Determine the formula of a hydrate
Problems #21 - 35

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

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

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

CoSO₄ = 0.00389 mol / 0.00389 mol = 1
H₂O = 0.02333 mol / 0.00389 mol = 6

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

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