### Determine the formula of a hydrateTen Examples

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:

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

Over the time I put the above examples (and the problems) together, 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

Bonus Example: 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 Na / 1 mol) = 1.187 g Na2CO3

This is how many moles 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