percent composition of hydrates 2020.notebook
TRANSCRIPT
Percent Composition of Hydrates 2020.notebook
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March 05, 2020
Jan 172:34 PM
Percent Composition of a Hydrate
Aim: What is a hydrate? How do we analyze the composition of a hydrate?
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What is a hydrate?Hydrate A salt compound that contains water molecules within it's crystal structure.
Hydrate formulas:
CuSO4●5 H2O 1 CuSO4 has 5 H2O molecules attached.
GFM of CuSO4 ● 5 H2O
Percent Composition of Hydrates 2020.notebook
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What is a hydrate?
Hydrate A salt compound that contains water molecules within it's structure.
Hydrate formulas:
CuSO4●5 H2O 1 CuSO4 molecule w/ 5 H2O molecules attached.
Hydrate symbol. The "dot" represents water molecules attached.
GFM of CuSO4 ● 5 H2O
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Heating a Hydrate
• When a hydrate is heated, the water molecules evaporate.
• The hydrate is now called an anhydrate or andhydrous salt it no longer contains water.
• mass of water= mass of hydrate mass of anhydrous salt
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Calculating the Percent of Water in a Hydrate by Heating
1. Calculate the percentage by mass of water in a hydrate using the percent composition by mass formula and the following information:
Mass of hydrate: 93.0 gMass of anhydrous salt remaining: 71.3 gMass of water lost: ______
(Experimental % of H2O in a Hydrate)
Jan 172:34 PM
Calculating the Percent of Water in a Hydrate by Heating
1. Calculate the percentage by mass of water in a hydrate using the percent composition by mass formula and the following information:
Mass of hydrate: 93.0 gMass of anhydrous salt remaining: 71.3 gMass of water lost: ______
(Experimental % of H2O in a Hydrate)
% H2O = mass H2O x 100
mass hydrate
% H2O = 21.7 g x 100 = 23.3%
93.0 g
mass hydrate= 93.0 g
mass anhydrate= 71.3 g
mass water= 93.0‐ 71.3
= 21.7 g
21.7 g
If the hydrates composition was LiClO4 ● 3 H2O then what is the theoretical % by mass of water in the hydrate?
GFM LiClO4 ● 3 H2O = 160.4 g/mol
% H2O = mass H2O x 100
mass hydrate
% H2O = 54.0 g x 100 = 34 %
160.4 g
To determine the theoretical % of water, you must analyze the formula using gfm values.
The experimental percent of water was only 23.3 % while the theoretical % of water was 34 % therefore, all of the water did not evaporate from the hydrate when it was heated.
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Calculating the Percent of Water in a Hydrate by Heating
2. A student heats a hydrate until the mass remains constant.
The student records the following data:Mass of hydrate before heating: 4.65 gMass of the hydrate after heating: 3.09 g
a. What was the mass of the water lost during heating?
b. What was the percent by mass of water in the hydrate before heating?
(Experimental % of H2O in a Hydrate)
Jan 172:34 PM
Calculating the Percent of Water in a Hydrate by Heating
2. A student heats a hydrate until the mass remains constant.
The student records the following data:Mass of hydrate before heating: 4.65 gMass of the hydrate after heating: 3.09 g
a. What was the mass of the water lost during heating?
b. What was the percent by mass of water in the hydrate before heating?
(Experimental % of H2O in a Hydrate)
4.65 g 3.09 g = 1.56 g
% H2O = mass H2O x 100 mass hydrate
% H2O = 1.56 g x 100 = 33.5% 4.65 g
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Percent of water in a Hydrate using GFM
4. What is the percent composition of water in the hydrate, Ba(ClO3)2 ● 7 H2O?
3. What is the percent composition of water in the hydrate, MnCl7 ● 4H2O?
(Theoretical % of H2O in a Hydrate)
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PRACTICE HYDRATE PROBLEMS3. What is the percent composition of water in the hydrate, MnCl7 ● 4H2O ?
4. What is the percent composition of water in the hydrate, Ba(ClO3)2 ●
7 H2O ?
GFM MnCl7● 4 H2O
Mn: 54.9 H: 8.0
Cl: 248.5 O: 64.0
375. 4 g/mol
GFM Ba(ClO3)2 ● 7 H2O
Ba: 137.3 H: 14.0
Cl: 71.0 O: 112.0
O: 96.0
430.3 g/mol
% H2O = 72.0 x 100 = 19.2 % 375.4
% H2O =126.0 x 100 = 29.3 % 430.3
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Water in a Hydrate (Experimental % of H2O)1. The initial weight of a hydrate is 5.06 grams. After heating, the final weight of the anhydrate is 4.50 grams. Determine the percent composition of water in the hydrate.
Water in a Hydrate by Formula AnalysisWhat is the percent by mass of water in FeCl3●6 H2O?
% H2O = mass H2O x 100 mass hydrate
% H2O = 0.56 g x 100 = 11.1% 5.06 g
mass hydrate= 5.06 gmass anhydrate= 4.50 gmass water= 5.064.50
= 0.56 g
GFM FeCl3●6 H2O Fe: 1 x 55.8= 55.8 H: 12 x 1.0= 12.0Cl: 3 x 35.5= 106.5 O: 6 x 16.0 = 96.0162.3 g + 108.0 g = 270.3 gsalt component water component hydrate
mass
% H2O = mass H2O x 100 mass hydrate
% H2O = 108.0 g x 100 270.3 g
% H2O = 39.95 % mass mass