Fifteen 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 MgCO_{3} and water:

MgCO_{3}---> 7.58 g / 84.313 g/mol = 0.0899 mol

H_{2}O ---> 8.09 g / 18.015 g/mol = 0.449 mol

3) Find a whole number molar ratio:

MgCO_{3}---> 0.0899 mol / 0.0899 mol = 1

H_{2}O ---> 0.449 mol / 0.0899 mol = 5MgCO

_{3}·5H_{2}O

**Example #2:** A hydrate of Na_{2}CO_{3} 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 Na_{2}CO_{3} and water:

Na_{2}CO_{3}---> 3.22 g / 105.988 g/mol = 0.0304 mol

H_{2}O ---> 1.09 g / 18.015 g/mol = 0.0605 mol

3) Find a whole number molar ratio:

Na_{2}CO_{3}---> 0.0304 mol / 0.0304 mol = 1

H_{2}O ---> 0.0605 mol / 0.0304 mol = 2Na

_{2}CO_{3}·2H_{2}Osodium 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 Na_{2}CO_{3} **·** nH_{2}O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na_{2}CO_{3} (value of n)?

**Solution:**

1) Some preliminary comments:

Ignore the water of hydration for a moment.Na

_{2}CO_{3}(s) + 2HCl(aq) ---> 2NaCl(aq) + CO_{2}(g) + H_{2}O(ℓ)The key is that there is a 1:1 molar ratio between Na

_{2}CO_{3}and CO_{2}. (Also, note that we assume that the gas is pure CO_{2}and that there is no water vapor whatsoever. All of the water stays as a liquid. We also assume that no CO_{2}dissolves in the water.)

2) Determine moles of CO_{2}:

0.6039 g / 44.009 g/mol = 0.013722 mol of CO_{2}

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 Na_{2}CO_{3} 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 water2.4723 g / 18.015 g/mol = 0.13724 mol of water

6) For every one Na_{2}CO_{3}, how many waters are there?

0.13724 mol / 0.013722 mol = 10Na

_{2}CO_{3}·10H_{2}O

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

**Example #4:** If 1.951 g BaCl_{2} **·** nH_{2}O yields 1.864 g of anhydrous BaSO_{4} after treatment with sulfuric acid, calculate n.

**Solution:**

1) Calculate mass of Ba in BaSO_{4}:

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

2) Calculate mass of anhydrous BaCl_{2} that contains 1.0968 g of Ba:

1.0968 g is to 137.33 g/mol as x is to 208.236 g/molx = 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 BaCl_{2} 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 BaCl_{2} set to a value of one:

BaCl_{2}---> 0.0080 mol / 0.0080 mol = 1

H_{2}O ---> 0.0160 mol/ 0.0080 mol = 2BaCl

_{2}·2H_{2}O

**Example #5:** Given that the molar mass of Na_{2}SO_{4} **·** nH_{2}O is 322.1 g/mol, calculate the value of n.

**Solution:**

1) The molar mass of anhydrous Na_{2}SO_{4} 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:

Na_{2}SO_{4}·10H_{2}O

**Example #6:** 4.92 g of hydrated magnesium sulphate crystals (MgSO_{4} **⋅** nH_{2}O) 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 MgSO_{4}:

2.40 g / 120.4 g/mol = 0.020 mol

4) Determine smallest whole-number ratio:

MgSO_{4}---> 0.020 / 0.020 = 1

H_{2}O ---> 0.14 / 0.020 = 7MgSO

_{4}⋅7H_{2}O

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 CoCl_{2} **⋅** 2H_{2}O 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:

CoCl_{2}⋅2H_{2}O ---> 165.8686 g (in one mole)

mass of two moles of water ---> 36.0296 gdecimal 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 ZnSO_{4} **⋅** 7H_{2}O 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 ZnSO_{4}⋅7H_{2}O ---> 287.5446 g

molar mass of ZnSO_{4}---> 161.441 gdecimal 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 gTo three sig figs, this is 1.44 g

**Example #9:** If 29.0 g of MgSO_{4} **⋅** 7H_{2}O 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, MgSO_{4} 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 MgSO_{4}is produced

3) Determine grams of anhydrate produced:

(0.11766157 mol) (120.366 g/mol) = 14.16245 gTo three sig figs, this is 14.2 g

**Example #10:** What is the formula of the hydrate formed when 66.3 g of Ga_{2}(SeO_{4})_{3} combines with 33.7 g of H_{2}O?

**Solution:**

1) Determine moles of each substance present:

Ga_{2}(SeO_{4})_{3}---> 66.3 g / 568.314 g/mol = 0.11666 mol

H_{2}O ---> 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 Ga_{2}(SeO_{4})_{3} is present:

1.8707 mol / 0.11666 mol = 16.035

3) The formula is as follows:

Ga_{2}(SeO_{4})_{3}⋅16H_{2}O

**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 CuSO_{4}. Calculate the formula of the hydrated compound.

**Solution:**

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

CuSO_{4}---> 74.7 g

H_{2}O ---> 25.3 g

2) Change to moles:

CuSO_{4}---> 74.7 g / 159.607 g/mol = 0.468 mol

H_{2}O ---> 25.3 g / 18.0 g/mol = 1.406 mol

3) Divide by the smallest:

CuSO_{4}---> 0.468 mol / 0.468 mol = 1

H_{2}O ---> 1.406 mol / 0.468 mol = 3

4) Formula:

CuSO_{4}·3H_{2}OThe most common CuSO

_{4}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 = 0.1290.129 * 100 = 12.9%

**Example #13:** 1.534 grams of BaCl_{2} **·** 2H_{2}O is heated to dryness. What will be the mass of BaCl_{2}(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 MgSO_{4} **⋅** 7H_{2}O 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 MgSO_{4}⋅7H_{2}O is divided by the molar mass of MgSO_{4}⋅7H_{2}O to give moles of MgSO_{4}⋅7H_{2}O.(b) The molar ratio of 1 to 1 is derived from this equation:

MgSO_{4}⋅7H_{2}O(s) ---> MgSO_{4}(s) + 7H_{2}O(ℓ)For every one mole of MgSO

_{4}⋅7H_{2}O heated, one mole of anhydrous MgSO_{4}is left behind when all the water is driven off.(c) The moles of MgSO

_{4}are multipled by the molar mass of MgSO_{4}to give grams of MgSO_{4}(the answer).

**Example #15:** When a hydrate of Na_{2}CO_{3} 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 Na_{2}CO_{3} is present. The molar mass of anhydrous Na_{2}CO_{3} 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 g125.933 g / 18.015 g = 6.99

5) The formula:

Na_{2}CO_{3}·7H_{2}O

**Solution #2:**

1) Assume 100 g of Na_{2}CO_{3} **·** nH_{2}O is present. Therefore, of the 100 grams:

Na_{2}CO_{3}= 45.7 g

H_{2}O = 54.3 g

2) Convert mass to moles:

Na_{2}CO_{3}---> 45.7 g / 105.988 g/mol = 0.43118 mol

H_{2}O ---> 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:

Na_{2}CO_{3}·7H_{2}O

**Bonus Example:** 3.20 g of hydrated sodium carbonate, Na_{2}CO_{3} **⋅** nH_{2}O was dissolved in water and the resulting solution was titrated against 1.00 mol dm¯^{3} hydrochloric acid. 22.4 cm^{3} of the acid was required. What is the value of n?

**Solution:**

1) Sodium carbonate dissolves in water as follows:

Na_{2}CO_{3}⋅nH_{2}O(s) ---> 2Na^{+}(aq) + CO_{3}^{2}¯(aq) + nH_{2}O(ℓ)

2) The addition of HCl will drive all of the CO_{3}^{2}¯ ion to form CO_{2} gas. One mole of carbonate ion will produce n moles of water.

CO_{3}^{2}¯ + 2H^{+}---> CO_{2}(g) + H_{2}O(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 Na_{2}CO_{3}.

(0.0112 mol Na_{2}CO_{3}) (105.988 g Na / 1 mol) = 1.187 g Na_{2}CO_{3}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 H_{2}O

6) Convert to moles of water

2.013 g H_{2}O x (1 mol/18.015 g) = 0.11174 mol H_{2}O

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

Na_{2}CO_{3}: 0.0112 mol / 0.0112 = 1

H_{2}O: 0.11174 mol / 0.0112 = 9.97 = 10Na

_{2}CO_{3}⋅10H_{2}O