Determine the formula of a hydrate
Fifteen Examples

Problems #1 - 10      Problems #11 - 20      Problems #21 - 35      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


Examples and Problems only, no solutions


Here's a worksheet with eight hydrate problems, all of which have (hand-written) solutions. The original location of this document is here.


Example #0: In exactly 1 mole of the hydrate CuSO4 5H2O, how many grams are present of (a) the hydrate, (b) the anhydrate, and (c) water.

Solution:

1) In one mole of the hydrate, there is present this:

one mole of CuSO4
five moles H2O

2) Determine the molar mass for the hydrate. Remember to include the five waters.

249.681 g

This is the answer to (a)

3) Determine the molar mass of the anhydrate:

159.607 g

This is the answer to (b)

Remember, the anhydrate does not have any water in it, hence the formula is CuSO4.

4) Determine the total mass of water in one mole of the hydrate:

(18.015 g/mol) (5 mol) = 90.074 g

This is the answer to (c).

You could have also done this:

249.681 g − 159.607 g = 90.074 g

Example #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?

Solution:

1) Determine mass of water driven off:

15.67 − 7.58 = 8.09 g of water

2) Determine moles of MgCO3 and water:

MgCO3 ---> 7.58 g / 84.313 g/mol = 0.0899 mol
H2O ---> 8.09 g / 18.015 g/mol = 0.449 mol

3) Find a whole number molar ratio:

MgCO3 ---> 0.0899 mol / 0.0899 mol = 1
H2O ---> 0.449 mol / 0.0899 mol = 5

MgCO3 · 5H2O


Example #2: A hydrate of Na2CO3 has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate.

Solution:

1) Determine mass of water driven off:

4.31 − 3.22 = 1.09 g of water

2) Determine moles of Na2CO3 and water:

Na2CO3 ---> 3.22 g / 105.988 g/mol = 0.0304 mol
H2O ---> 1.09 g / 18.015 g/mol = 0.0605 mol

3) Find a whole number molar ratio:

Na2CO3 ---> 0.0304 mol / 0.0304 mol = 1
H2O ---> 0.0605 mol / 0.0304 mol = 2

Na2CO3 · 2H2O

sodium carbonate dihydrate

Comment: sodium carbonate forms three hydrates and the above is not one of them. This is a problem probably crafted so that you cannot look up possible answers via the InterTubez®. Just sayin'.


Example #3: When you react 3.9267 grams of Na2CO3 · nH2O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na2CO3 (value of n)?

Solution:

1) Some preliminary comments:

Ignore the water of hydration for a moment.

Na2CO3(s) + 2HCl(aq) ---> 2NaCl(aq) + CO2(g) + H2O(ℓ)

The key is that there is a 1:1 molar ratio between Na2CO3 and CO2. (Also, note that we assume that the gas is pure CO2 and that there is no water vapor whatsoever. All of the water stays as a liquid. We also assume that no CO2 dissolves in the water.)

2) Determine moles of CO2:

0.6039 g / 44.009 g/mol = 0.013722 mol of CO2

3) Use the 1:1 molar ratio referenced above:

This means that the HCl reacted with 0.013722 mole of sodium carbonate.

4) How many grams of Na2CO3 is that?

0.013722 mol times 105.988 g/mol = 1.4544 g

5) Determine grams, then moles of water

3.9267 g − 1.4544 g = 2.4723 g of water

2.4723 g / 18.015 g/mol = 0.13724 mol of water

6) For every one Na2CO3, how many waters are there?

0.13724 mol / 0.013722 mol = 10

Na2CO3 · 10H2O

Comment: this is one of the three sodium carbonate hydrates that exist.


Example #4: If 1.951 g BaCl2 · nH2O yields 1.864 g of anhydrous BaSO4 after treatment with sulfuric acid, calculate n.

Solution:

1) Calculate mass of Ba in BaSO4:

(1.864 g) (137.33 g/mol / 233.39 g/mol) = 1.0968 g

2) Calculate mass of anhydrous BaCl2 that contains 1.0968 g of Ba:

1.0968 g is to 137.33 g/mol as x is to 208.236 g/mol

x = 1.663 g

3) Calculate mass of water in original sample:

1.951 g − 1.663 g = 0.288 g

4) Calculate moles of anhydrous BaCl2 and water:

1.663 g / 208.236 g/mol = 0.0080 mol
0.288 g / 18.015 g/mol = 0.0160 mol

5) Express the above ratio in small whole numbers with BaCl2 set to a value of one:

BaCl2 ---> 0.0080 mol / 0.0080 mol = 1
H2O ---> 0.0160 mol/ 0.0080 mol = 2

BaCl2 · 2H2O


Example #5: Given that the molar mass of Na2SO4 · nH2O is 322.1 g/mol, calculate the value of n.

Solution:

1) The molar mass of anhydrous Na2SO4 is:

142.041 g/mol

2) The mass of water in one mole of the hydrate is:

322.1 g − 142.041 g = 180.059 g

3) Determine moles of water:

180.059 g / 18.0 g/mol = 10 mol

4) Write the formula:

Na2SO4 · 10H2O

Example #6: 4.92 g of hydrated magnesium sulphate crystals (MgSO4 nH2O) gave 2.40 g of anhydrous magnesium sulfate on heating to a constant mass. Determine the value of n.

Solution:

1) Mass of water:

4.92 g − 2.40 g = 2.52 g

2) Moles of water:

2.52 g / 18.0 g/mol = 0.14 mol

3) Moles of anhydrous MgSO4:

2.40 g / 120.4 g/mol = 0.020 mol

4) Determine smallest whole-number ratio:

MgSO4 ---> 0.020 / 0.020 = 1
H2O ---> 0.14 / 0.020 = 7

MgSO4 7H2O


During the time I assembled the examples (and the problems) above and below, I never addressed the following type of problem, one in which the full formula of the hydrate is known and you are asked how much anhydrate remains after driving off the water. When I realized the lack, I put three examples here (as opposed to their own separate file) and included another in the problems.

There are three different ways you can approach this type of problem. What follows is an example of each way.


Example #7: A 241.3 gram sample of CoCl2 2H2O is heated to dryness. Find the mass of anhydrous salt remaining.

Solution using percent water:

1) Determine the percentage of water in cobalt(II) chloride dihydrate:

CoCl2 2H2O ---> 165.8686 g (in one mole)
mass of two moles of water ---> 36.0296 g

decimal percent of water in the hydrate ---> 36.0296 g / 165.8686 g = 0.217218

2) Determine mass of water in 241.3 g of the hydrate:

(241.3 g) (0.217218) = 52.4147 g

3) Determine mass of anhydrous salt remaining after heating to dryness:

241.3 g − 52.4147 g = 188.9 g (to four sig figs)

Example #8: A 2.56 g sample of ZnSO4 7H2O is heated to dryness. Determine the anhydrous mass remaining after all the water has been driven off.

Solution using percent anhydrate:

1) Determine percentage of anhydrous zinc sulfate in zinc sulfate heptahydrate:

molar mass of ZnSO4 7H2O ---> 287.5446 g
molar mass of ZnSO4 ---> 161.441 g

decimal percent of anhydrate in the hydrate ---> 161.441 / 287.5446 = 0.561447 (keep some extra digits)

2) Determine mass of anhydrate in 2.56 g of the hydrate:

(2.56 g) (0.561447) = 1.4373 g

To three sig figs, this is 1.44 g


Example #9: If 29.0 g of MgSO4 7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Solution using 1:1 molar ratio:

1) Convert the 29.0 g of magnesium sulfate heptahydrate to moles:

29.0 g / 246.4696 g/mol = 0.11766157 mol

2) Note that, after heating, you end up with anhydrous magnesium sulfate, MgSO4 and you will have remaining the same number of moles of the anhydrous product as you had moles of the hydrated reactant. In other words, there is a 1:1 molar ratio between the hydrated compound used and the anhydrous compound produced.

0.11766157 mol of MgSO4 is produced

3) Determine grams of anhydrate produced:

(0.11766157 mol) (120.366 g/mol) = 14.16245 g

To three sig figs, this is 14.2 g


Example #10: What is the formula of the hydrate formed when 66.3 g of Ga2(SeO4)3 combines with 33.7 g of H2O?

Solution:

1) Determine moles of each substance present:

Ga2(SeO4)3 ---> 66.3 g / 568.314 g/mol = 0.11666 mol
H2O ---> 33.7 g / 18.015 g/mol = 1.8707 mol

2) We want to know who many moles of water are present when one mole of Ga2(SeO4)3 is present:

1.8707 mol / 0.11666 mol = 16.035

3) The formula is as follows:

Ga2(SeO4)3 16H2O

Example #11: A student determined that the percent of water in a hydrate was 25.3%. The formula of the anhydrous compound was determined to be CuSO4. Calculate the formula of the hydrated compound.

Solution:

1) Assume 100. g of the hydrate is present. That means this:

CuSO4 ---> 74.7 g
H2O ---> 25.3 g

2) Change to moles:

CuSO4 ---> 74.7 g / 159.607 g/mol = 0.468 mol
H2O ---> 25.3 g / 18.0 g/mol = 1.406 mol

3) Divide by the smallest:

CuSO4 ---> 0.468 mol / 0.468 mol = 1
H2O ---> 1.406 mol / 0.468 mol = 3

4) Formula:

CuSO4 · 3H2O

The most common CuSO4 hydrate is the pentahydrate, but the trihydrate does exist.


Example #12: 0.572 grams of a hydrate is heated to dryness, ending with 0.498 grams of anhydrous compound. What is the percentage of water by mass in the hydrate?

Solution:

(0.572 g − 0.498 g) / 0.572 g = 0.129

0.129 * 100 = 12.9%


Example #13: 1.534 grams of BaCl2 · 2H2O is heated to dryness. What will be the mass of BaCl2(s) that remains?

Solution:

1) Determine moles of hydrate:

1.534 g / 244.2636 g /mol = 0.0062801 mol

2) 0.0062801 mol of the anhydrate remain:

(0.0062801 mol) (208.233 g/mol) = 1.308 g

Example #14: 31.0 g of MgSO4 7H2O is thoroughly heated. What mass of anhydrous magnesium sulfate will remain?

Solution:

1) Here's a dimensional analysis solution:

  1 mol   1 mol   120.3676 g  
31.0 g x  –––––––––  x  –––––  x  –––––––––  = 15.1 g (to three sig figs)
  246.4746 g   1 mol   1 mol  

2) Explanation of the steps:

(a) 31.0 g of MgSO4 7H2O is divided by the molar mass of MgSO4 7H2O to give moles of MgSO4 7H2O.

(b) The molar ratio of 1 to 1 is derived from this equation:

MgSO4 7H2O(s) ---> MgSO4(s) + 7H2O(ℓ)

For every one mole of MgSO4 7H2O heated, one mole of anhydrous MgSO4 is left behind when all the water is driven off.

(c) The moles of MgSO4 are multipled by the molar mass of MgSO4 to give grams of MgSO4 (the answer).


Example #15: When a hydrate of Na2CO3 is heated until all the water is removed, it loses 54.3 percent of its mass. Determine the formula of the hydrate.

Solution #1:

1) Let us assume one mole of the hydrated Na2CO3 is present. The molar mass of anhydrous Na2CO3 is 105.988 g/mol.

2) The hydrate sample lost 54.3% of its mass (all water) to arrive at 105.988 g. This means that the 105.988 g is 45.7% of the total mass.

3) We can now write a ratio and proportion:

105.988 g   x
–––––––  =  –––––––
45.7   100

x = 231.921 g (this is the molr mass of the hydrate since one mole of it was present at the start)

4) Determine mass, then moles of water:

231.921 − 105.988 = 125.933 g

125.933 g / 18.015 g = 6.99

5) The formula:

Na2CO3 · 7H2O

Solution #2:

1) Assume 100 g of Na2CO3 · nH2O is present. Therefore, of the 100 grams:

Na2CO3 = 45.7 g
H2O = 54.3 g

2) Convert mass to moles:

Na2CO3 ---> 45.7 g / 105.988 g/mol = 0.43118 mol
H2O ---> 54.3 g / 18.015 g/mol = 3.014155 mol

3) Set up a ratio and proportion:

0.43118 mol   1
–––––––––––  =  –––––––
3.014155 mol   n

n = 6.99

4) The formula:

Na2CO3 · 7H2O

Bonus Example #1: Two different experiments were performed. Each one was designed to determine the value of 'n' in CoCl2 · nH2O

Experiment one: 0.256 g sample of CoCl2 · nH2O 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.

Experiment two: 0.416 g of CoCl2 · nH2O 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 0.145 g of cobalt(III) oxide.

Determine the value of 'n' from each experiment.

Solution to exp 1:

1) AgCl is precipitated:

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

2) How many grams of CoCl2 reacted? A standard mass-mass stoichiometry solution follows.

0.308 g AgCl   1 mol AgCl   1 mol CoCl2   129.8392 g CoCl2  
–––––––––––  x  ––––––––––––––  x  ––––––––––  x  –––––––––––––––  = 0.1395 g CoCl2
    143.3212 g AgCl   2 mol AgCl   1 mol CoCl2  

3) How many grams of water were in the original sample?

0.256 g − 0.1395 g = 0.1165 g H2O

4) Convert grams of water to moles of water:

0.1165 g / 18.015 g/mol = 0.0064668 mol

5) Determine moles of water per one mole of the anhydrate:

n = 0.0064668 mol / 0.0010745 mol CoCl2 = 6

6) The 0.0010745 mol came from the first three parts of the dimensional analysis in step 2 (the 0.308 divided by 143.3212 then divided by 2).

Solution to exp 2:

1) The two relevant chemical equations:

CoCl2 + 2NaOH ---> Co(OH)2 + 2NaCl

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

2) The overall reaction is:

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

3) A mass-mass stoichiometry solving technique is used the mass of anhydrous CoCl2 originally present:

0.145 g Co2O3   1 mol Co2O3   4 mol CoCl2   129.8392 g CoCl2  
–––––––––––––  x  ––––––––––––––  x  ––––––––––  x  ––––––––––––––  = 0.22701 g CoCl2
    165.8646 g Co2O3   2 mol Co2O3   1 mol CoCl2  

4) Determine grams of water in the original sample:

0.416 g − 0.22701 g = 0.18899 g

5) Convert grams of water to moles of water:

0.18899 g / 18.01532 g/mol = 0.0104907 mol H2O

6) Determine moles of water per one mole of the anhydrate:

n = 0.0104907 mol / 0.0017484 mol = 6

7) The 0.0017484 mol comes from the first three parts of the dimensional analysis in step 3 (the 0.145 divided by 165.8646 and then the result divided by 2 and then multiplied by 4).


Bonus Example #2: 3.20 g of hydrated sodium carbonate, Na2CO3 nH2O was dissolved in water and the resulting solution was titrated against 1.00 mol dm¯3 hydrochloric acid. 22.4 cm3 of the acid was required. What is the value of n?

Solution:

1) Sodium carbonate dissolves in water as follows:

Na2CO3 nH2O(s) ---> 2Na+(aq) + CO32¯(aq) + nH2O(ℓ)

2) The addition of HCl will drive all of the CO32¯ ion to form CO2 gas. One mole of carbonate ion will produce n moles of water.

CO32¯ + 2H+ ---> CO2(g) + H2O(ℓ)

3) Determine moles of HCl and from that moles of carbonate:

MV = moles

(1.00 mol/L) (0.0224 L) = 0.0224 mole of HCl

Two moles of HCl react for every one mole of carbonate. Therefore:

0.0224 mole / 2 = 0.0112 mol of carbonate

4) Determine the mass of 0.0112 mol of Na2CO3.

(0.0112 mol Na2CO3) (105.988 g Na2CO3 / 1 mol) = 1.187 g Na2CO3

This is how many grams of anhydrous sodium carbonate dissolved.

5) Mass of hydrated salt − mass of anhydrous salt = mass of water

3.20 g − 1.187 g = 2.013 g H2O

6) Convert to moles of water

2.013 g H2O x (1 mol/18.015 g) = 0.11174 mol H2O

7) Determine smallest whole-number ratio between sodium carbonate and water:

Na2CO3: 0.0112 mol / 0.0112 = 1
H2O: 0.11174 mol / 0.0112 = 9.97 = 10

Na2CO3 10H2O


Problems #1 - 10      Problems #11 - 20      Problems #21 - 35      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