### Determine the formula of a hydrate: fifteen problems

Problem #11: 6.9832 g of FeSO4 xH2O is dissolved in water acidified with sulfuric acid. The solution is made up to 250. cm3. 25.00 cm3 of this solution required 25.01 cm3 of 0.0200 M KMnO4 to titrate completely. Calculate x.

Solution:

1) Fe2+ gets oxidized by the MnO4-. The products are Fe3+ and Mn2+

Fe2+ ---> Fe3+ + e-
5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O

8H+ + 5Fe2+ + MnO4- ---> 5Fe3+ + Mn2+ + 4H2O

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 cm3. That will get me to grams of FeSO4 in the solution. A subtraction will give me the water in the hydrate.

moles KMnO4 ---> (0.0200 mol/L) (0.02501 L) = 0.000500 mol

for every one mole of permanganate, five moles of ferrous are oxidized.

0.000500 mol of MnO4¯ oxidizes 0.0025 of ferrous ion

3) That's the moles in 25.00 cm3. We originally had 250. cm3, so 0.0250 mol total of dissolved FeSO4 xH2O

0.0250 moles of anhydrous FeSO4 weighs ---> 0.025 mol times 151.906 g/mol = 3.79765 g

water 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 0.17683

in smallest whole number terms where the FeSO4 part is 1:

1 to 7.0732

5) Close enough for this:

FeSO4 7H2O

Problem #12: A 81.4 gram sample of BaI2 2H2O 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:

BaI2 2H2O(s) ---> BaI2(s) + 2H2O(g)

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

 1 mol BaI2 ⋅ 2H2O 1 mol BaI2 391.1 g BaI2 81.4 g BaI2 ⋅ 2H2O x ––––––––––––––––– x ––––––––––––––– x –––––––––– = 74.5 g BaI2 427.1 g BaI2 ⋅ 2H2O 1 mol BaI2 ⋅ 2H2O 1 mol BaI2

3) A brief explanation of the steps:

(a) 81.4 g BaI2 2H2O ---> the starting mass

(b) (1 mol BaI2 2H2O / 427.1 g BaI2 2H2O) ---> divide by the molar mass of BaI2 2H2O

(c) (1 mol BaI2 / 1 mol BaI2 2H2O) ---> the 1:1 molar ratio

(d) (391.1 g BaI2 / 1 mol BaI2) ---> multiply by the molar mass of BaI2

Problem #13: If the hydrated compound UO2(NO3)2 9H2O is heated gently, the water of hydration is lost. If you heat 4.05 g of the hydrated compound to dryness, what mass of UO2(NO3)2 will remain?

Solution:

Calculate how many moles of UO2(NO3)2 9H2O you have in 4.05 g

When you heat it, you have only UO2(NO3)2 remaining. 1 mole of UO2(NO3)2 9H2O leaves 1 mole of UO2(NO3)2.

Multiply moles of UO2(NO3)2 by the molar mass of UO2(NO3)2 to get the mass.

Problem #14: Solid copper(II) chloride forms a hydrate of formula CuCl2 xH2O. 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:

CuCl2 ---> 17.917 − 16.221 = 1.696 g
H2O ---> 18.360 − 17.917 = 0.443 g

2) Determine moles of each:

CuCl2 ---> 1.696 g / 134.452 g/mol = 0.012614 mol
H2O ---> 0.443 g / 18.015 g/mol = 0.02459 mol

3) Divide by smallest:

CuCl2 ---> 0.012614 mol / 0.012614 mol = 1
H2O ---> 0.02459 mol / 0.012614 mol = 1.95

x = 2

CuCl2 2H2O

Problem #15: How many grams of water and anhydrous salt would you get when heating 9.42 g of Fe(NO3)3 9H2O?

Solution:

1) Determine mass percentages of Fe(NO3)3 and H2O:

mass of one mole of Fe(NO3)3 9H2O ---> 403.9902 g

decimal percent by mass of Fe(NO3)3 ---> 241.857 g / 403.9902 g = 0.59867

decimal 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 CuSO4 nH2O 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 H2O

0.25 g H2O / 18 g/mol = 0.014 mol

0.445 g CuSO4 / 159.5 g/mol = 0.0028 mol

0.0028 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:

0CuSO4 1H2O

The correct technique is to multiply by 5 so as to get a 1 in front of the CuSO4:

0.2CuSO4 1H2O times 5 equals this:

CuSO4 5H2O

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 MgSO4 nH2O. 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 MgSO4 is 120.366 g/mol

There are 4 moles of O in MgSO4

The grams of O in one mole of MgSO4 is 63.9976 g

Let x = grams of oxygen from H2O

therefore, grams of H2O is this:

(x) (18.015 grams of H2O/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 MgSO4 nH2O)

(111.951 g) (1.000 mole H2O/15.9994 g O) = 6.997 moles H2O

MgSO4 7H2O

Problem #18: A sample of hydrated magnesium sulphate, MgSO4 · nH2O, is found to contain 51.1% water. What is the value of n?

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

MgSO4 ---> 48.9 g
H2O ---> 51.1 g

2) Convert the masses to moles:

MgSO4 ---> 48.9 g / 120.366 g/mol = 0.406261 mol
H2O ---> 51.1 g / 18.015 g/mol = 2.836525 mol

3) Divide through by smallest:

MgSO4 ---> 0.406261 mol / 0.406261 mol = 1
2.836525 mol / 0.406261 mol = 6.98

4) Empirical formula:

MgSO4 · 7H2O

Solution #2:

51.1% is water so 48.9% must be MgSO4

Assume one mole of MgSO4 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

MgSO4 · 7H2O

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 mol

We 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 MCl2 · 2H2O. In an experiment, the following data were determined:

mass of the hydrate: 1.000 g
mass of H2O: 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 H2O in 1 mol MCl2 · 2H2O.

Therefore 0.185 g is to 36.0 g/mol as 0.815 is to x

x = 158.6 g/mol <--- this is the formula weight of MCl2

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

Two Cl weigh 71.0 g

Therefore M is 87.6 g/mol (based on 158.6 − 71.0)

M is strontium

Problem #21: A 0.256 g sample of CoCl2 yH2O 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:

CoCl2 + 2Ag+ ---> 2AgCl + Co2+

The CoCl2 to AgCl molar ratio is 1:2

2) Determine moles of CoCl2 that were in solution:

(0.308 g AgCl) / (143.3212 g AgCl/mol) x (1 mol CoCl2 / 2 mol AgCl) = 0.0010745 mol CoCl2

3) Convert moles of CoCl2 to grams:

(0.0010745 mol CoCl2) x (129.8392 g CoCl2/mol) = 0.1395 g CoCl2

4) Determine mass, then moles, of water in original CoCl2 sample:

(0.256 g total) − (0.1395 g CoCl2) = 0.1165 g H2O

(0.1165 g H2O) / (18.01532 g H2O/mol) = 0.0064667 mol H2O

5) Determine value of y:

y = (0.0064667 mol H2O) / (0.0010745 mol CoCl2) = 6

Problem #22: A sample of 0.416 g of CoCl2 yH2O 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 (Co2O3). The mass of cobalt(III) oxide formed was 0.145 g. What is the value of y?

Solution:

1) The following reactions take place:

CoCl2 + 2NaOH ---> Co(OH)2 + 2NaCl
4Co(OH)2 + O2 ---> 2Co2O3 + 4H2O

2) The combined reaction is:

4CoCl2 + 8NaOH + O2 ---> 8NaCl + 2Co2O3 + 4H2O

Note 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 CoCl2 that dissolved:

(0.145 g Co2O3) / (165.8646 g Co2O3 / mol) x (4 mol CoCl2 / 2 mol Co2O3) = 0.0017484 mol CoCl2

4) Determine mass of CoCl2:

(0.0017484 mol CoCl2) x (129.8392 g CoCl2 / mol) = 0.22701 g CoCl2

5) Determine mass, then moles, of water in the solid CoCl2 before dissolving:

(0.416 g total) − (0.22701 g CoCl2) = 0.18899 g H2O

(0.18899 g H2O) / (18.01532 g H2O/mol) = 0.0104905 mol H2O

6) Determine value of y:

y = (0.0104905 mol H2O) / (0.0017484 mol CoCl2) = 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 FeCl3 are left after the 2 grams of H2O are driven off.

2) Determine moles of each:

FeCl3 ---> 3 g / 162.204 g/mol = 0.018495 mol
H2O ---> 2 g / 18.015 g/mol = 0.11102 mol

3) Divide both values by the FeCl3 value.

FeCl3 ---> 0.018495 mol / 0.018495 mol = 1
H2O ---> 0.11102 mol / 0.018495 mol = 6

This gives us a molar ratio between FeCl3 and H2O of 1 to 6

4) The formula of the hydrate is FeCl3 6H2O

Problem #24: Cupric chloride, CuCl2, dehydrates when heated. If 0.235 g of CuCl2 xH2O gives 0.185 g of CuCl2 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 CuCl2:

0.185 g / 134.452 g/mol = 0.001376 mol of anhydrous CuCl2

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 CuCl2

0.001376 mol / 0.001376 mol = 1

0.0028 mol / 0.001376 mol = 2.03

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

CuCl2 2H2O

Solution #2:

CuCl2 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 CuCl2 xH2O. 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:

CuCl2 ---> 2.69 g
H2O ---> 3.41 − 2.69 = 0.72 g

2) Moles:

CuCl2 ---> 2.69 g / 134.452 g/mol = 0.02
H2O ---> 0.72 g / 18.0 g/mol = 0.04

3) Get lowest whole-number mole ratio:

CuCl2 ---> 0.02 / 0.02 = 1
H2O ---> 0.04 / 0.02 = 2

CuCl2 2H2O

Solution #2:

1) Calculate percent composition of the hydrate

CuCl2 ---> 2.69 g / 3.41 g = 78.9%
H2O ---> 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:

CuCl2 ---> (100 g) (0.789) = 78.9 g
H2O ---> (100 g) (0.211) = 21.1 g

CuCl2 ---> 78.9 g / 134.5 g/mol = 0.5866 mol
H2O ---> 21.1 g / 18.0 g/mol = 1.17 mol

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

CuCl2 ---> 0.5866 / 0.5866 = 1
H2O ---> 1.17 / 0.5866 = 2

CuCl2 2H2O

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 MgCl2:

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 MgCl2 to water is:

1 : 11.3

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

MgCl2 · 12H2O

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.