Stoichiometry Formulas and Mole

The Mole Chemical reactions involve atoms and molecules. The ratios with which elements combine depend on the number of atoms not on their mass. Individual atoms and molecules are extremely small. Hence a larger unit is appropriate for measuring quantities of matter. A mole is equal to exactly the number of atoms in exactly 12.0000 grams of carbon 12. This number is known as Avogadro’s constant. 1 mole is equal to 6.02 x 1023 particles. One gram of H atoms contains 6.02 x 1023 atoms.

Definitions of the Mole 1 mole of a substance has 6.02 x 1023 particles of that substance. 1 mole of atoms has a mass equal to the atomic weight in grams. Examples 1 mole of atoms of O is the number of atoms in 8g of O 1 mole of atoms of Na is the number of atoms in 23g of Na 1 mole H2O is the number of molecules in 18g H2O 1 mole H2 is the number of molecules in 2g H2. 1 mole of any gas at STP is equal to 22.4 dm3. STP is defined as 0 oC and 1 atmosphere of pressure.

The Molar and Formula Mass The Molar mass is the mass of one mole of a substance The units for Molar Mass are g mol-1. The formula mass is the sum of atomic masses in a formula. If the formula is a molecular formula, then the formula mass may also be called a molecular mass. Example: H2O From the Periodic Table - Atomic Masses: H =1.00797, O = 15.999 The formula mass = 2(1.00797)+15.999 = 18.015

Formula Mass and the Mole The atomic mass of Carbon 12 is exactly 12.00000. 1 atomic mass unit = 1/12 of the atomic mass of carbon 12. The periodic table gives the average atomic mass for an element relative to Carbon 12. 1 mole of a substance is 6.02 x 1023 particles. The mole of atomic mass units is equal to 1.000 gram.

Example 1: Calculating the Molar Mass of a Compound Calculate the formula mass or Molar Mass of Na3PO4. Atom # Atomic Mass Total Na 3 X 23.0 = 69.0 P 1 X 31.0 31.0 O 4 X 16.0 64.0 164.0 Therefore the molar mass is 164.0 g mol-1

Example 2: Find the mass of 2.50 moles of Ca(OH)2 Find the molar mass of Calcium hydroxide and multiply by 2.50 mol The molar mass of Ca(OH)2 is 1 Ca 1 x 40.08 = 40.08 2 O 2 x 16.00 = 32.00 2 H 2 x 1.01 = 2.02 Molar Mass = 74.10 g mol-1 2.50 mol x 74.10 g mol-1 = 185.25 g

Calculating Moles Moles = Mass The number of moles in a given mass of a substance can be determined by dividing the mass by the molar mass Moles = Mass Molar Mass

Example 3: Find the number of moles in 44.46 grams of Ca(OH)2 Find the molar mass of and divide it into the given mass From the previous example the molar mass of calcium hydroxide is 74.10 gmol-1. 44.46 g Ca(OH)2 . = 0. 6000 mol 74.10 g mol-1 Ca(OH)2

Calculating Mass From Moles The mass of a quantity of a substance can be found by multiplying the number of moles by the molar mass Mass = Moles X Molar Mass

Example 4 Calculating Mass from Moles Calculate the mass of 2.50 moles of Na3PO4 Mass = Moles X Molar Mass = 2.50 mol x 164.0 g mol-1 409 g

Percentage Composition To find the percentage of each element in a compound it is necessary to compare the total mass of each element with the formula mass. The percent by mass of each element in a compound is equal to the percentage that its atomic mass is to the formula mass. Example: Calculate the percentage of oxygen in potassium chlorate, KClO3 Atomic masses: K = 39.09, Cl = 35.45 and O = 16.00. Formula mass = 39.09 + 35.45+ 3(16.00) = 122.54 Percent Oxygen = (3(16.00)/122.54) (100) = 39.17%

Example Calculate the percentage by mass of each element in potassium carbonate, K2CO3 First calculate the formula mass for K2CO3 . 2 Potassium atoms K 2 x 39.10 = 78.20 1 carbon atom C 1 x 12.01 = 12.01 3 Oxygen atoms O 3 x 16.00 = 48.00 Total = 138.21 To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass Percent of Potassium K = 78.20 X 100 =56.58 % 138.21 Percent of Carbon C = 12.01 X 100 = 8.69 % 138.21 Percent of Oxygen O = 48.00 X 100 = 34.73 %

Empirical Formula Determination The empirical formula is the simplest whole number ratio of atoms of each element in a substance Example: CH2Ois the empirical formula for methanal (CH2O), ethanoic acid (C2H4O2) and glucose (C6H12O6) To find the empirical formula of a compound: Divide the amount of each element (either in mass or percentage) by its atomic mass. This calculation gives you moles of atoms for each element that appears in the formula Convert the results to small whole number ratios. Often the ratios are obvious. If they are not divide all of the other quotients by the smallest quotient

Example Analysis of a certain compound showed that 32.356 grams of compound contained 0.883 grams of hydrogen, 10.497 grams of Carbon, and 27.968 grams of Oxygen. Calculate the empirical formula of the compound. First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula Hydrogen H = 0.883 g = 0.874 mol 1.01 g mol-1 Carbon C = 10.497 g = 0.874 mol 12.01 g mol-1 Oxygen O = 27.968 g = 1.748 mol 16.00 g mol-1 Analysis of the ratios shows that the first two are identical and that the third is twice the other two. Therefore the ratio of H to C to O is 1 to 1 to 2. The empirical formula is HCO2

Molecular Formula Gives the actual number of atoms of each element in a molecule To calculate the molecular formula from the empirical formula it is necessary to know the molecular (molar) mass. Add up the atomic masses in the empirical formula to get the factor Divide this number into the molecular formula mass. If the number does not divide evenly you probably have a mistake in the empirical formula or its formula mass Multiply each subscript in the empirical formula by the factor to get the molecular formula

Molecular Formula Example Example: Suppose the molecular mass of the compound in the previous example, HCO2 is 90.0. Calculate the molecular formula. The empirical formula mass of is 1 H 1.0 x 1 = 1.0 1 C 12.0 x 1 = 12.0 2 O 16.0 x 2 = 32.0 Total 45.0 45 is exactly half of the molecular mass of 90. So the formula mass of HCO2 is exactly half of the molecular mass. So the molecular formula is double that of the empirical formula or H2C2O4

Structural formula Shows the arrangement of atoms and bonds within a molecule. Two compounds with same molecular formula but different structural formula are known as isomers or structural isomers.

Part 2: Stoichiometry Problems Mass-Mass Problems Mass-Volume

Stoichiometry Problems Stoichiometry problems involve the calculation of amounts of materials in a chemical reaction from known quantities in the same reaction The substance whose amount is known is the given substance The substance whose amount is to be determined is the required substance

Mass to Mass Problems Goal: To calculate the mass of a substance that appears in a chemical reaction from the mass of a given substance in the same reaction. The given substance is the substance whose mass is known. The required substance is the substance whose mass is to be determined.

Steps in a Mass to Mass Problem Find the formula masses for the given and the required substances Convert the given mass to moles by dividing it by the molar mass Multiple the given moles by the mole ratio to get the moles of the required substance Multiple the number of moles of the required substance by its molar mass to get the mass of the required substance

Summary of Mass Relationships

Example 1 Mass-Mass Problem Glucose burns in oxygen to form CO2 and H2O according to this equation: C6H12O6 + 6 O2  6 CO2 + 6 H2O How many grams of CO2 are produced from burning 45.0 g of glucose?

Example 2 Mass-Mass Problem What mass of Barium chloride is required to react with 48.6 grams of sodium phosphate according to the following reaction: 2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl

Example 2 = 92.6 g of BaCl2 2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl What mass of Barium chloride is required to react with 48.6 grams of sodium phosphate according to the following reaction 2 Na3PO4 + 3BaCl2  Ba3(PO4)2 + 6 NaCl Molar Masses: Na3PO4 = 3(23.0)+31.0+4(16.0) =164 g mol-1 BaCl2 = 137.3 +2(35.5) = 208.3 g mol-1 48.6g Na3PO4 x 3 mol BaCl2 208.3 g mol-1 BaCl2 164.0 g mol-1 Na3PO4 2 mol Na3PO4 = 92.6 g of BaCl2

Example 3 What mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen according to the following reaction : C2H5OH + 3 O2  2 CO2 + 3 H2O

Example 3 C2H5OH + 3 O2  2 CO2 + 3 H2O = 191.3 g CO2 What mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen according to the following reaction : C2H5OH + 3 O2  2 CO2 + 3 H2O Molar Masses: C2H5OH = 2(12) +6(1)+ 16 = 46 CO2 = 12 + 2(16) = 44.0 100.0 g C2H5OH x 2 mol CO2 X 44.0 g mol-1 CO2 46.0 g mol-1 1 mol C2H5OH = 191.3 g CO2

Mass to Volume Problems

Mass to Volume Problems Goal: To calculate the volume of a gas that appears in a chemical reaction from the mass of a given substance in the same reaction. Remember 1 mole of any gas at STP is equal to 22.4 dm3. STP is defined as 0 oC and 1 atmosphere of pressure.

Steps in a Mass to Volume Problem Find the gram formula masses for the given substance. Convert the given mass to moles by dividing it by the molar mass Multiple the given moles by the mole ratio to get the moles of the required substance Multiple the number of moles of the required substance by the molar volume, 22.4 dm3 mol-1, to get the volume of the required substance. This procedure is only valid if the required substance is a gas. It does not work for solids, liquids, or solutions.

Example 1 Mass-Volume Problem Sucrose burns in oxygen to form CO2 and H2O according to this equation: C12H22O11 + 12 O2  12 CO2 + 11 H2O What volume of CO2 measured at STP is produced from burning 100 g of sucrose?

Example 2 Mass-Volume Problem What volume of carbon dioxide gas would be produced by reacting 25.0 g of sodium carbonate with hydrochloric acid according to the following reaction: Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O = 5.28 dm3 of CO2

Summary of Stoichiometric Relationships

Solutions and Stoichiometry Many times the reactants and/or products of chemical reactions are water solutions. In these cases the concentration of the solution must be determined in order to determine amounts of reactants or products The concentration of a solution is a measure of the amount of solute that is dissolved in a given amount of solution

Molarity The most common concentration unit is Molarity Molarity (M) = Moles of solute dm3 of solution

Molarity Calculations How many grams of NaOH are required to prepare 250 cm3 of 0.500 M solution? Molar Mass of NaOH = 23+16+1 = 40.0 g/mol 250 cm3 = 0.250 dm3 (0.500 mol) x (40.0 g) x (0.250 dm3 ) = 5.00 g ( dm3 ) x ( mol )

Molarity Calculations Calculate the concentration of a NaCl solution that contains 24.5 g of NaCl in 250 cm3 of solution. Molar mass of NaCl = 23.0 + 35.5 = 58.5 (24.5 g NaCl) X 1 = 1.67 M (58.5 g mol-1 ) (0.250 dm3 )

Stoichiometry Calculations Involving Solutions 1 Copper metal reacts with nitric acid according to the following reaction: 8 HNO3 (aq) + 3 Cu  3 Cu(NO3)2 (aq) + 4 H2O (l) + 2 NO (g) What volume of 8.00 M HNO3 would be required to consume a copper penny whose mass is 3.08 grams? (3.08 g Cu ) (8 mol HNO3) (1 dm3) ( 1000 cm3) (63.55 g mol-1 Cu ) (3 mol Cu) = 16.2 cm3

Stoichiometry Calculations Involving Solutions 2 15.0 cm3 of a 0.500 M AgNO3 solution is required to precipitate the sodium chloride in 10 cm3 of a salt solution. What is the concentration of the solution? AgNO3 (aq) + NaCl (aq) AgCl (s) +KNO3 (aq) Molar Mass NaCl = 23.0 + 35.5 = 58.5 g/mol 0.500 mol AgNO3 X 0.0150 dm3 x 1 mol NaCl X 58.5 g mol-1 NaCl dm3 1 mol AgNO3 = 0.439 g of NaCl 0.439 g of NaCl x 1 58.5 g mol-1 0.0100 dm3 = 0.75 mol dm-3 or 0.75 M

Cookie Recipe Recipe Ingredients 1 cube butter 1 cup canola oil 2 cups white sugar 1 egg 1 teaspoon vanilla extract 1/2 teaspoon salt 1 teaspoon baking soda 4 1/2 cups all-purpose flour 1 cup oatmeal 1 (12 ounce) package chocolate chips Makes 24 cookies In my cupboard I have: 5 cubes butter 8 cups canola oil 8 cups white sugar 12 eggs 20 teaspoons vanilla extract 1 pound salt 40 teaspoons baking soda 45 cups all-purpose flour 30 cups oatmeal 5 (12 ounce) packages chocolate chips 5 pounds of dog biscuits How many cookies I can make with out going to the store?

Limiting Reagent Although we have been basing our calculations thus far on only one of the reactants in a chemical reaction, the reaction will only occur if we have all of the reactants The mole ratio determines how much of each reactant we need for the reaction Often we have an excess of one of the reactantsThen not all of that reactant will be used up. There will be some left over. It is known as the excess reagent. The other reactant will be used up and it will determine the amount of product we can form. It is known as the limiting reagent

Limiting Reagent To determine which of the reagents is the limiting reagent Calculate the number of moles of each reactant Multiply first reactant by the appropriate mole ratio to get the number of moles of the second reactant that you need. Compare the amount of the second reactant you have to the amount you need . If you have more than you need it is in excess and the first reactant is the limiting reagent If you have less of the second reactant than you need it becomes the limiting reagent Use the number of moles of the limiting reagent to calculate the required quantity in the problem

Limiting Reagent Example 1 Barium chloride reacts with potassium phosphate as follows: 3  BaCl2 (aq)  +  2  K3PO4(aq)  à  6 KCl (aq)    +  Ba3(PO4)2 (s) Calculate the mass of barium phosphate that could be formed when a solution containing 10.00 g of potassium phosphate is added to a solution containing 12.00 g of barium chloride. Molar mass potassium phosphate = 3(39.10) + (30.97) + 4(16.00) = 212.27 g mol-1 Molar mass barium chloride = (137.34) + 2(35.45) = 208.24 g mol-1 Molar mass barium phosphate = 3(137.34)+ 2(30.97)+(8)(16.00) = 601.96 g mol-1 Moles barium chloride = 12.00g / 208.24 g mol-1 = 0.05762 mol Moles potassium phosphate = 10.00g / 212.27 g mol-1 = 0.04711 mol The mole ratio is 3 mol BaCl2 to 2 mol K3PO4. While there are more moles of BaCl2 than K3PO4, It is not 1.5 times greater. Therefore BaCl2 is the limiting reagent and all other calculations will be based on barium chloride. (0.05762 mol BaCl2) (1mol Ba3(PO4)2) (601.96 g mol-1 Ba3(PO4)2 ) ( 3 mol BaCl2) = 11.56 g of Ba3(PO4)2

Percentage Yield Chemical reactions frequently do not proceed to completion. The amount of product recovered is often less than would be predicted from stoichiometric calculations. In these situations it is helpful to calculate a percent yield.       

Percent Yield The Theoretical Yield is defined as the amount of product(s) calculated using Stoichiometry calculations The Actual Yield is the amount of product that can actually be recovered when the reaction is done in a lab. The Percent Yield is calculated as follows Actual yield x 100 Theoretical yield

Percent Yield Iron  reacts with copper sulfate in a single replacement reaction as follows                 Fe (s)  +  CuSO4 (aq)   FeSO4 (aq)  + Cu  (s) 30.00 grams of iron metal  were added to excess were added to excess copper sulfate dissolved in a water solution. 22.50 grams of copper were recovered.  Calculate the theoretical yield of copper in this experiment . 1. First calculate the theoretical yield (30.00 g Fe) (1 mol Cu ) (63.55 g mol-1 Cu) (55.85 g mol-1 Fe) (1 mol Fe ) 2. Divide the actual yield by the theoretical yield and multiply by 100 = 34.14 g Cu 22.50 g Cu 34.14 g Cu X 100 = 65.91%