Presentation on theme: "Solutions Solute Solvent Solutions are homogenous mixtures. When a substance is mixed with a liquid and it disintegrates into sub microscopic particles."— Presentation transcript:
Solutions Solute Solvent Solutions are homogenous mixtures. When a substance is mixed with a liquid and it disintegrates into sub microscopic particles such as molecules atoms or ions, the mixture is a homogenous mixture and therefore a solution. The substance dissolved in the other. The substance present in excess that dissolves the solute.
http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html Example of the dissolving process of an ionic compound in water because it can dissolve most ionic and polar compounds. Why is water called the universal solvent?
Vocabulary Solubility: amount of substance that can be dissolved in a certain solvent. Diluted: Contains small amount of solute for the given amount of solution. Concentrated: Contains large amounts of solute for the given amount of solution. Diluted solution Concentrated solution
Saturated solution: It cannot contain more solute at the given temperature. Supersaturated solution: Contains more solute that the amount it can contain at the given temperature. This is an unstable solution and the excess of solute will eventually separate, with any disturbance, or with the use of a seed crystal, from the solution forming a precipitate. This process is called crystallization http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/11solutionspag e/crystallization_from_super.htm
Is the amount of substance contained within a given volume of solution. c = Amount of substance Solution volume in dm 3 Concentration:
Molarity is the concentration of a solution in moles of solute per liter of solution. M = n V L MOLARITY or M = n V dm 3
Other forms of concentration used are: % composition (parts per hundred) and ppm (parts per million) ppm is used when the concentration is very small (like the addition of fluorine to tap water) ppm = mg dm -3 % by mass = g solute X 100 g solution
Sample Problems 11.5g NaOH X g NaOH Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution. What do we need to find first? The moles of NaOH… mol NaOH 1 Molar mass Na: 22.99 O: 16.00 H: 1.01 40.00g/mol 40.00 M = n = V (L) = 0.288 mol NaOH 1.50L solution = 0.192 mol L -1
Sample Problems 11.5g NaOH X g NaOH What is the concentration in ppm of the solution in the previous problem?. What do we need to find first? The mg of NaOH… mg NaOH 1,000 1 ppm = mg = V (L) = 11,500 mg NaOH 11,500 NaOH 1.50L solution = 7,670 ppm Solve the problems in your notes
Dilution When a solution is prepared dissolving a concentrated solution the dilution formula can be used. Formula: V c M c = V d M d Remember that you can only use this formula for dilution
Stoichiometry Stoichiometry refers to the use of a balanced equation to calculate the mass of products and/or reactants. Example: How many moles and what mass of carbon dioxide and water can be obtained when 5.00g of ethane (C 2 H 6 ) react with enough oxygen (Combustion reaction).
1.Write and balance the equation for the reaction 2. Convert the known (given) mass of the reactant or product to moles of the substance 3. Use the balanced equation to set the appropriate mole ratio 4. Use the ratio to calculate the number of moles of the desired reactant or product 5. Convert from moles back to grams if required by the problem.Steps Solve the problems in your notes
Limiting reactant is the reactant that is totally consumed during the reaction and therefore controls the amount of products that can be formed. Mg(s) + FeSO 4 (aq) ----- MgSO 4 (aq) + Fe(s) Excess reactant is the reactant that doesn’t react completely during the reaction and therefore there is an excess of it remaining after the reaction. 2.00g Which one will react completely? Limiting and excess reactants
Mg(s) + FeSO 4 (aq) ----- MgSO 4 (aq) + Fe(s) 2.00g Mg X 1 mol Mg 24.31g Mg = 0.0823 mole Mg 0.0823 mole Mg X 1 mole FeSO 4 1 mol Mg = 0.0823 mol FeSO 4 2.00g FeSO 4 X 1mole FeSO 4 151.92 g FeSO 4 = 0.0132 moles FeSO 4 We need 0.0832 moles FeSO 4 to react with 2.00 g of Mg and we have only 0.0132 moles of FeSO 4 Which reactant is going to be totally consumed in the reaction?Example Moles of FeSO 4 needed to react completely with 0.0823 moles Mg Moles of FeSO 4 that we have
FeSO 4 is the limiting reactant It will be totally consumed during this reaction It determines the amount of products that can be obtained Mg is the reactant in excess. This means that at the end of the reaction there will be some magnesium left. Do prob.15 p 177. Students do 16 How much iron (Fe) will be obtained? (explain) How much Mg will be left after the reaction? (explain) Answer: 0.320g Mg Answer: 0.737g Fe
Molarity (review) Molarity is the concentration of a solution in moles of solute per liter of solvent. M = n V in dm 3 M= Molarity N= number of moles of solute V L = Volume in liters of solution. Do prob.17and 18 p 179.
Percent YIELD The real amounts of products formed during a chemical reaction are usually different from the expected amounts. The real amount of one product obtained (experimental) is called the ACTUAL YIELD The expected amount, the amount of products calculated through the stoichiometry of the reaction, from the limiting reactant, is the THEORETICAL YIELD Do prob.19and 20 p 182. Percent Yield = Actual yield Theoretical Yield X 100 %
combustion reaction to find the Empirical formula A 0.496 g sample of an unknown hydrocarbon was completely burned in oxygen. The sample produced 1.56 g of carbon dioxide and 0.638 g of water. a) How many moles of carbon dioxide were formed? b) How many moles of water were formed? c) What is the empirical formula of the hydrocarbon? d) A 26.0 g sample of another unknown compound containing only carbon and hydrogen was burned in excess oxygen and 88g of CO 2 were produced. What is the possible molecular formula of this compound?