Determine the formula of a hydrate
Problems #11 - 20


Go to the five hydrate examples

Go to hydrate problems #1 - 10

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Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data


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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:

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

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

leads to:

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

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

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 minus 3.79765 g = 3.18555 g

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

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

Close enough for this:

FeSO4 7H2O

Is this the correct answer?


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:

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

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

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

81.4 g BaI2 2H2O x (1 mol BaI2 2H2O / 427.1 g BaI2 2H2O) x (1 mol BaI2 / 1 mol BaI2 2H2O) x (391.1 g BaI2 / 1 mol BaI2) = 74.5 g BaI2

Brief Explanation:

1) 81.4 g BaI2 2H2O ---> the starting mass

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

3) (1 mol BaI2 / 1 mol BaI2 2H2O) ---> the 1:1 molar ratio

4) (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 mg 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: Here is a Yahoo Answers solution to a waters of hydration problem which I answered. Another person answered it as well and their approach differed from mine. You may want to examine it to see the differences.


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 times 0.59867 = 5.64 g
water ---> 9.42 g minus 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 Yahoo Answers because the person gave a correct solution in the question. That is, a correct solution right up until the end of the problem.

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 times (18.015 grams of H2O/15.9994 grams O) = 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 times (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?

Solution #1 (the traditional way):

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 minus 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

A water of hydration problem on Yahoo Answers.


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

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 can occur.


Go to the five hydrate examples

Go to hydrate problems #1 - 10

Return to Mole Table of Contents

Calculate empirical formula when given mass data

Calculate empirical formula when given percent composition data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data


Look at a list of only the questions.