Determine the formula of a hydrate
Five Examples


Go to hydrate problems #1 - 10

Go to hydrate problems #11 - 20

Return to Mole Table of Contents

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


Look at a list of only the questions.


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 minus 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 minus 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 minus 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 exists.


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 times (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 minus 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

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+ + CO32- + 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 x (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 minus 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


Look at a list of only the questions.


Go to hydrate problems #1 - 10

Go to hydrate problems #11 - 20

Return to Mole Table of Contents

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