### Determine the formula of a hydrate: ten problems

Problem #1: Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystalized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the hydrated salt obtained was 7.21 g. What hydrate is it?

Solution:

1) The amount of water in the hydrate is:

7.21 g − 4.78 g = 2.43 g

2) The moles of anhydrous LiClO4 and water are:

LiClO4 ---> 4.78 g / 106.39 g/mol = 0.044929 mol
H2O ---> 2.43 g / 18.015 g/mol = 0.13489 mol

3) Determine whole number ratio:

LiClO4 ---> 0.044929 mol / 0.044929 mol = 1
H2O ---> 0.13489 mol / 0.044929 mol = 3

LiClO4 · 3H2O

Problem #2: A substance was found to have the following percentages by mass: 23% zinc; 11% sulfur; 22% oxygen; 44% water. What is the empirical formula?

Solution:

1) Assume 100 g of the compound is present, then find the moles of each:

Zn ---> 23/65.4 = 0.3517
S ---> 11/32 = 0.34375
O ---> 22/16 = 1.375
H2O ---> 44/ 18 = 2.444

2) Divide the smallest number into the others. The answers will not be exact but enough to tell the formula:

Zn ---> 0.3517 / 0.34375 = 1
S ---> 0.344 / 0.34375 = 1
O ---> 1.375 / 0.34375 = 4
H2O ---> 2.44 / 0.34375 = 7

The formula is ZnSO4 · 7H2O

Problem #3: 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) Calculate moles of anhydrous barium chloride:

4.26 g / 208.236 g/mol = 0.020458 mol

2) Calculate moles of water:

5.00 − 4.26 = 0.74 g

0.74 g / 18.015 g/mol = 0.041077 mol

3) Determine whole number ratio:

0.041077 / 0.020458 = 2

4) Formula is:

BaCl2 · 2H2O

Problem #4: A 1.98 g sample of a cobalt(II) chloride hydrate is heated over a burner. When cooled, the mass of the remaining dehydrated compound is found to be 1.55 g. What is the formula for the original hydrate? How can you make sure that all of the water of hydration has been removed?

Solution:

1) Determine mass of water driven off:

1.98 g − 1.55 g = 0.43 g

2) Determine moles of anhydrous CoCl2 and H2O:

CoCl2 ---> 1.55 g / 129.839 g/mol = 0.01194 mol
H2O ---> 0.43 g / 18.015 g/mol = 0.0239 mol

3) Look for lowest whole-number ratio:

CoCl2 ---> 0.01194 mol / 0.01194 mol = 1
H2O ---> 0.0239 mol / 0.01194 mol = 2

4) Formula is:

CoCl2 · 2H2O

5) How can you make sure that all of the water of hydration has been removed?

After weighing the anhydrous CoCl2, you would continue to heat it. Then, you would weigh it again. If the two weights are in agreement, then you are done heating. If the two weights disagree, you continue heating and weighing until you gets weights that agree. In some cases, where an extra amount of care must be taken, you would want three straight weighings that were in agreement.

Also, weighs being in agreement does not mean that they are exactly the same. The standards for being in agreement might vary from one instructor to the next, so make sure to consult with your lab teacher on this point.

Problem #5a: A solution was made by dissolving 52.0 g of hydrated sodium carbonate in water and making it up to 5.00 dm3 of solution. The concentration of the solution was determined to be 0.0366 M. Determine the formula of hydrated sodium carbonate.

Solution:

1) Moles of hydrated sodium carbonate in 5.00 liters:

(0.0366 mol/L) (5.00 L) = 0.183 mol

2) Molecular weight of hydrated sodium carbonate:

52.0 g / 0.183 mol = 284.153 g/mol

3) Mass of water in one mole of hydrate:

284.153 − 105.988 = 178.165 g

(105.988 is molar mass of anhydrous sodium carbonate)

4) Moles of water in one mole of hydrate:

178.165 g / 18.018 g/mol = 9.9 mol

5) Formula of hydrated sodium carbonate:

Na2CO3 · 10H2O

Problem #5b: A solution was made by dissolving 71.50 g of hydrated sodium carbonate in water and making it up to 5.00 L of solution. The concentration of the solution was found to be 0.04805 M. What is the water of hydration for this hydrate of sodium carbonate?

Solution:

1) Mass of sodium carbonate in 5.00 L:

(0.04805 mol/L) (5.00 L) = x / 105.988 g/mol

x = 25.4636 g

2) Mass of water in 71.50 g of hydrated Na2CO3:

71.50 g − 25.4636 g = 46.0364 g

3) Moles of each:

moles of water ---> 46.0364 g / 18.015 g/mol = 2.55545 mol
moles of Na2CO3 ---> 25.4636 g / 105.988 g/mol = 0.24025 mol

4) Smallest whole-number ratio:

The ratio we want in smallest whole-numbers is this ---> 0.24025 mol to 2.55545 mol

That is a 1 to 10 ratio.

Na2CO3 · 10H2O

Problem #6: Determine the formula and name for the hydrate: 73.42% ammonium phosphate and 26.58% water.

Solution:

1) Assume 100 grams of the compound is present. Therefore:

73.42 g of (NH4)3PO4
26.58 g of H2O

2) Determine the moles of each compound:

73.42 g / 149.0858 g/mol = 0.492468 mol
26.58 g / 18.015 g/mol = 1.475437 mol

3) We want to know how many moles of H2O are present for every one mole of (NH4)3PO4:

1.475437 mol / 0.492468 mol = 2.996 = 3

4) Formula and name:

(NH4)3PO4 · 3H2O

ammonium phosphate trihydrate.

Problem #7: 5.00 g of borax (Na2B4O7 · 10H2O) was heated to remove the water. What is the mass of anhydrous sodium tetraborate that remains?

Solution:

The hydrate's molecular weight is 381.365 g/mol

The total amount of water in the molecular weight is 180.148 g.

180.148 / 381.365 = 0.47238 (decimal amount of the hydrate that is water)

(5.00 g) (0.47238) = 2.3619 g (this is the water lost)

5.00 g − 2.3619 g = 2.6381 g of the anhydrous Na2B4O7 remaining

Round off to three sig figs: 2.64 g

Problem #8: A sample of hydrate lost 14.75% of its original weight during heating. Determine the number of moles of hydration per mole of anhydrous substance if the molecular weight of the anhydrate is 208 grams/mole.

Solution:

Let's assume we had one mole of the hydrate present. We know that the one mole of anhyrate weighs 208 grams and represents 85.25% of the weight.

208 is to 85.25 as x is to 100

x = 244 (the molar mass of the hydrate)

244 − 208 = 36 (the mass of water in one mole of hydrate)

36/18 = 2 (two moles of hydration per mole of anhydrous substance)

Problem #9: 1.33 g of hydrated ethanedioic acid (H2C2O4 nH2O) were dissolved in distilled water and the solution made up to 250.0 mL in a graduated flask. 25.0 mL of this solution were titrated by 21.1 mL of 0.100M NaOH. Calculate the number of moles of water of crystallization in the hydrated ethanedioic acid.

Equation: H2C2O4 + 2NaOH ---> Na2C2O4 + 2H2O

Solution:

(0.100 mol/L) (0.0211 L) = 0.00211 mol of NaOH used in titration

There is a 1 to 2 molar ratio between H2C2O4 and NaOH

0.00211 mol / 2 = 0.001055 mol of H2C2O4 in the 25.0 mL that was titrated.

25.0 mL is to 0.001055 mol as 250.0 mL is to x

x = 0.01055 mol <--- this is how many moles of H2C2O4 nH2O dissolved in the 250.0 mL

1.33 g / 0.01055 mol = 126.066 g/mol <--- the molar mass of H2C2O4 nH2O

126.066 − 90.0338 = 36.0322 g <--- the mass of water in one mole of H2C2O4 nH2O

36.0 / 18.0 = 2 <--- moles of water in one mole of H2C2O4 nH2O

H2C2O4 2H2O

Problem #10: A solution was made by dissolving 71.5 g of hydrated sodium carbonate in water and making up to 5.00 dm3 of solution. The concentration of a 25.0 cm3 portion was determined to be 0.04805 M. Use the information to find the waters of hydration of the hydrated sodium carbonate.

Solution:

1) Mass of dissolved Na2CO3 in 25.0 cm3:

(0.04805 mol/L) (0.0250 L) = x / 105.988 g/mol

x = 0.12732 g

2) Mass of dissolved Na2CO3 in 5.00 dm3

0.12732 g is to 0.0250 L as x is to 5.00 L

x = 25.464 g

3) Water in 71.5 g of hydrated Na2CO3:

71.5 g − 25.464 g = 46.036 g

4) Determine molar ratio:

moles of water ---> 46.036 g / 18.015 g/mol = 2.555426 mol
moles of Na2CO3 ---> 25.464 g / 105.988 g/mol = 0.2402536 mol

The ratio we want in smallest whole-numbers is this ---> 0.2402536 mol to 2.555426 mol

That is a 1 to 10 ratio

5) The formula of the hydrate is:

Na2CO3 10H2O

Bonus Problem: A certain quantity of sodium carbonate decahydrate was heated to remove the water. The mass of the anhydrous compound that remained was 2.764 g. What was the original mass of the hydrated sodium carbonate?

Solution:

2.764 g / 105.988 g/mol = 0.0260784 mol

The Na2CO3 and H2O are in a 1:10 molar ratio in the decahydrate. Therefore:

(0.0260784 mol) (10) = 0.260784 mol of H2O

(0.260784 mol) (18.015 g/mol) = 4.698 g <--- mass of the water

Total mass of the hydrated compound:

2.764 + 4.698 = 7.462 g