Determine the formula of a hydrateProblems #11 - 20

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:

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

Bonus Problem: What is the correct chemical formula for the hydrated crystal Mn(NO3)4 · nH2O if analysis shows 511 g of Mn(NO3)4 and 334 g of H2O?

Solution:

1) Determine moles of each:

511 g / 302.954 g/mol = 1.6867 mol
334 g / 18.015 g/mol = 18.54 mol

2) Seek smallest whole-number ratio:

1.6867 mol / 1.6867 mol = 1
18.54 mol1.6867 mol = 11

Mn(NO3)4 · 11H2O