Analytical Chemistry Examples and Solutions Using Percentage by Volume
Dive into practical examples and solutions in analytical chemistry by Dr. Jamal Ahmed Abdel Barry, focusing on percentage by volume calculations. Learn how to determine the percent by volume of ethanol in solutions, prepare isopropyl alcohol solutions, and more. Explore scenarios like diluting solutions and calculating molarity for a comprehensive understanding.
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Analytical Chemistry Analytical Chemistry By Dr. Jamal Ahmed Abdel Barry Professor in Clinical Biochemistry
Percent solutions (cont.)- B. Percent by volume (v/v) (usually a liquid - liquid) Volume of solute x 100 Volume % Solute = Volume of solution
Example # 1: What is the percent by volume of Ethanol (C2H6O) in the final solution when 20.0 mL EtOH are diluted to a volume of 250 mL ?
Volume of solute Volume % Solute = x 100 Volume of solution 20.0 mL EtOH x 100 X = 250 mL solution X = 8.00% EtOH (v/v)
Example # 2: Isopropyl alcohol can be prepared as a 90% Isopropyl (v/v) solution. If you dilute 25.0 mL of this solution to 100 mL with water, what will the final percent (v/v) be ?
Volume of solute Volume of solution Volume % Solute = x 100 90% = X mL Isopropyl 25.0 mL solution x 100 X = 22.5 mL Isopropyl 22.5 mL Isopropyl = 22.5% Isopropyl (v/v) 100 mL of solution
22.5 mL Isopropyl 100 mL of solution = 22.5% Isopropyl (v/v)
Example # 3: A 750 mL bottle of 40% Ethanol (v/v). How many mL of C2H6O can be distilled from this solution (assume 100%efficiency of the still ) ? What is the volume of water that will remain ?
Volume of solute Volume of solution Volume % Solute = x 100 X mL Ethanol 40% = x 100 750 mL solution X = 300 mL Ethanol Therefore, 450 mL of water remain.
Example # 4: The density of an Ethanol is 0.789 g/mL. What is the Molarity of the solution in the previous example ?
1. Using the density of EtOH, determine how many grams of EtOH are in the bottle. 2. Convert grams of EtOH to moles; and, convert of water to Liters. mL 3.Set up a proportion to determine Molarity.
X g 300 mL 0.789 g 1 mL X =236.7g EtOH =
236.7g EtOH 750 mL sol n. How much H2O ? Convert grams to Moles. Convert mL of water to Liters of water.
236.7g EtOH 5.14 Mol EtOH L of sol n X = 0.750 X 750 mL sol n. What is the molarity of EtOH in this solution ? 6.85 M
Percent solutions (cont.)- C. Percent mass/volume (m/v) (usually a solid-liquid) Mass of solute (g) x 100 % Mass / Volume = Solution volume (mL)
Example # 1: A solution contains 4.3g of Iron (II) sulfate in 0.591 liters of solution. What is the percent (m/v) of the solution ?
Mass of solute (g) % Mass / Volume = x 100 Solution volume (mL) 4.30 g 591 mL x 100 % Mass / Volume = % (m/v) = 0.728% (m/v) Iron (II) sulfate
What is the molarity of the above solution ? # of moles of solute Liter of solution Molarity = M = Iron (II) sulfate = FeSO4 = 151.9046 g/mol Therefore: 4.3 g FeSO4 = mol FeSO4 0.0283 X
0.0283 mol FeSO4 0.591 Liters of sol n 0.0479 M Molarity =
What is the molality of the previous solution ? (Assume addition of the 4.3g of solute does not effect the volume of solvent. Density of sol n = 1 g/mL) Moles of solute Kg of solvent Molality = m = 0.0283 mol 0.5867 Kg Moles of solute Kg of solvent = 0.0482 m =
Example # 2: How many liters of 2.00% glucose (w/v) solution can you prepare with 40.0 grams of Glucose ? 40.0g 2.00% = 0.0200 = X mL X = 2000 mL = 2.00 L
Example # 3: How many grams of Chloride are in 5.00 liters of seawater if the molal concentration of Chloride is 0.568m ? (The density of this seawater (the solvent) = 1.024 g/mL.) What is the molarity of this solution ? What is the percent (m/v) of this solution ?
A solution of HCl has a density of 1.1 6g/mL, and is labeled as 31.45% by mass (m/m). What is the molality of the solution? 1. Find the moles of solute (HCl) 2. 3. Find the mass of the solvent (water) Substitute values into the formula to solve.
Dilutions with Normality: What if you wished to dilute a more concentrated Normal solution to a specific concentration. How would you do it ? NiVi = NfVf
Example # 1: A lab requires 500 mL of 0.20 N Sulfuric acid. You have a significant volume of 4.0 N H2SO4. How do you prepare the desired solution ?
NiVi = NfVf 0.20 N x 0.500 L = 4.0 N x X X = 0.025 L Dilute 25 mL of 4.0 N Sulfuric acid to 500 mL.
Example # 2: A lab requires 870 mL of 2.0 N Potassium hydroxide. You have a significant volume of 3.0 N KOH. How do you prepare the desired solution ?
NiVi = NfVf 2.0 N x 0.870 L = 3.0 N x X X = 0.58 L Dilute 580 mL of 3.0 N Potassium hydroxide to 870 mL.
You can determine the Normality of an unknown solution by adding a specific volume of a solution of known Normality until the mixture is neutral. Nacid x Volumeacid = Nbase x Volumebase Both volumes must be the same unit.
Example # 1: If 17 mL of 2.0 N HCl neutralizes 50 mL of Ca(OH)2, what is the Normality of the base ? Nacid x Volumeacid = Nbase x Volumebase 2.0 N x 17 mL= X N Ca(OH)2 x 50 mL X = 0.68 N Ca(OH)2
Example # 2: A 25.0 mL sample of Ba(OH)2 solution was neutralized by 45.3 mL of 0.150 N HCl. What is the Normality of the Ba(OH)2 ? Nacid x Volumeacid = Nbase x Volumebase 0.150 N x 45.3 mL= X N Ba(OH)2 x 25 mL X = 0.272 N Ba(OH)2
