g

mass of crucible and anhydrous copper(II) chloride = 17.917 g

From these data, determine the value of x and write the complete formula for hydrated copper(II) chloride.

**Solution:**

1) Determine the mass of the anhydrate and of the water that was lost:

CuCl_{2}---> 17.917 − 16.221 = 1.696 g

H_{2}O ---> 18.360 − 17.917 = 0.443 g

2) Determine moles of each:

CuCl_{2}---> 1.696 g / 134.452 g/mol = 0.012614 mol

H_{2}O ---> 0.443 g / 18.015 g/mol = 0.02459 mol

3) Divide by smallest:

CuCl_{2}---> 0.012614 mol / 0.012614 mol = 1

H_{2}O ---> 0.02459 mol / 0.012614 mol = 1.95x = 2

CuCl

_{2}⋅2H_{2}O

**Problem #15:** How many grams of water and anhydrous salt would you get when heating 9.42 g of Fe(NO_{3})_{3} **⋅** 9H_{2}O?

**Solution:**

1) Determine mass percentages of Fe(NO_{3})_{3} and H_{2}O:

mass of one mole of Fe(NO_{3})_{3}⋅9H_{2}O ---> 403.9902 gdecimal percent by mass of Fe(NO

_{3})_{3}---> 241.857 g / 403.9902 g = 0.59867decimal percent by mass of water ---> 1 − 0.59867 = 0.40133

2) Determine masses of water and anhydrous salt:

anhydrous salt ---> (9.42 g) (0.59867) = 5.64 g

water ---> 9.42 g − 5.64 g = 3.78 g

**Problem #16:** Heating 0.695 g CuSO_{4} **⋅** nH_{2}O gives a residue of 0.445 g. Determine the value of n.

**Solution:**

I copied this problem from an "answers" website because the person gave a correct solution in the question. That is, a correct solution right up until the end of the solution.

0.695 − 0.445 = 0.25 g of H_{2}O0.25 g H

_{2}O / 18 g/mol = 0.014 mol0.445 g CuSO

_{4}/ 159.5 g/mol = 0.0028 mol0.0028 mol / 0.014 mol = 0.2

Here's the error:

So since there's no such thing has 0.2 of a molecule I round it to a whole number which in this case is 0 to get:0CuSO

_{4}⋅1H_{2}OThe correct technique is to multiply by 5 so as to get a 1 in front of the CuSO

_{4}:0.2CuSO_{4}⋅1H_{2}O times 5 equals this:CuSO

_{4}⋅5H_{2}O

Comment: the student who made this mistake failed to see the coefficients of 0.2 and 1 as representing moles, focusing only on the ratio as molecules. It is perfectly fine to have a molar ratio of 0.2 to 1, which then becomes 1 to 5, which is the lowest whole-number ratio of moles.

**Problem #17:** Epsom salt is MgSO_{4} **⋅** nH_{2}O. The hydrate was found to contain 71.4% oxygen. Calculate the number of water molecules associated with each formula unit of magnesium sulfate hydrate.

**Solution:**

The molar mass of MgSO_{4}is 120.366 g/molThere are 4 moles of O in MgSO

_{4}The grams of O in one mole of MgSO

_{4}is 63.9976 gLet x = grams of oxygen from H

_{2}Otherefore, grams of H

_{2}O is this:(x) (18.015 grams of H

_{2}O/15.9994 grams O)which is 1.126x

grams of oxygen divided by total weight of compound equals 0.714 (from information given in the problem)

(63.9976 + x) / (120.366 + 1.126x) = 0.714

63.9976 + x = 0.714[120.366 + 1.126x]

63.9976 + x = 85.94 + 0.804x

0.196x = 21.9424

x = 111.951 grams (of oxygen in entire MgSO

_{4}⋅nH_{2}O)(111.951 g) (1.000 mole H

_{2}O/15.9994 g O) = 6.997 moles H_{2}OMgSO

_{4}⋅7H_{2}O

**Problem #18:** A sample of hydrated magnesium sulphate, MgSO_{4} **·** nH_{2}O, is found to contain 51.1% water. What is the value of n?

**Solution #1 (the traditional way):**

1) Assume 100 g of the compound is present. This allows the percentages to be easily converted to masses:

MgSO_{4}---> 48.9 g

H_{2}O ---> 51.1 g

2) Convert the masses to moles:

MgSO_{4}---> 48.9 g / 120.366 g/mol = 0.406261 mol

H_{2}O ---> 51.1 g / 18.015 g/mol = 2.836525 mol

3) Divide through by smallest:

MgSO_{4}---> 0.406261 mol / 0.406261 mol = 1

2.836525 mol / 0.406261 mol = 6.98

4) Empirical formula:

MgSO_{4}·7H_{2}O

**Solution #2:**

51.1% is water so 48.9% must be MgSO_{4}Assume one mole of MgSO

_{4}is present. This represents 48.9% of the hydrate.120.366 is to 48.9 as x is to 51.1

x = 125.78 g <--- mass of water in the hydrate

125.78 g / 18.015 g/mol = 6.98

MgSO

_{4}·7H_{2}O

**Problem #19:** If a 9.15 g sample of a hydrated salt produced 6.50 g of anhydrous salt (309.650 g/mol) and 2.65 g of water (18.015 g/mol), what is the molecular mass of the hydrated salt?

**Solution:**

6.50 g / 309.650 g/mol = 0.021 mol

2.65 g / 18.015 g/mol = 0.147 molWe want to know how many moles of water are present when one mole of the anhydrous salt is present.

0.147 is to 0.021 as x is to 1

x = 7

molecular mass = 309.650 g/mol + [(7) (18.015 g/mol)] = 435.755 g/mol

**Problem #20:** A hydrated compound has the formula MCl_{2} **·** 2H_{2}O. In an experiment, the following data were determined:

mass of the hydrate: 1.000 g

mass of H_{2}O: 0.185 g

Determine the identity of element M from these results.

**Solution:**

1) The mass of the anhydrate is:

1.000 − 0.185 = 0.815 g

2) There are 2 mol H_{2}O in 1 mol MCl_{2} **·** 2H_{2}O.

Therefore 0.185 g is to 36.0 g/mol as 0.815 is to xx = 158.6 g/mol <--- this is the formula weight of MCl

_{2}

3) Determine the atomic weight of M and identify it:

Two Cl weigh 71.0 gTherefore M is 87.6 g/mol (based on 158.6 − 71.0)

M is strontium

**Problem #21:** A 0.256 g sample of CoCl_{2} **⋅** yH_{2}O was dissolved in water, and excess silver salt was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. What is the value of y?

**Solution:**

1) The chemical reaction of interest:

CoCl_{2}+ 2Ag^{+}---> 2AgCl + Co^{2+}The CoCl

_{2}to AgCl molar ratio is 1:2

2) Determine moles of CoCl_{2} that were in solution:

(0.308 g AgCl) / (143.3212 g AgCl/mol) x (1 mol CoCl_{2}/ 2 mol AgCl) = 0.0010745 mol CoCl_{2}

3) Convert moles of CoCl_{2} to grams:

(0.0010745 mol CoCl_{2}) x (129.8392 g CoCl_{2}/mol) = 0.1395 g CoCl_{2}

4) Determine mass, then moles, of water in original CoCl_{2} sample:

(0.256 g total) − (0.1395 g CoCl_{2}) = 0.1165 g H_{2}O(0.1165 g H

_{2}O) / (18.01532 g H_{2}O/mol) = 0.0064667 mol H_{2}O

5) Determine value of y:

y = (0.0064667 mol H_{2}O) / (0.0010745 mol CoCl_{2}) = 6

**Problem #22:** A sample of 0.416 g of CoCl_{2} **⋅** yH_{2}O was dissolved in water, and an excess of sodium hydroxide (NaOH) was added. The cobalt hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide (Co_{2}O_{3}). The mass of cobalt(III) oxide formed was 0.145 g. What is the value of y?

**Solution:**

1) The following reactions take place:

CoCl_{2}+ 2NaOH ---> Co(OH)_{2}+ 2NaCl

4Co(OH)_{2}+ O_{2}---> 2Co_{2}O_{3}+ 4H_{2}O

2) The combined reaction is:

4CoCl_{2}+ 8NaOH + O_{2}---> 8NaCl + 2Co_{2}O_{3}+ 4H_{2}ONote that the first equation was multiplied through by 4 before adding. This was done so as to allow for the 4Co(OH)

_{2}to be cancelled.

3) Determine moles of CoCl_{2} that dissolved:

(0.145 g Co_{2}O_{3}) / (165.8646 g Co_{2}O_{3}/ mol) x (4 mol CoCl_{2}/ 2 mol Co_{2}O_{3}) = 0.0017484 mol CoCl_{2}

4) Determine mass of CoCl_{2}:

(0.0017484 mol CoCl_{2}) x (129.8392 g CoCl_{2}/ mol) = 0.22701 g CoCl_{2}

5) Determine mass, then moles, of water in the solid CoCl_{2} before dissolving:

(0.416 g total) − (0.22701 g CoCl_{2}) = 0.18899 g H_{2}O(0.18899 g H

_{2}O) / (18.01532 g H_{2}O/mol) = 0.0104905 mol H_{2}O

6) Determine value of y:

y = (0.0104905 mol H_{2}O) / (0.0017484 mol CoCl_{2}) = 6

**Problem #23:** When 5 g of iron(III) chloride hydrate are heated, 2 g of water are driven off . Find the chemical formula of the hydrate.

**Solution:**

1) 3 grams of FeCl_{3} are left after the 2 grams of H_{2}O are driven off.

2) Determine moles of each:

FeCl_{3}---> 3 g / 162.204 g/mol = 0.018495 mol

H_{2}O ---> 2 g / 18.015 g/mol = 0.11102 mol

3) Divide both values by the FeCl_{3} value.

FeCl_{3}---> 0.018495 mol / 0.018495 mol = 1

H_{2}O ---> 0.11102 mol / 0.018495 mol = 6This gives us a molar ratio between FeCl

_{3}and H_{2}O of 1 to 6

4) The formula of the hydrate is FeCl_{3} **⋅** 6H_{2}O

**Problem #24:** Cupric chloride, CuCl_{2}, dehydrates when heated. If 0.235 g of CuCl_{2} **⋅** xH_{2}O gives 0.185 g of CuCl_{2} on heating, what is the value of x?

**Solution #1:**

1) For the water:

0.235 minus 0.185 = 0.050 g of water driven off.0.050 g / 18.0 g/mol = 0.0028 mol of water

2) For the CuCl_{2}:

0.185 g / 134.452 g/mol = 0.001376 mol of anhydrous CuCl_{2}

3) We now want to see if there is a small whole number ratio between the two amounts of moles. Let's divide each number of the number of moles of CuCl_{2}

0.001376 mol / 0.001376 mol = 10.0028 mol / 0.001376 mol = 2.03

4) To a reasonable level of accuracy, this is a 2 to 1 ratio. The answer is this:

CuCl_{2}⋅2H_{2}O

**Solution #2:**

CuCl_{2}has a molecular weight of 134.3 and water weighs 18.0.The ratio 235/185 expresses the ratio of the weight of hydrate to anhydrate

Let X equal the number of water molecules. Therefore:

235/185 = [134.3 + X(18.0)] / 134.3

Solve for X to the nearest integer.

**Problem #25:** An hydrate of copper (II) chloride has the formula CuCl_{2} **⋅** xH_{2}O. The water in a 3.41 g sample of the hydrate was driven off by heating. The remaining sample had a mass of 2.69 g. Find the number of waters of hydration (x) in the hydrate.

**Solution #1:**

1) Mass:

CuCl_{2}---> 2.69 g

H_{2}O ---> 3.41 − 2.69 = 0.72 g

2) Moles:

CuCl_{2}---> 2.69 g / 134.452 g/mol = 0.02

H_{2}O ---> 0.72 g / 18.0 g/mol = 0.04

3) Get lowest whole-number mole ratio:

CuCl_{2}---> 0.02 / 0.02 = 1

H_{2}O ---> 0.04 / 0.02 = 2CuCl

_{2}⋅2H_{2}O

**Solution #2:**

1) Calculate percent composition of the hydrate

CuCl_{2}---> 2.69 g / 3.41 g = 78.9%

H_{2}O ---> 0.72 g / 3.41 g = 21.1 %

2) Assume 100 g of substance is present. Calculate mass of the anhydrate and water, then moles of each:

CuCl_{2}---> (100 g) (0.789) = 78.9 g

H_{2}O ---> (100 g) (0.211) = 21.1 gCuCl

_{2}---> 78.9 g / 134.5 g/mol = 0.5866 mol

H_{2}O ---> 21.1 g / 18.0 g/mol = 1.17 mol

3) Divide by the smaller to get whole number mole ratio:

CuCl_{2}---> 0.5866 / 0.5866 = 1

H_{2}O ---> 1.17 / 0.5866 = 2CuCl

_{2}⋅2H_{2}O

**Bonus Problem** A hydrate of magnesium chloride is present and the following data is collected:

mass of crucible = 22.130 grams

mass of crucible + hydrate = 25.290 grams

mass of crucible and contents after heating = 23.491 grams

What is the complete formula of this hydrate?

**Solution:**

1) Mass of hydrate:

25.290 g − 22.130 g = 3.160 g

2) Mass of anhydrate:

23.491 g − 22.130 g = 1.181 g

3) Water lost:

3.160 g − 1.181 g = 1.979 g

4) Moles MgCl_{2}:

1.181 g / 95.211 g/mol = 1.24 mol

5) Moles water lost:

1.979 g / 18.015 g/mol = 0.109853 mol

6) Molar ratio of MgCl_{2} to water is:

1 : 11.3

7) Within fairly reasonable experimental error, the formula of the hydrate is:

MgCl_{2}·12H_{2}O

The Wiki page for magnesium chloride shows hydrates with 12, 8, 6, 4 and 2 waters of hydration exist. The one that exists at room temperature is the one with 6. The one with 12 loses 4 waters of hydration above −16.4 °C. So, while the problem above does not occur near room temperature, it theoretically could occur.

In any event, it doesn't matter. It's simply a problem for a student to study.