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Moles! Whatcha talkn’ about a mole? First…a dozen. 12 cookies = 12 bagels = 12 cans = 12 footballs = 12 dance moves = 1 dozen 1 dozen 1 dozen 1 dozen 1 dozen As scientist we need to determine something to quantify atoms, molecules, particles, or something very, very small As scientist we don't work with dozen....we work with something called a mole! Back to original question: "What the heck is a mole?!!" Same concept as a dozen but larger....much larger. If I want to have a mole of cookies…. 6.022 x 1023 times!! I mole of cookies = 6.022 x 1023 cookies What about a mole of dance moves? 6.022 x 1023 times!! I mole of dance moves= 6.022 x 1023 dance moves Mole in Chemistry? Obviously in chemistry you don’t have cookies and dance moves….we have Atoms Molecules Particles Formula units (ionic compounds) Electrons Abbreviation for mole is mol Moles! 1 mole of atoms = 6.022 x 1023 atoms 1 mole of molecules = 6.022 x 1023 molecules 1 mole of particles = 6.022 x 1023 particles 1 mole of formula units = 6.022 x 1023 formula units 1 mole of electrons = 6.022 x 1023 electrons 6.022 x 1023 is called Avogadro number (named after Amadeo Avogadro) So what does this mean? Converting….DA!!! Yeah! 1. How many moles are in 4.56 x 1024 atoms of hydrogen? 4.56 x 1024 atoms 1 mole 6.022 x 1023 atoms = 7.57 moles 2. How many formula units are in 8.92 moles of NaCl? 8.92 moles 6.022 x 1023 formula units 1 mole = 5.37 x 1024 formula units More moley practice… 1. How many atoms of oxygen are in 5.00 moles of carbon dioxide (CO2)? 5.00 mol CO2 6.022 x 1023 molecules CO2 2 atoms O 1 mole of CO2 = 6.02 x 1024 atoms of O 1 molecule CO2 Why moles, not atoms? 1 atom of Al = 4.48 x 10-23 grams Using moles to work with easier numbers 1 mole of Al = 26.98 grams Which is called molar mass ◦ Mass (in grams) of a mole Guess what….another conversion Yeah!! Boom! No doubt! Molar Mass 1. Determine the molar mass of Sucrose (C12H22O11) C – 12(12.01) = 342.29 g/mol H – 22(1.008) O – 11(15.999) 2. Determine the molar mass of Ca(NO3)2 Ca – 1(40.078) N – 2(14.0067) = 164.09 g/mol O – 6(15.999) Converting Moles to grams or grams to moles All conversions come from molar mass…aka periodic table 1. How many gramsMolar of aluminum are in 4.89 mass of Al moles of Al? 4.89 mol of Al 26.98 grams of Al 1 2. = 132 g of Al mol of Al Molar mass How many moles of CaCl2 are in 13.5 g of of CaCl CaCl2? 2 13.5 g of CaCl2 1 mol of CaCl2 110.984 g of CaCl2 = 0.122 mol of CaCl2 Practice 1. How many moles are in 25.67 grams of nickel? 0.4374 mol of Ni 2. How many grams are in 0.234 moles of lead(II) iodide (PbI2)? 108 grams of PbI2 More Practice 1. How many atoms are in 14.0 grams of manganese? 1.53 x 1023 atoms of Mn 2. How many molecules are in 46.7 grams of nitrogen gas? 1.00 x 1024 molecules of N2 3. How many oxygen atoms are in 3.45 grams of CO2? 9.44 x 1022 atoms of O Molarity What are these bottles we keep working with numbers and a big M? Well, Molarity is how scientist describe concentration of a solution Molarity = how many moles per 1 liter of solution Molarity What does this mean? For example: 2.00 M HCl 2.00 moles of HCl per 1 liter OR 2.00 moles of HCl = 1 L That’s right another conversion factor!! Yeah! Other examples 1. 0.500 M of Pb(NO3)2 0.500 moles of Pb(NO3)2 = 1 L 2. 1.56 M of NaCl 1.56 moles of NaCl = 1 L Using molarity…All about the DA 1. A 250. mL container has a solution of NaCl with a molarity of 1.54 M. How many moles of NaCl is present? 250. mL 1L 1.54 mol 1000 mL 2. 1 How many grams L = 0.385 mol of NaCl Molar mass ofNaCl NaCl would of 58.443 g 1 mol that contain? = 22.5 g of NaCl Why finding mass? You need the mass… - how much you need to measure in the lab (because you can’t measure moles on a scale) - make accurate concentrations of your solutions Practice 1. How many moles are present in 500.0 mL of 0.800 M solution of HCl? 500. 0 mL 1L 1000 mL 2. 0.800 mol = 0.400 mol of HCl 1L mass mL of How many grams are present Molar in 250. of Pb(NO3)2 0.500 M solution of Pb(NO3)2? 250. 0 mL 1L 1000 mL 0.500 mol 1L 331.21 g Pb(NO3)2 1 mol = 41.4 g of Pb(NO3)2 More Practice… 1. How many grams are needed to make 4.00 L of 0.250 M solution of NaCl? = 58.4 g of NaCl 2. How many grams are needed to make 2.50 L of 0.500 M solution of Na2CO3? = 132 g of Na2CO3 More Practice… Complete the table: Formula Grams Dissolved NaCl 4.00 grams KOH ? Moles Dissolved ? 2.56 x 10-3 Volume of Molarity Solution 1.50 L ? 1.80 L ? Moles NaCl = 0.0684 mol Grams KOH = 0.144 g Molarity of NaCl = 0.0456 M Molarity of KOH= 0.00142 M Making Solutions 1. Obtain correct volumetric flask Determine the correct number of grams needed to measure 3. Add the massed out compound to flask 4. Rinse weigh boat with DI into flask 5. Add DI to etched line on flask 2. Example 1. How many grams are needed Molar to make mass 250 mL of 0.250 M solution of CuSO4of ? CuSO4 250. 0 mL 1L 1000 mL 0.250 mol 1L 159.608 g CuSO4 1 mol = 9.98 g of CuSO4 -Mass out 9.98 grams of copper(II) sulfate -Hot dog weigh boat and use DI water to add CuSO4 to 250 mL volumetric flask -Add DI water to 250 mL mark -Add stir bar and stir till all dissolves -Add solution to appropriate containers Percent Mass and Determining Compound Formulas Find percent of each element in the compound Define Molecular and Empirical Formulas Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol %C= 24.02 g/mol 46.07 g/mol x 100 = 52.14% of C Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol %H= 6.048 g/mol 46.07 g/mol x 100 = 13.13% of H Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol %O= 15.999 g/mol 46.07 g/mol x 100 = 37.73% of O Mass Percent Practice 1. Find the percent of each element in CuSO4. Cu = (1)63.546 = 63.546 S = (1)32.065 = 32.065 O = (4)15.999 = 63.996 %Cu = 39.81% %S = 20.09% %O = 40.10% 159.607 g/mol 2. Find the percent of each element in C2H2. H = (2)1.008 = 2.016 C = (2)12.01 = 24.02 26.04 g/mol %H = 7.743% %C = 92.26% Empirical and Molecular Formulas Look at C2H2 (Acetylene gas) Acetylene + O2 + pumpkin = awesome! What is the ratio between hydrogen and carbon atoms? 2 atoms of H = 1 H :1 C 2 atoms of C What about the mole ratio between hydrogen and carbon? Same: 1 H : 1 C Why? 2 atoms O 1 mol 2 atoms H 1 mol 6.022 x 3.32 x 10-24 1023 moles H atoms 6.022 x 1023 atoms 3.32 x 10-24 moles O What does this tell us? A chemical compound’s formula can be determined by finding the moles of each element. Yeah!! Moles! I Moles This formula is called Empirical (strikes back) formula The empirical formula only tells you the ratio between elements… For example: C2H2 and C6H6 both have the same empirical formula… CH To find the correct formula (molecular formula)…you need the molar mass of the compound to be given. Example: Given molar mass of 78.1 g/mol and empirical formula of CH… Molar mass of CH = 13.02 g/mol (Periodic table) It takes 6 times of 13.02 to get 78.1… therefore the empirical formula (CH) gets multiplied by 6 and thus C 6 H6 Molecular Formula The exact formula for the compound. Ex. C4H10 or P4O8 Empirical Formula The reduced formula for the compound. Ex. C2H5 or PO2 The empirical can be the molecular just as MgCl2 or Fe2O3 Determine Empirical Formula Must find moles of each element (molar mass) 2. Find ratios between each element (large over smallest) Problem: A compound is made up of sulfur and oxygen. 25.0 grams each of sulfur and oxygen are present. Find the empirical formula. 1. 25.0 g S 1 mol of S 32.06 g S 25.0 g O = 0.780 mol of S 1.56 mol of O = 2 O :1 S 0.780 mol of S = 1.56 mol of O 15.999 g O Formula: SO2 1 mol of O Determine Molecular Formula Must find molar mass of empirical formula (periodic table) 2. Find ratio between given molecular molar mass and empirical molar mass (large over small) Problem: A compound is made up of sulfur and oxygen. 25.0 grams each of sulfur and oxygen are present. The molar mass of the compound is 128.12 g/mol. Determine the molecular formula. 1. Empirical Formula: SO2 Molar Mass of SO2 = 64.06 g/mol 128.12 g/mol = 2 :1 64.06 g/mol Molecular Formula: S2O4 Determine Empirical/Molecular Formula 1. 2. 3. 4. 5. Must find moles of each element (molar mass) Find ratios between each element (large over smallest) to get empirical formula Find molar mass of the empirical formula Find the ratio between molar mass Multiply the ratio to each element in the empirical formula Example 1 An unknown compound contains both hydrogen and carbon has a molecular molar mass of 26.04 g/mol. The compound contains 1.85 grams of hydrogen and 22.15 grams of carbon. Determine the empirical and molecular formulas. 1.85 g H 1 mol of H 1.008 g H 22.15 g C 1 mol of C 12.01 g C = 1.84 mol of H 1.84 mol of H = 1 H :1 C 1.84 mol of C = 1.84 mol of C Emp Formula: CH Molar mass of CH = 13.02 g/mol 26.04 g/mol 13.02 g/mol = 2:1 Mol. Formula: C2H2 Example 2 An unknown compound contains both sulfur and oxygen has a molecular molar mass of 160.1 g/mol. The compound contains 40.0% sulfur andAlert!!***** 60.0% oxygen Determine the *****Math empirical and molecular formulas. 40.0 g S 1 mol of S Attention: This is a math ALERT!!! = 1.25 mol of S 32.06 g S 3.75 mol of O = 3 O :1 S Treat percentage just like grams….it doesn’t 1.25 mol of S 1 mol of O 60.0 g O matter how= 3.75 much sample you have!! mol of O 15.999 g O Emp. Formula: SO3 Therefore have %...make Molar mass of SOif3 you = 80.06 g/mol 160.1 g/mol 80.06 g/mol = 2:1 it grams! Mol. Formula: S2O6