Properties of Solutions II PDF

Properties of Solutions II Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU 11 Percent Composition One can find the percentage of the mass of each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (MW of the compound) x 100 21...

Properties of Solutions II Dr. Alya A. Arabi College of Medicine and Health Sciences UAEU 11 Percent Composition One can find the percentage of the mass of each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (MW of the compound) x 100 21 The percentage of carbon in ethane (C2H6) is… MW of C2H6 is 2x12 + 6x1 = 30 g/mol %C = (2)(12.0 amu) (30.0 amu) 24.0 amu x 100 = 30.0 amu = 80.0% 31 Percent Composition Practice Is NaCl 50.0% Na by weight? No MW of Na is 23.0 g/mole and Cl is 35.5 g/mole MW Na X 100 In fact, % Na = MW NaCl 23 X 100 = 39.3% (35.5 + 23) 41 Practice • % oxygen in CaCO3? 3(16) (48) = X 100 = 47.9520%  48.0%O [40.1 + 12.0 + 3(16)] (100.1) • Calculate the mass of Mg in 4.00 g of MgO First: Calculate % Mg in MgO 24.3 = 60.2977%  60.3%Mg (24.3 + 16) Second: Calculate g of Mg in 4.00 g of MgO 0.603(4.00 g ) = 2.41gMg 51 Empirical Formula The simplest formula indicating the mole ratio of elements in a compound. Only integers are allowed, no decimal. • Examples: – H2O2→ HO – C6H6→CH – N2O4→? 61 – CO2 → ? Practice: Solving Empirical Formuladetermined from gram composition • A compound contains 0.90 g Ca and 1.60 g Cl, what is its empirical formula? 1.60 0.90 = 0.045 moles Cl = 0.022 moles Ca 35.5 40.1 Divide by the smallest number 0.045 moles Cl 2 0.022 moles Ca 0.022 moles Ca/0.022 moles Ca = 1 CaCl2 71 Try this… What is the empirical formula for a chemical compound with 0.556 g C and 0.0933 g H? Units!! 0.0933 0.0556 = 0.0933molesH = 0.046molesC 1 12.0 0.0933 moles H 2 0.046 moles C CH2 (empirical formulae are not necessarily molecules that exist in 81 life) A compound containing carbon, hydrogen, and oxygen has the following composition: 68.12 g C, 13.73 g H, and 18.15 g O, Determine its empirical formula. Solution: Mole= m/MW Use mass percent to calculate mole and mole ratio of C:H:O Mole of C = 68.12 g x (1 mol C/12.01 g) = 5.672 mol C Mole of H = 13.73 g x (1 mol H/1.008 g) = 13.62 mol H Mole of O = 18.15 g x (1 mol O/16.00 g) = 1.134 mol O Divide all moles by mole of O (smallest value) to get simple ratio: 5.672 mol C/1.134 mol O = 5; 13.62 mol H/1.134 mol O = 12, and 1.134 mol O/1.134 mol O = 1; Mole ratio: 5C:12H:1O ➔ Empirical formula = C5H12O 91 What is the empirical formula given the following? • K = 0.26 moles • N = 0.25 moles • O = 0.78 moles KNO3 101 Ways of Expressing Concentration moles solute Molarity = liters of solution • Calculate the molarity of a solution prepared by dissolving 9.8 moles of solid NaOH in enough water to make 3.62 L of solution? Molarity (M) = 9.8 mol/3.62 L = 2.7 mol/L • You dissolve 152.5 g CuCl2 (134.45 g/mol) in Water to make a solution with a final volume of 2.25 L. What is its molarity? n=m/M= 152.5 g / 134.45 g/mol = 1.13425 mol 1.13425 mol/2.25 L = 0.5041 mol/L 111 https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_( 121 CK-12)/16%3A_Solutions/16.11%3A_Molality 131 Ways of expressing extremely dilute concentration Mass Percentage, ppm, and ppb • This is a way of expressing very dilute concentrations of substances, e.g. average amount of iron in blood can be reported in ppm. • Per cent, %, means out of a hundred. • parts per million, ppm, means out of a million. • ppm usually describes the concentration of something in water or soil. One ppm is equivalent to 1 milligram of something per liter of water (mg/l) or 1 milligram of something per kilogram soil (mg/kg). mass of component in solution ppm of component =  106 total mass of solution Ways of Expressing Concentration Mass Percentage, ppm, and ppb mass of component in solution ppm of component =  106 total mass of solution • Parts per million (ppm) can be expressed as 1 mg of solute per kilogram of solution. – If the density of the solution is 1g/mL, then 1 ppm is equivalent to 1 mg solute per liter of solution. • Parts per billion (ppb) are 1 g of solute per kilogram of solution. mass of component in solution ppb of component =  109 total mass of solution What would part per trillion, ppt, equal to? 161 A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. (a) What is the mass percentage of solute in this solution? Solution (a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan: We can calculate the mass percentage by using the equation discussed earlier. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve: Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%. 171 PRACTICE EXERCISE Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. Answers: 2.91% 1.50/(1.50+50.0)*100 181 (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+ What is the concentration of Zn2+ in ppm? Analyze: In this case we are given the number of micrograms of solute. Because 1 g is 1  10–6 g, 5.4 g = 5.4  10–6 g. Plan: We calculate the ppm using the following equation Solve: 191