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P E3 (pg of ) Formula of a Hydrate, Lecture notes of Chemistry

University of Advancing TechnologyChemistry

Determine the formula for a lead(II) acetate hydrate Pb(C2H3O2)2 ... 1.21 g − 0.17 g decrease (water removed) = 1.04 g of the anhydrate (Pb(C2H3O2)2).

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P E3 (pg ! of !) Formula of a Hydrate Name_______________________Per____1 3
If you are having difficulty please make use of the examples provided in NS F.4 Show your work clearly and carefully on an
extra piece of paper so that you can refer to it later to check for mistakes, use for review and/or share your expertise with others.
1. Consider aluminum nitrate nonohydrate. Al(NO3)3 • ____ H2O
a. Calculate the theoretical % of water this hydrate.
b. If you were given 10.0 g of this hydrate, when heated to dryness, what would be the remaining mass of anhydrate?
2. Bert and Ernie were working with an unknown hydrate. They each had their own individual samples.
a. Bert’s sample weighed 1.46 g and then after heating the dried salt (the anhydrate) had a mass of 1.25 g, calculate the %
of water in the hydrate.
b. Ernie’s hydrate sample weighed 3.62 g, what water can he expect to remove from his sample?
3. Determine the formula for a calcium sulfate hydrate CaSO4 • ____ H2O that was determined to be 44 % water.
4. Determine the formula for a lead(II) acetate hydrate Pb(C2H3O2)2 • ____H2O that had an original mass of 1.21 g and when
heated, its mass decreased by 0.17 g
5. 12.25 g of a hydrate compound was heated and the anhydrate weighed 8.93 g. The anhydrate was further analyzed and
determined to be 2.06 g iron and 1.55 g nitrogen, and the rest oxygen. Determine the formula of this hydrate. What is the
name of this hydrate?
6. A hydrate was determined to be 44.9 % water. The anhydrate was analyzed to be 37.9% nickel, 20.7% sulfur, and 41.4%
oxygen. Determine the formula of this hydrate.
7. Determine the formula of a hydrate that was only 17.97 % copper, 1.13 % hydrogen, 9.05 % oxygen, and the rest iodine.
Assume that the anhydrate is a made of just copper and iodine.
8. The iron(II) chloride hydrate is heated to remove its water, the water removed is nearly the same mass as the anhydrate itself.
What is the formula of this hydrate? FeCl2 • ____ H2O
This problem is a bit tricky…give it a try after you are feeling confident.
9. A chromium, sulfur, oxygen hydrate was was analyzed and found to be 14.59% chromium, 18.00% sulfur, 62.87% oxygen
and 4.54% hydrogen. Determine the formula of this hydrate.
pf3

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P E3 (pg! 1 of! 3 ) Formula of a Hydrate Name_______________________Per____

If you are having difficulty please make use of the examples provided in NS F.4 Show your work clearly and carefully on an extra piece of paper so that you can refer to it later to check for mistakes, use for review and/or share your expertise with others.

  1. Consider aluminum nitrate nono hydrate. Al(NO 3 ) 3 • ____ H 2 O a. Calculate the theoretical % of water this hydrate. b. If you were given 10.0 g of this hydrate, when heated to dryness, what would be the remaining mass of anhydrate?
  2. Bert and Ernie were working with an unknown hydrate. They each had their own individual samples. a. Bert’s sample weighed 1.46 g and then after heating the dried salt (the anhydrate) had a mass of 1.25 g, calculate the % of water in the hydrate. b. Ernie’s hydrate sample weighed 3.62 g, what water can he expect to remove from his sample?
  3. Determine the formula for a calcium sulfate hydrate CaSO 4 • ____ H 2 O that was determined to be 44 % water.
  4. Determine the formula for a lead(II) acetate hydrate Pb(C 2 H 3 O 2 ) 2 • ____H 2 O that had an original mass of 1.21 g and when heated, its mass decreased by 0.17 g
  5. 12.25 g of a hydrate compound was heated and the anhydrate weighed 8.93 g. The anhydrate was further analyzed and determined to be 2.06 g iron and 1.55 g nitrogen, and the rest oxygen. Determine the formula of this hydrate. What is the name of this hydrate?
  6. A hydrate was determined to be 44.9 % water. The anhydrate was analyzed to be 37.9% nickel, 20.7% sulfur, and 41.4% oxygen. Determine the formula of this hydrate.
  7. Determine the formula of a hydrate that was only 17.97 % copper, 1.13 % hydrogen, 9.05 % oxygen, and the rest iodine. Assume that the anhydrate is a made of just copper and iodine.
  8. The iron(II) chloride hydrate is heated to remove its water, the water removed is nearly the same mass as the anhydrate itself. What is the formula of this hydrate? FeCl 2 • ____ H 2 O This problem is a bit tricky…give it a try after you are feeling confident.
  9. A chromium, sulfur, oxygen hydrate was was analyzed and found to be 14.59% chromium, 18.00% sulfur, 62.87% oxygen and 4.54% hydrogen. Determine the formula of this hydrate.

P E3 (pg! 2 of! 3 ) Formula of a Hydrate ANSWERS

  1. First you need to write the correct formula for the salt: Al(NO 3 ) 3 • 9 H 2 O a. Al(NO 3 ) 3 26.98 + 3(14.01) + 9(16) = 213.01 and 9 H 2 O 9[2(1.01) + 16.0] = 162.18 The hydrate total = 375. ! b. 43.2 % water means that the hydrate is 56.8 % anhydrate. Thus! 2. Although we do not know the identity of this hydrate, we can calculate the percent of its parts, water and anhydrate. a. 1.46 – 1.25 = 0.21 g water! b.!
  2. If hydrate is said to be 44% water, then we know that 56% is the anhydrate. CaSO 4 •? H 2 O CaSO 4 40.08 + 32.07 + 4(16) = 136.15 g/mol water = 18.02 g/mol Thus CaSO 4 • 6 H 2 O
  3. 1.21 g − 0.17 g decrease (water removed) = 1.04 g of the anhydrate (Pb(C 2 H 3 O 2 ) 2 ) Thus Pb(C 2 H 3 O 2 ) 2 •? H 2 O Pb(C 2 H 3 O 2 ) 2 207.2 + 4(12.01) + 6(1.01) + 4(16) = 325.29 g/mol water = 18.02 g/mol Voilà Pb(C 2 H 3 O 2 ) 2 • 3H 2 O 5. In this problem it is important to recognize that you must determine the empirical formula of the anhydrate before proceeding on to the formula of the entire hydrate. Voilà FeN 3 O 9 •? H 2 O Next you must proceed on to the hydrate part of the problem. 12.25 g − 8.93 g = 3.32 g of water FeN 3 O 6 55.85 + 3(14.01) + 9(16) = 241.88 g/mol water = 18.02 g/mol FeN 3 O 6 • 5 H 2 O tetra hydrate
  4. 18 g
  5. 19 g × 100 = 43. 2 % water 10 g × 0. 568 = 5. 68 g ofAnhydrate
  6. 21 g
  7. 46 g × 100 = 14. 4 % water
  8. 62 g × 0. 144 = 0. 521 g ofWater 56 g (%) × 1 mol
  9. 15 g = 0. 411 mol

44 g (%) × 1 mol

  1. 02 g = 2. 44 mol
  1. 04 g × 1 mol
  2. 29 g = 0. 003197 mol

0. 17 ×

1 mol

  1. 02 g = 0. 009433 mol

2. 06 ×

1 mol

  1. 85 g = 0. 0369 mol

1. 55 ×

1 mol

  1. 01 g = 0. 111 mol

5. 32 ×

1 mol 16 g = 0. 3325 mol

  1. 93 g × 1 mol
  2. 88 g = 0. 0369 mol

3. 32 ×

1 mol

  1. 02 g = 0. 1842 mol