1 / 14

Hydrate Calculations

Hydrate Calculations. Hydrates. Hydrates are ionic compounds that have crystalline structures involving H 2 O In MgSO 4 . 7H 2 O, there are 7 water molecules present for every 1 MgSO 4 .

archie
Télécharger la présentation

Hydrate Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Hydrate Calculations

  2. Hydrates • Hydrates are ionic compounds that have crystalline structures involving H2O • In MgSO4. 7H2O, there are 7 water molecules present for every 1 MgSO4. • Scientists do not always know how many H2O molecules are present, but this can be determined through calculations

  3. Anhydrous Compounds • Anhydrous Compounds do not have water molecules incorporated • Instead they have the ability to absorb water from the local environment.

  4. Applications • Hydrates can be used in Fire Retardation • Anhydrates can be used as dessicants (ie. Keep the atmosphere dry) • Anhydrates can indicate the presence of water.

  5. Example 1: Calculate the percentage of water in Na2S2O3. 5H2O Mass of water Mass of hydrate %H2O = x 100 = x 100 = 36.30 % 5(18.02 g/mol) 248.22 g/mol

  6. Example 2: Calculate the mass of water in 215 g ofNa2S2O3. 5H2O Solution 1: Use the percentage 36.30% (from the previous example). In Na2S2O3. 5H2O, there is 36.30% of water. In 215 g, there is x 215 g = 78.0 g of water 36.30 100

  7. Solution 2: = = m(H2O) = 78.0 g M(H2O) M (Total) m(H2O) M(Total) m(H2O) 215 g 5(18.02 g/mol) 248.22 g/mol

  8. Example 3: A hydrate of barium chloride (BaCl2.x H2O) has a mass of 1.500g. When heated to drive off the water, the residue (BaCl2) has a mass of 1.279 g Step 1: Calculate the percentage of water in the hydrate. BaCl2.x H2O (s)  BaCl2 (s) + x H2O 1.500 g 1.279g 1.500g – 1.279g = 0.221 g % H2O = x 100 = x 100 = 14.7% Mass of water Mass of hydrate 0.221 g 1.500 g

  9. Step 2: Calculate the formula of the hydrate Mole Calculation: Using n=m/M BaCl2 H2O n = n= = 0.006140 mol = 0.0123 mol 1.279g 208.3 g/mol 0.221 g 18.02 g/mol

  10. Step 3: Ratio Calculation: Divide by the smallest number of moles to figure out the ratio between the atoms BaCl2 H2O = = =1 = 2 Therefore the formula is BaCl2 . 2H2O 0.006140 mol 0.006140 mol 0.0123 mol 0.006140 mol

  11. Example 4: A hydrate of barium hydroxide (Ba(OH)2.x H2O) has a mass of 50.0g. When heated to drive off the water, the residue (Ba(OH)2) has a mass of 27.2 g Step 1: Calculate the percentage of water in the hydrate. Ba(OH)2.x H2O (s)  Ba(OH)2 (s) + x H2O 50.0 g 27.2g 50.0g – 27.2g = 22.8 g % H2O = x 100 = x 100 = 45.6% Mass of water Mass of hydrate 22.8 g 50.0 g

  12. Step 2: Calculate the formula of the hydrate Mole Calculation: Using n=m/M Ba(OH)2 H2O n = n= = 0.159 mol = 1.27 mol 27.2g 171.3 g/mol 22.8 g 18.02 g/mol

  13. Step 3: Ratio Calculation: Divide by the smallest number of moles to figure out the ratio between the atoms Ba(OH)2 H2O = = =1 = 8 Therefore the formula is Ba(OH)2 . 8H2O 0.159 mol 0.159 mol 1.27 mol 0.159 mol

More Related