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# Composition of Substances and Solutions CHEM 1127Q 001

UCONN

GPA 3.9

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This 4 page Class Notes was uploaded by Caitrín Hall on Sunday March 6, 2016. The Class Notes belongs to CHEM 1127Q 001 at University of Connecticut taught by Fatma Selampinar (TC), Joseph Depasquale (PI) in Spring 2016. Since its upload, it has received 46 views. For similar materials see General Chemistry in Chemistry at University of Connecticut.

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Date Created: 03/06/16

Chapter 3: Composition of Substances and Solutions 3.1 Formula Mass and the Mole Concept Formula mass – sum of average atomic masses of all atoms in the substance’s formula Covalent formulas represent # and types of atoms in a molecule; formula mass = molecular mass Ionic compounds contain cations and anions but do not represent the composition of a discrete molecule; formula mass is not molecular mass Measured in amu The Mole – the amount of substance containing the sam12number of discrete entities as the numbers of atoms in a sample of pure C weighing exactly 12 g An amount unit like pair or dozen Avogadro’s number (N ) = 6.02 x 10 23entities composing a mole A Molar mass – mass in grams of 1 mole of that substance (g/mol) o Same numerical value of atomic/molecular mass of a substance 3.2 Determining Empirical and Molecular Formulas Percent Composition – percentage by mass of each element in the compound To find % comp. by each element, divide the experimentally derived mass of each element by the overall mass of the compound, then convert to a percentage o Ex) 12.04 g of a compound containing 7.34 g carbon, 1.85 g hydrogen, and 2.85 g nitrogen. What is the percent composition? To determine % comp. from formula mass, consider 1 mol of given compound and use its molar mass to calculate the percentage of each of its elements o Ex) find the % comp. of C 9 8 4 To determine empirical formula, use the given masses to find moles of each element, divide each by the lesser number of moles, multiply ratio (if necessary) to get the smallest possible whole number subscripts o Ex) find empirical formula of a compound with composition of 27.29% C and 72.71% O Derivation of Molecular Formulas Compare compound’s molecular or molar mass to its empirical formula mass Molar mass/empirical formula mass = n formula units/molecule Multiply each subscript of the empirical formula by n (A B ) = A B x y n nx nx 3.3 Molarity Solutions Concentration – relative amount of a given solution component Solvent – medium in which other components are dissolved; has a significantly greater concentration that that of other components Aqueous solution – a solution in which water is the solvent Solute – component of a solution present at a lower concentration than solvent o Dilute – relatively low concentration o Concentrated – relatively high concentration Molarity (M) – number of moles of solute in exactly 1 liter of solvent M = mol solute/L solution o Moles and volumes can be determined from molar concentrations o Molar concentrations can be determined from mass of solute o Mass of solute in given volume of solution can be determined from molarity Dilution of Solutions Dilution – the process whereby the concentration of a solution is lessened by the addition of solvent Common means of preparing solutions of desired concentration According to molarity, the molar amount of solute is equal to the solution’s molarity and volume in liters n = ML M1 1=M L 2 2 More general dilution equation: C1V =1 V 2 2 *C = concentration & V = volume* o Rearrange equation to find concentration of a diluted solution (solve for 2 ); volume of a diluted solution (solve fo2 V ); volume of a concentrated solution needed for dilution (solve for 1 ) 3.4 Other Units for Solution Concentrations Mass percentage – the ratio of the component’s mass to the solution’s mass (mass of component/mass of solution) x 100% Percent mass %mass, percent weight %weight, weight/weight percent (w/w)% Calculation of percent by mass o Ex) A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid? Divide mass of glucose by mass of sample Calculations using Mass Percentage o “Concentrated” hydrochloric acid is an aqueous solution of 37.2% HCl that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L of this solution? Use solution density given to find solution’s volume and mass, then use the given mass percentage to calculate solute mass Volume percentage = (volume solute/volume solution) x 100% Calculations using volume percentage o Ex) Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol? Given amount of solution x given volume% x given density Mass-volume percentage – ratio of a solute’s mass to the solution’s volume; (m/v) 6 Parts per million (ppm) = (mass solute/mass solution) x 10 ppm Parts per billion (ppb) = (mass solute/mass solution) x 10 ppb 9

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