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Chapter 3 Chapter 3 Molar Calculation of Molar

 Molar mass =  Calculate the molar mass of the following

Mass in grams of one of any element, Magnesium nitrate, Mg(NO 3)2 numerically equal to its atomic weight 1 Mg = 24.3050  Molar mass of can be determined from the 2 N = 2x 14.0067 = 28.0134 and molar masses of elements 6 O = 6 x 15.9994 = 95.9964 Molar mass of Mg(NO 3)2 = 148.3148 g Each H 2O contains 2 H and 1 O Calcium , CaCO 3 1 Ca = 40.078 Each mole of H O molecules contains 2 moles of H 2 1 C = 12.011 and 1 mole of O 3 O = 3 x 15.9994 One mole of O atoms corresponds to 15.9994 g Molar mass of CaCO 3 = 100.087 g Two moles of H atoms corresponds to 2 x 1.0079 g

Sum = molar mass = 18.0152 g H 2O per mole (II) , FeSO 4 Molar mass of FeSO 4 = 151.909 g

Chapter 5 Chapter 5 Solutions

 : a homogenous mixture in which the components are evenly distributed in each other  The properties and behavior of solutions often depend not only on the type of solute but also on the concentration of the solute.  Solute: the component of a solution that is dissolved in another substance

 Concentration : the amount of solute dissolved in a given quantity of or solution  Solvent: the medium in which a solute dissolved to form a solution – many different concentration units • (%, ppm, g/L, etc)

 Aqueous: any solution in which water is the solvent – often expressed as Molarity

1 Chapter 5 Chapter 5 Solution Concentrations Solution Concentrations  Molarity = moles of solute per liter of solution Designated by a capital M (mol/L)  Molarity can be used as a conversion factor.

 The definition of molarity contains 3 quantities: 6.0 M HCl 6.0 moles of HCl per liter of solution. Molarity = moles of solute of solution in liters 9.0 M HCl 9.0 moles of HCl per liter of solution.  If you know two of these quantities, you can find the third.

Chapter 5 Chapter 5 Solution Concentrations Solution Preparation

 Determine the molarity of each solution Example: How many moles of HCl are present in 2.5 L of 2.50 L of solution containing 1.25 mol of solute 0.10 M HCl? Molarity = moles of solute Molarity = moles of solute volume of solution in liters volume of solution in liters Given: 2.5 L of soln 225 mL of solution containing 0.486 mole of solute 0.10M HCl Use molarity as a Find: mol HCl conversion factor 100. mL of solution containing 2.60 g of NaCl

Strategy: Mol HCl = 2.5 L soln x 0.10 mol HCl 1 L of soln g →→→ mol →→→ molarity = 0.25 mol HCl

2 Chapter 5 Chapter 5 Solution Preparation Solution Preparation  Solutions of exact concentrations are prepared by dissolving the proper amount of solute in the correct Example: What volume of a 0.10 M NaOH solution is amount of solvent – to give the desired final volume needed to provide 0.50 mol of NaOH? Determine the proper amount of solute

Given: 0.50 mol NaOH Use M as a 0.10 M NaOH conversion factor Find: volsoln

How is the final volume measured accurately? Vol soln = 0.50 mol NaOH x 1 L soln 0.10 mol NaOH = 5.0 L solution

Chapter 5 Chapter 5 Solution Preparation Solution Preparation

Given: 250.0 mL solution, 1.00 M CuSO Example: How many grams of CuSO 4 are needed to 4 Find: g CuSO prepare 250.0 mL of 1.00 M CuSO 4? 4 Strategy: Given: 250.0 mL solution mL  L  mol  grams Conversion factors: 1.00 M CuSO 4 Find: g CuSO Molarity, molar mass 4 g CuSO4 = 250.0 mL soln x 1 L x 1.00 mol 1000 mL 1 L soln

Strategy: x 159.6 g CuSO4 mL  L  mol  grams 1 mol

= 39.9 g CuSO4

3 Chapter 5 Chapter 5 Solution Preparation Solution Preparation  Describe how to prepare the following:

500. mL of 1.00 M FeSO 4 Strategy: Steps involved in preparing solutions from mL  L  mol  grams pure solids

g FeSO 4 = 500.0 ml x 1 L x 1.00 mol x 151.909 g 1000 ml 1 L 1 mol

= 75.9545 g

100. mL of 3.00 M

250. mL of 0.100 M NaCl

Chapter 5 Chapter 5 Solution Preparation Solution Preparation  Solutions of exact concentrations can also be  Steps involved in preparing solutions from pure solids prepared by diluting a more concentrated solution of the solute to the desired concentration – Calculate the amount of solid required – Weigh out the solid – Place in an appropriate volumetric flask – Fill flask about half full with water and mix. – Fill to the mark with water and invert to mix.

4 Chapter 5 Chapter 5 Solution Preparation Solution Preparation

A given volume of a stock solution contains  Many laboratory chemicals such as acids are a specific number of moles of solute. purchased as concentrated solutions (stock solutions). – 25 mL of 6.0 M HCl contains 0.15 mol HCl (How do you know this???)  12 M HCl

12 M H 2SO 4 If 25 mL of 6.0 M HCl is diluted with 25 mL  More dilute solutions are prepared by taking a of water, the number of moles of HCl certain quantity of the stock solution and diluting present does not change. it with water. – Still contains 0.15 mol HCl

Chapter 5 Chapter 5 Solution Preparation Solution Preparation

Moles solute moles solute  When a solution is diluted, the concentration of the new solution = can be found using: before dilution after dilution Mc•Vc = moles = Md•Vd

Although the number of moles of solute Where, Mc = concentration of concentrated solution (mol/L) does not change, the volume of solution Vc = volume of concentrated solution

does change. Md = concentration of diluted solution (mol/L)

Vd = volume of diluted solution The concentration of the solution will change since Molarity = mol solute volume solution

5 Chapter 5 Chapter 5 Solution Preparation Solution Preparation

Example: What is the concentration of a solution prepared Mc •Vc = Md •Vd by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL? 6.00 M x 25.0 mL = Md x 50.0 mL

Given: Vc = 25.0 mL

Mc = 6.00 M Mc x Vc = Md x Vd M = 6.00 M x 25.0 mL = 3.00 M V = 50.0 mL d d 50.0 mL

Find: Md Note: Vc and Vd do not have to be in liters, but they must be in the same units.

Chapter 5 Chapter 5 Solution Preparation Concentrations of Acids  Describe how to prepare 500. mL of 0.250 M NaOH solution using a 6.00 M NaOH solution.  The pH scale pH = -log [H +] Given: Mc = 6.00 M pH of vinegar = -log (1.6 x 10 -3 M) = - (-2.80) = 2.80 Md = 0.250 M pH of pure water = -log (1.0 x 10 -7 M) = - (-7.00) = 7.00 Vd = 500.0 mL pH of blood = -log (4.0 x 10 -8 M) = - (-7.40) = 7.40

Find: Vc pH of = -log (1.0 x 10 -11 M) = - (-11.00) = 11.00 Mc•Vc = Md•Vd  Lower the concentration of H +, higher the pH

acidic substance, pH< 7  What volume of 2.30 M NaCl should be diluted to Basic substance, pH >7 give 250. mL of a 0.90 M solution? neutral, pH = 7

6 Chapter 5 Chapter 5 Concentrations of Acids Concentrations of Acids

 On serial dilution of acid solution, the pH increases because the concentration of H + ions decreases with  Calculate pH of 0.0065 M HCl solution. dilution

Concentration pH 0.1M HCl 1 0.01 M HCl 2 0.001 M HCl 3  Calculate the concentration of H + ion in a solution of pH 7.5.  The H + concentration of a solution of known pH can be calculated using the following equation:

[H +] = 10 -pH

Chapter 5 Chapter 5 Solution Preparation Review Solution Preparation Review  If you dissolve 9.68 g of potassium in 1.50 L,  What volume of 8.00 M should be diluted to what is the final ? produce 0.500 L of 0.250 M solution?

 How many grams of are contained in (dissolved in) 45.0 mL of 3.00 M solution?

 What’s the pH of the final solution?

7 Chapter 5 Chapter 3 Solution Preparation Review Conversion Factors  What volume of 2.06 M contains 322 g of the solute?

Number of particles Moles Mass

Avogadro’s number Molar 6.022 x 10 23 mass

Chapter 3 Chapter 3 Molar Conversions Molar Conversions

 Determine the following:  Determine the following:

The moles of potassium atoms in a 50.0 g sample The moles of FeCl 3 in a 50.0 g sample

1 mol Grams x = moles grams

The mass of MgCl 2 in a 2.75 mole sample

The mass of Mg in a 1.82 mole sample

grams Moles x = grams 1 mol

8 Chapter 3 Chapter 3 Molar Conversions Molar Ratios  glucose = C H O (Molar mass = 180.1 g) 6 12 6  The relative number of moles of each element in a substance can be used as a conversion factor How many moles of glucose are contained in 70.0 g of C H O ? called the molar ratio . 6 12 6 1 mol Grams x = moles grams  Molar ratio = moles element A How many moles of oxygen are contained in 70.0 g mole of substance of C H O ? 6 12 6 or

How many grams of H are contained in 70.0 g of  Molar ratio = moles element A C6H12 O6? moles element B

Chapter 3 Chapter 3 Molar Ratios Molar Ratios

 C6H12 O6:  H2O: – Molar Ratio = 6 moles of C – Molar Ratio = 2 moles of H mole C6H12 O6 mole H 2O – Molar Ratio = 12 moles H – Molar Ratio = 1 mole O mole C6H12 O6 mole H 2O

– Molar Ratio = 2 moles H – Molar Ratio = 6 moles O mole O mole C6H12 O6

– Molar Ratio = 6 moles C = 1 mol C 12 moles H 2 moles H

9 Chapter 3 Chapter 3 Molar Conversions Molar Conversions

 Molar ratios can be used to determine the Given: 70.0 g of C 6H12 O6 number of moles of a particular element in a Find: Moles of O given substance. Strategy:

 How many moles of oxygen are contained in Molar mass Molar ratio 70.0 g of C H O ? 6 12 6 grams of glucose →→→ moles of glucose →→→ moles of oxygen

Molar Moles of O = 70.0 g of C 6H12 O6 x 1mol C 6H12 O6 x 6 mol O Moles Ratio Moles 180.1 g mol C6H12 O6 C6H12 O6 O = 2.33 mol O

Chapter 3 Chapter 3 Molar Conversions Molar Conversions

 Once the number of moles of a substance present is  How many grams of H are contained in 70.0 g of known, we can use: C6H12 O6? – Molar mass to find the number of grams – Avogadro’s number to find the number of atoms, ions, or molecules Moles C H O Moles 6 12 6 A Molar Ratio Molar Ratio Grams Molar Moles NA Atoms H mass H H Grams Molar Moles NA Atoms B mass B B

10 Chapter 3 Chapter 3 Molar Conversions Molar Conversions

 How many moles of nitrogen are contained in 70.0 g of C H NO ? Given: 70.0 g of C 6H12 O6 6 5 3 Find: gram H

 How many grams of oxygen are contained in 1.5 moles g H = 70.0 g C 6H12 O6 x 1mol C 6H12 O6 x 12 mol H x 1.0079g of C H NO ? 180.1 g 6 5 3 mol C6H12 O6 mol H

=4.70g H  How many atoms of C are contained in 70.0 g of

C6H5NO 3? 23 # H = 70.0 g C 6H12 O6 x 1mol C 6H12 O6 x 12 mol H x 6.023 x 10 atoms

180.1 g mol C6H12 O6 mol H = 28.1 x 10 23 atoms of H

Chapter 3 Chapter 3 Molar Conversions Empirical and Molecular Formulas

 Empirical formula: smallest whole-number ratio of atoms How many moles of H + ions are present in 2.5 present in a compound moles of H SO ? 2 4  Molecular formula: actual number of each type of atom present in a given compound

Given: 2.5 moles H 2SO 4 Conversion factor:  Many molecular compounds have different empirical and Find: moles H + molar ratio molecular formulas Molecular Formula Empirical Formula

C6H6 CH Mol H + = 2.5 mol H SO x 2 mol H + 2 4 C6H12 O6 CH 2O 1 mol H 2SO 4 Mol H + = 5.0 mol H + N2H4 NH 2

N2O4 NO 2

11 Chapter 3 Chapter 3 Percent Composition Percent Composition

 Empirical formulas are generally obtained by determining Calculate the % composition of H 2O (i.e find %H and %O). the percent composition of a compound: First, find the FW of H O:  Percent composition: 2 the percentage of the mass contributed by each element in a substance FW = 2(1.0079 amu) + 1(15.9994 amu) = 18.0152 amu  % Element X = (# atoms of X)(AW) x 100% FW of compound % H = 2(1.0079 amu) x 100% = 11.2% H 18.0152 amu Formula Weight (FW): the sum of the atomic weights of all of the atoms in a chemical formula % O = 1(15.9994 amu) x 100% = 88.8 % O Note: Molar mass is a mass in grams that is numerically the 18.0152 amu same as the formula weight

Chapter 3 Chapter 3 Mass Percentage Calculating Empirical Formulas

 What is the mass percentage of each element in (CH N O)? 4 2  Percent composition data is commonly used to determine the empirical formula of a compound. FW = 12.011 amu + 4(1.0079 amu) + 2 (14.0067 amu) + 1(15.9994 amu) Empirical formula: smallest whole-number ratio of atoms present in a compound = 60.0554 amu Percent composition: the percentage of the mass contributed by each element % C = 1(12.011 amu) x 100% = 20.00% C in a substance 60.0554 amu

% H = 4(1.0079 amu) x 100% = 6.71 % H  Four steps: 60.0554 amu percent to mass % N = 2(14.0067amu) x 100% = 46.65% C mass to mole 60.0554 amu divide by smallest

% O = 1(15.9994 amu) x 100% = 26.64 % H multiple ‘til whole 60.0554 amu

12 Chapter 3 Chapter 3 Calculating Empirical Formulas Calculating Empirical Formulas

 A sample of has the following mass composition: 74.03% C; 8.70% H; 17.27% N. What is the empirical Step 2: Mass to moles formula of nicotine? # moles C = 74.03 g C x 1 mole C = 6.164 mol C 12.01 g C Step 1: % to mass # moles H = 8.70 g H x 1 mole H = 8.631 mol H If you assume 100.00 g of substance, then 1.008 g H 74.03 % C  74.03 g C # moles N = 17.27 g N x 1 mole N = 1.234 mol N 8.70 % H  8.70 g H 14.00 g N

17.27 % N  17.27 g N

Chapter 3 Chapter 3 Calculating Empirical Formulas Calculating Empirical Formulas

Step 4: Multiply ‘til Whole Step 3: Divide by smallest (this step is necessary only when step 3 (this gives the molar ratio of the elements) does not give whole number molar ratios) C: 6.164 moles C = 4.995 C: 6.164 moles C = 4.995 x 1 = 5.000 mol C 1.234 moles N 1.234 moles C H: 8.631 mol H = 6.994 H: 8.631 mol H = 6.994 x 1 = 7.000 mol H 1.234 mol N 1.234 mol C

N: 1.234 mol N = 1.000 N: 1.234 mol N = 1.000 x 1 = 1.000 mol N 1.234 mol N 1.234 mol C

13 Chapter 3 Chapter 3 Calculating Empirical Formulas Calculating Empirical Formulas

An iron compound contains 69.943 % Fe and Use these numbers to write the empirical 30.057 % O. Calculate its empirical formula. formula:

5.000 mol C Step 1: % to mass 7.000 mol H C5H7N 1.000 mol N – 69.943 % Fe  69.943 g Fe – 30.057 % O  30.057 g O

Chapter 3 Chapter 3 Calculating Empirical Formulas Calculating Empirical Formulas

Step 2: Mass to Moles Step 3: Divide by smallest

mol Fe = 69.943 g Fe x mol Fe = 1.2524 mol Fe O: 1.8786 mol O = 1.5000 55.845 g Fe 1.2524 mol Fe mol O = 30.057 g O x mol O = 1.8786 mol O 15.9994 g O Fe: 1.2524 mol Fe = 1.0000 1.2524 mol Fe

14 Chapter 3 Chapter 3 Calculating Empirical Formulas Using Empirical Formulas

 A sample of nicotine has the following mass Step 4: Multiply ‘til whole composition: 74.03% C; 8.70% H; 17.27% N  The molar mass of nicotine is 162.23 g  What is the molecular formula of nicotine? FeO 1.5 doesn’t make sense.

1.5000 mol O x 2 = 3.000 mol O Fe 2O3 1.0000 mol Fe x 2 = 2.000 mol Fe

Chapter 3 Chapter 3 Using Empirical Formulas to Find Using Empirical Formulas to Find Molecular Formulas Molecular Formulas Step 1:  Steps: Empirical formula of nicotine: C H N Find the empirical formula 5 7 Step 2: Formula Weight = 81.06 Calculate the formula weight for the empirical formula. Molar mass = Molecular weight = 162.23 Step 3: Get the whole number ratio between MW and Ratio= MW = 162.23 = 2.00 FW FW 81.06 Step 4: Multiply the subscripts in the empirical formula Molecular formula: C H N = C H N by the whole number ratio. (5x2) (7x2) (1x2) 10 14 2

15 Chapter 3 Chapter 4 Using Mass Percent Data Combustion Analysis

 glycol analyzes as 38.70 %C, and 9.74 %H A 1.125 g sample of a hydrocarbon was burned to produce with the remainder being oxygen. 3.447 g of CO and 1.647 g of H O. Determine the empirical  Determine the empirical formula of 2 2 formula.

Chapter 3 Empirical and Molecular Formulas

Molecular weight : Mass spectrometer

% C, H, O, N : Combustion analysis

Empirical and Molecular Formulas of an unknown compound

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