Determine the formula of a hydrate
Problems #21 - 35

Fifteen Examples      Problems #1 - 10      Problems #11 - 20      Return to Mole Table of Contents

Determine empirical formula when given mass data

Determine 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

Determine molecular formula using the Ideal Gas Law

Hydrate lab calculations


The example and problem questions only, no solutions


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 this:

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


Problem #26: A sample of hydrated copper(II) sulfate weighs 124.8 g. The sample has been determined to contain 31.8 g of copper(Il) ions and 48.0 g of sulfate ions. Determine how many molecules of water of crystallization are present in the sample and thus deduce the chemical formula of hydrated copper(ll) sulfate.

Solution:

31.8 g + 48.0 g = 79.8 g (the mass of the anhydrous CuSO4)

124.8 g − 79.8 g = 45.0 g (the mass of water in the hydrate)

45.0 g / 18.0 g/mol = 2.5 mol (of water)

79.8 g / 160 g/mol = 0.50 mol (of anhydrous CuSO4)

2.5 mol / 0.50 mol = 5 (molecules of water per one formula unit of CuSO4)

CuSO4 5H2O


Problem #27: A hydrated salt is found to have the empirical formula CaN2H8O10. What is its dot formula?

Solution:

This is the general formula for a substance that is hydrated:
X nH2O

where X is the salt and it has n molecules of water attached. The term "dot formula" comes from the use of the "" to show the water attachment.

For this problem then, CaN2H8O10, all of the hydrogen will be found in the hydrating water. Since water is H2O, there will be n = 8/2 = 4 molecules of water. This requires 4 oxygen atoms; the remaining 6 are part of the salt.

Now we have CaN2O6 4H2O. If you look up salts containing calcium you'll find calcium nitrate, Ca(NO3)2. You can also look at the salt ionically. Balancing the charges will put you on the path to Ca(NO3)2.

This is the dot formula for hydrated calcium nitrate:

Ca(NO3)2 4H2O

Problem #28: You have 1.023 g of hydrated CoSO4. After heating and driving off all the water, the mass of the anhydride is 0.603 g. Determine the hydrate's chemical formula (number of waters attached).

Solution:

1) Mass of H2O removed after heating = mass of hydrate − mass of anhydrate

1.023 g − 0.603 g = 0.42 g
2) Find the moles of CoSO4 and H2O from the given masses:
moles of CoSO4 = 0.603 g / 154.9958 g/mol = 0.00389 mol
moles of H2O = 0.42 g / 18 g/mol = 0.02333 mol
3) Divide moles of CoSO4 and H2O with the lowest number of moles among the two, which is 0.00389 mol.
CoSO4 = 0.00389 mol / 0.00389 mol = 1
H2O = 0.02333 mol / 0.00389 mol = 6
4) Empirical Formula:
CoSO4 6H2O

Problem #29: A solution of FeSO4 xH2O has a concentration of 0.220 mol dm¯3 and a mass concentration of 61.16 g dm¯3. Find the value of x.

Solution:

The 61.16 grams (mass conc.) represents 0.220 moles (molar conc.) of solute.

61.16 is to 0.220 as x is to 1

x = 278 <--- this is the mass of one mole of FeSO4 xH2O

278 − 151.906 = 126.094 g <--- that's the mass of water in one mole of FeSO4 xH2O

The 151.906 is the mass of one mole of anhydrous FeSO4

126.904 g / 18.015 g/mol = 7 mol

FeSO4 7H2O


Problem #30: A 1.281 g sample of anhydrous Na2SO4 is exposed to the atmosphere and found to gain 0.391 g in mass. What is the percent, by mass, of Na2SO4 10H2O in the resulting mixture of anhydrous Na2SO4 and the decahydrate?

Solution:

0.391 g / 18.0 g/mol = 0.02172 mol

The moles of water is represented by 10. The anhydrous Na2SO4 would be in a 1:10 molar ratio with the water.

0.02172 mol / 10 = 0.002172 mol of Na2SO4

(0.002172 mol) (142.041 g/mol) = 0.3085 g of Na2SO4 "bonded" with the water

0.3085 g + 0.391 g = 0.6995 g of the hydrate

1.281 + 0.391 = 1.672 g <--- new total mass of hydrate/anhydrate mixture

(0.6995 / 1.672) (100) = 41.8% (to three sig figs)


Problem #31: A hydrate is found to have the following percent composition: 48.8% MgSO4 and 51.2% H2O. What is the formula of the hydrate?

Solution:

1) Assume 100 g of the compound is present. This allows us to change percents to mass. For example, in 100 g of the compound, 48.8 gams of MgSO4 will be present.

2) Now, calculate moles:

MgSO4 ---> 48.8 g / 120.366 g/mol = 0.40543 mol
H2O ---> 51.2 g / 18.015 g/mol = 2.8421 mol

3) Look for a whole number ratio:

MgSO4 ---> 0.40543 / 0.40543 = 1
H2O ---> 2.8421 / 0.40543 = 7

MgSO4 7H2O


Problem #32: A student dissolves 9.86 g Epsom salt (a MgSO4 hydrate) into 50.0 mL water. From this solution the student isolated 9.30 g of anhydrous BaSO4 (s) which is formed in the following reaction. BaSO4 is very insoluble in water, and esentially all sulfate ions will precipitate as BaSO4 (s). Assume handling losses are small.

MgSO4(aq) + BaCl42(aq) ---> MgCl2(aq) + BaSO4(s)

(a)  Determine the formula of the MgSO4 hydrate (Epson Salt)
(b) Calculate the % by mass of H2O in the hydrate

Solution:

1) Moles of BaSO4:

9.30 g / 233.391 g/mol = 0.0398473 mol

2) From the chemical reaction, BaSO4 and MgSO4 are in a 1:1 molar ratio, so this many moles of MgSO4 formed:

0.0398473 mol

3) Grams of MgSO4

(0.0398473 mol) (120.366 g/mol) = 4.79626 g

4) Water in the hydrate:

9.86 g − 4.79626 g = 5.06374 g

5) Moles of water:

5.06374 g / 18.015 g/mol = 0.2810846 mol

6) We want to know a whole number ratio between MgSO4 and H2O where MgSO4 is set to 1.

MgSO4 ---> 0.0398473 / 0.0398473 = 1
H2O ---> 0.2810846 / 0.0398473 = 7

MgSO4 7H2O

7) Percent by mass of water:

(5.06374 g / 9.86 g) (100) = 51.356% (round off to 51.4% for three sig figs)

Problem #33: A 5.00 g sample of hydrated barium chloride, BaCl2 · nH2O, is heated to drive off the water. After heating, 4.26 g of anhydrous barium chloride, BaCl2, remains. What is the value of n in the hydrate's formula?

Solution:

1) How much water was driven off?

5.00 − 4.26 = 0.74 g

2) Convert mass to moles for both BaCl2 and H2O:

BaCl2 ---> 4.26 g / 208.23 g/mol = 0.020458 mol
H2O ---> 0.74 g / 18.015 g/mol = 0.041077 mol

3) Find lowest whole-number ratio:

BaCl2 ---> 0.020458 mol / 0.020458 mol = 1
H2O ---> 0.041077 mol / 0.020458 mol = 2

4) The answer is 2. Although not asked for, here is the formula:

BaCl2 · 2H2O

Problem #34: When 1.40 g of a hydrated metal sulphate, XSO4 · 7H2O is heated to constant mass, 0.68 g of the anhydrous salt remains. Find the atomic mass of the metal in the subtance. Identify the metal.

Solution:

1) Determine mass of water lost:

1.40 g − 0.68 g = 0.72 g

2) How many moles is this?

0.72 g / 18.015 g/mol = 0.0399667 mol

I think 0.0400 is close enough.

3) XSO4 and H2O are in a 1:7 molar ratio. Determine how many moles of XSO4 are present:

0.0400 / 7 = 0.0057143 mol

4) Determine the molecular weight of XSO4:

0.68 g / 0.0057143 mol = 119 g/mol

5) Remove the weight of the sulfate:

119 − 96.06 = 22.94 g/mol

With a resonable amount of experimental error, X is magnesium.


Problem #35: 8.70 grams of hydrated potassium carbonated was heated to constant mass. The salt lost 1.80 g of water. What is the formula of the hydrated carbonate?

Solution:

mass of K2CO3 ---> 8.70 − 1.80 = 6.90 g

6.90 g / 138.2 g/mole = 0.050 mole of K2CO3

1.8 g H2O / 18.0 g/mole = 0.10 mole of H2O

K2CO3 · 2H2O


Bonus Problem: A 0.572 g sample of a compound containing only thorium , sulfate and water of hydration. After heating to drive off the water of hydration, the remainer weights 0.414 g . A second sample weighting 0.249 g is dissolved in water and excess aqueous barium chloride is added to precipitate 0.198 g of barium sulfate . Calculate the empirical formula of the compound.

Solution:

1) Determine mass of water driven off in first sample:

0.572 g − 0.414 g = 0.158 g

2) Determine mass of water that would have been driven off from a second sample weighting 0.249 g:

(0.249 g) (0.158 g / 0.572 g) = 0.06878 g

3) Determine the mass of sulfate in the second sample:

(0.198 g BaSO4) (96.06 g sulfate/mol) / (233.4 g BaSO4/mol) = 0.08149 g sulfate

4) Determine the mass of thorium in the second sample:

0.249 g − (0.08149 g + 0.06878 g) = 0.09873 g

5) Determine moles:

thorium ---> 0.09873 g / 232.038 g/mol = 0.00042549 mol
sulfate ---> 0.08149 g / 96.06 g/mol = 0.000848324 mol
water ---> 0.06878 g / 18.015 g/mol = 0.00381793 mol

6) Divide through by smallest:

thorium ---> 0.00042549 mol / 0.000848324 mol = 0.5
sulfate ---> 0.000848324 mol / 0.000848324 mol = 1
water ---> 0.00381793 mol / 0.000848324 mol = 4.5

7) Double to get whole numbers:

Th(SO4)2 9H2O

Fifteen Examples      Problems #1 - 10      Problems #11 - 20      Return to Mole Table of Contents

Determine empirical formula when given mass data

Determine 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

Determine molecular formula using the Ideal Gas Law

Hydrate lab calculations