### Determine the formula of a hydrate

Problem #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

Problem #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'.

Problem #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 + 2HCl ---> CO2 + 2NaCl + H2O

The key is that there is a 1:1 molar ratio between Na2CO3 and CO2

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.

Problem #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
0.288 g / 18.015 g/mol = 0.0160

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

1 : 2

BaCl2 · 2H2O

Problem #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

Problem #6: 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 minus 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 #7: 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 = 0.352
S ⇒ 11/32 = 0.344
O ⇒ 22/16 = 1.375
H2O ⇒ 44/ 18 = 2.44

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

Zn ⇒ 0.352 / 0.344 = 3
S ⇒ 0.344 / 0.344 = 1
O ⇒ 1.375 / 0.344 = 3
H2O ⇒ 2.44 / 0.344 = 7

The formula is ZnSO4 · 7H2O

Problem #8: 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 minus 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 #9: 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 minus 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 #10a: 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 times 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 #10b: 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 minus 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 #11: 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 / 10.185 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 #12: 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 times 0.47238 = 2.3619 g (this is the water lost)

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

Round off to three sig figs: 2.64 g

Problem #13: 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 #14: 1.33g 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 minus 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 #15: 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 minus 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

Problem #16: 6.9832 g of FeSO4 xH2O is dissolved in water acidified with sulfuric acid. The solution is made up to 250. cm3. 25.00 cm3 of this solution required 25.01 cm3 of 0.0200 M KMnO4 to titrate completely. Calculate x.

Solution:

Fe2+ gets oxidized by the MnO4-. The products are Fe3+ and Mn2+

Fe2+ ---> Fe3+ + e-
5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O

8H+ + 5Fe2+ + MnO4- ---> 5Fe3+ + Mn2+ + 4H2O

The key is the 5 to 1 molar ratio between ferrous ion and permanganate. What I want to get is the number of moles of ferrous ion in the 25.00 cm3. That will get me to grams of FeSO4 in the solution. A subtraction will give me the water in the hydrate.

moles KMnO4 ---> (0.0200 mol/L) (0.02501 L) = 0.000500 mol

for every one mole of permanganate, five moles of ferrous are oxidized.

0.000500 mol of MnO4- oxidizes 0.0025 of ferrous ion

That's the moles in 25.00 cm3. We originally had 250. cm3, so 0.0250 mol total of dissolved FeSO4 . xH2O

0.0250 moles of anhydrous FeSO4 weighs ---> 0.025 mol times 151.906 g/mol = 3.79765 g

water in the hydrate ---> 6.9832 g minus 3.79765 g = 3.18555 g

moles of water ---> 3.18555 g / 18.015 g/mol = 0.17683 mol

I want this molar ratio:

0.0250 to 0.17683

in smallest whole number terms where the FeSO4 part is 1:

1 to 7.0732

Close enough for this:

FeSO4 7H2O

Problem #17: A 81.4 gram sample of BaI2 2H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

Solution:

Dehydration of the hydrated barium iodide salt is shown by this reaction:

BaI2 2H2O(s) ---> BaI2(s) + 2H2O(g)

The following "dimensional analysis" set-up uses the concepts of mass-mass stoichiometry:

81.4 g BaI2 2H2O x (1 mol BaI2 2H2O / 427.1 g BaI2 2H2O) x (1 mol BaI2 / 1 mol BaI2 2H2O) x (391.1 g BaI2 / 1 mol BaI2) = 74.5 g BaI2

Brief Explanation:

1) 81.4 g BaI2 2H2O ---> the starting mass

2) (1 mol BaI2 2H2O / 427.1 g BaI2 2H2O) ---> divide by the molar mass of BaI2 2H2O

3) (1 mol BaI2 / 1 mol BaI2 2H2O) ---> the 1:1 molar ratio

4) (391.1 g BaI2 / 1 mol BaI2) ---> multiply by the molar mass of BaI2

Problem #18: If the hydrated compound UO2(NO3)2 9H2O is heated gently, the water of hydration is lost. If you heat 4.05 mg of the hydrated compound to dryness, what mass of UO2(NO3)2 will remain?

Solution:

Calculate how many moles of UO2(NO3)2 9H2O you have in 4.05 g

When you heat it, you have only UO2(NO3)2 remaining. 1 mole of UO2(NO3)2 9H2O leaves 1 mole of UO2(NO3)2.

Multiply moles of UO2(NO3)2 by the molar mass of UO2(NO3)2 to get the mass.

Problem #19: Here is a Yahoo Answers solution to a waters of hydration problem which I answered. Another person answered it as well and their approach differed from mine. You may want to examine it to see the differences.

Problem #20: How many grams of water and anhydrous salt would you get when heating 9.42 g of Fe(NO3)3 9H2O?

Solution:

1) Determine mass percentages of Fe(NO3)3 and H2O:

mass of one mole of Fe(NO3)3 9H2O ---> 403.9902 g

decimal percent by mass of Fe(NO3)3 ---> 241.857 g / 403.9902 g = 0.59867

decimal percent by mass of water ---> 1 - 0.59867 = 0.40133

2) Determine masses of water and anhydrous salt:

anhydrous salt ---> 9.42 g times 0.59867 = 5.64 g
water ---> 9.42 g minus 5.64 g = 3.78 g

Problem #21: Heating 0.695 g CuSO4 nH2O gives a residue of 0.445 g. Determine the value of n.

Solution:

I copied this problem from Yahoo Answers because the person gave a correct solution in the question. That is, a correct solution right up until the end of the problem.

0.695 - 0.445 = 0.25 g of H2O

0.25 g H2O / 18 g/mol = 0.014 mol

0.445 g CuSO4 / 159.5 g/mol = 0.0028 mol

0.0028 mol / 0.014 mol = 0.2

Here's the error:

So since there's no such thing has 0.2 of a molecule I round it to a whole number which in this case is 0 to get:

0CuSO4 1H2O

The correct technique is to multiply by 5 so as to get a 1 in front of the CuSO4:

0.2CuSO4 1H2O times 5 equals this:

CuSO4 5H2O

Comment: the student who made this mistake failed to see the coefficients of 0.2 and 1 as representing moles, focusing only on the ratio as molecules. It is perfectly fine to have a molar ratio of 0.2 to 1, which then becomes 1 to 5, which is the lowest whole-number ratio of moles.

Problem #22: Epsom salt is MgSO4 nH2O. The hydrate was found to contain 71.4% oxygen. Calculate the number of water molecules associated with each formula unit of magnesium sulfate hydrate.

Solution:

The molar mass of MgSO4 is 120.366 g/mol

There are 4 moles of O in MgSO4

The grams of O in one mole of MgSO4 is 63.9976 g

Let x = grams of oxygen from H2O

therefore, grams of H2O is this:

x times (18.015 grams of H2O/15.9994 grams O) = 1.126x

grams of oxygen divided by total weight of compound equals 0.714 (from information given in the problem)

(63.9976 + x) / (120.366 + 1.126x) = 0.714

63.9976 + x = 0.714[120.366 + 1.126x]

63.9976 + x = 85.94 + 0.804x

0.196x = 21.9424

x = 111.951 grams (of oxygen in entire MgSO4 nH2O)

111.951 g times (1.000 mole H2O/15.9994 g O) = 6.997 moles H2O

MgSO4 7H2O

Problem #23: A sample of hydrated magnesium sulphate, MgSO4 · nH2O, is found to contain 51.1% water. What is the value of n?

1) Assume 100 g of the compound is present. This allows the percentages to be easily converted to masses:

MgSO4 ---> 48.9 g
H2O ---> 51.1 g

2) Convert the masses to moles:

MgSO4 ---> 48.9 g / 120.366 g/mol = 0.406261 mol
H2O ---> 51.1 g / 18.015 g/mol = 2.836525 mol

3) Divide through by smallest:

MgSO4 ---> 0.406261 mol / 0.406261 mol = 1
2.836525 mol / 0.406261 mol = 6.98

4) Empirical formula:

MgSO4 · 7H2O

Solution #2:

51.1% is water so 48.9% must be MgSO4

Assume one mole of MgSO4 is present. This represents 48.9% of the hydrate.

120.366 is to 48.9 as x is to 51.1

x = 125.78 g <--- mass of water in the hydrate

125.78 g/18.015 g/mol = 6.98

MgSO4 · 7H2O

Problem #24: If a 9.15 g sample of a hydrated salt produced 6.50 g of anhydrous salt (309.650 g/mol) and 2.65 g of water (18.015 g/mol), what is the molecular mass of the hydrated salt?

Solution:

6.50 g / 309.650 g/mol = 0.021 mol
2.65 g / 18.015 g/mol = 0.147 mol

We want to know how many moles of water are present when one mole of the anhydrous salt is present.

0.147 is to 0.021 as x is to 1

x = 7

molecular mass = 309.650 g/mol + (7 * 18.015 g/mol) = 435.755 g/mol

Problem #25: A water of hydration problem on Yahoo Answers.

Bonus Problem A hydrate of magnesium chloride is present and the following data is collected:

mass of crucible = 22.130 grams
mass of crucible + hydrate = 25.290 grams
mass of crucible and contents after heating = 23.491 grams

What is the complete formula of this hydrate?

Solution:

1) mass of hydrate:

25.290 g - 22.130 g = 3.160 g

2) mass of anhydrate:

23.491 g - 22.130 g = 1.181 g

3) water lost:

3.160 g - 1.181 g = 1.979 g

4) moles MgCl2:

1.181 g / 95.211 g/mol = 1.24 mol

5) moles water lost:

1.979 g / 18.015 g/mol = 0.109853 mol

6) molar ratio of MgCl2 to water is:

1 : 11.3

Within fairly reasonable experimental error, the formula of the hydrate is:

MgCl2 · 12H2O

The Wiki page for magnesium chloride shows hydrates with 12, 8, 6, 4 and 2 waters of hydration exist. The one that exists at room temperature is the one with 6. The one with 12 loses 4 waters of hydration above -16.4 °C. So, while the problem above does not occur near room temperature, it can occur.