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Chapter 3

Chapter 3. Stoichiometry. Atomic Mass Unit (u). Do you remember how small the mass of a proton and a neutron was? Proton = 1.67262 x 10 -24 grams Neutron 1.67493 x 10 -24 grams Scientists can’t work with these numbers! Therefore, they have come up with a plan:

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Chapter 3

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  1. Chapter 3 Stoichiometry

  2. Atomic Mass Unit (u) Do you remember how small the mass of a proton and a neutron was? Proton = 1.67262 x 10-24 grams Neutron 1.67493 x 10-24 grams Scientists can’t work with these numbers! Therefore, they have come up with a plan: They measure the mass of one atom in something called the Atomic Mass Units (u) An atom from one element was chosen as a standard, and the other elements were compared with it. Carbon was chosen as the standard for the atomic mass scale, it has a mass number of 12. Carbon has 6 protons and 6 neutrons in its nucleus. Therefore, one Carbon atom has a mass of 12 atomic mass units. Each proton and neutron are almost equal to 1 u (atomic mass unit). Now we can just refer to them as almost being 1 u! An atomic mass unit is defined to be 1/12 the mass of the carbon 12 nucleus.

  3. Average Atomic Mass If every proton and neutron are almost equal to 1u, then why do elements such as chlorine have an atomic mass of 35.5 u? Remember Isotopes?They occur in nature! For this reason, they will be effect the mass of atoms because they will be heavier than the normal atoms. We use the average mass of all atoms found in nature for the Periodic Table of Elements. Therefore, you will see atomic masses of 35.5 u when elements have a high amount of isotopes that occur in nature. To find the average atomic mass: Multiply the mass of each isotope by its abundance. The resulting products are then added together and the total is divided by the total abundance to get the weighted average.

  4. The Mole and Avogadro’s Number The mole (mol) Is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 grams of the carbon-12 atom. This number is called Avogadro’s number(NA), the value is approximately: NA = 6.022 x 1023 atoms in 12 grams of carbon-12 (or 1 atom of carbon-12).

  5. Molar MassOne mole = 6.022 x 10^23 atoms. Each element’s atomic mass, in amu (u) is equal to the mass (in grams) of one mole, this is called the Molar Mass.Example: sodium (Na) has 22.99 u and it’s molar mass is 22.99 grams. Therefore 1 mole of sodium equals 22.99 grams.

  6. Molecular Mass In the same way that we can find the molar mass of an element, we can also find the molecular mass of a molecule. We simply figure out how many of each different types of atoms we have within the molecule and then we add the atomic masses for each different type of atom together. Example: To find the molecular mass of water, H2O, is: H x 2 = 1.008 x 2 = 2.016 grams/mole O x 1 = 16.00 x 1= 16.00 grams/mole The total molecular mass is: 18.016 grams/mole +

  7. Percent Composition by Mass of an Element inside of a Compound Percent composition of an element = n x molar mass of element Molar mass of a compound x 100 % (n) = Number of moles of the element Example: H2O2 (molar mass = 34.02 g/mol) is calculated as: % H = 2 x 1.00 g % O = 2 x 16.00 g 34.02 g 34.02 g x 100% = 94.06 % O x 100% = 5.926 % H

  8. Using Percent Composition Backwards Knowing that we can calculate % Composition for an element off of the molecular formula or the empirical formula, what else can we do with it? We can use it backwards in order to determine the empirical formula of a compound off of know elemental %s for a compound. Why do we want to do that? Remember our mass spectra read outs? They tell us the % composition for every element within a compound. From this we could determine the empirical formula and figure out what unknown we have!!! Use the following POEM to help you: % to grams (just erase % symbol and replace with g) Grams to moles (Use molar mass) Divide by smole (smallest mole) X till whole (till you get a whole #)

  9. Let’s Try It Out!! Ascorbic Acid (Vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.94 % carbon (C), 4.58 % hydrogen (H), and 54.50 % Oxygen (O) by mass. Determine it’s empirical formula. Step 1: Assume that 100% = 100 grams. Write out the grams of each element. Step 2: Calculate the number of moles of each element in the compound by using the molar masses of each element. Step 3: Divide each mole of each element by the smallest amount of moles from one of the elements. Step 4: Create an empirical formula off of step 3. But, if you still do not have whole numbers, find the lowest whole number(s) that you can create by multiplying the decimal number(s) by the lowest whole number in order to get whole number product(s). At this point, create an empirical formula off of your whole numbers.

  10. Carbon = 40.92 g; Hydrogen = 4.58 g; Oxygen = 54.50 g Moles of Carbon = 40.92 g x 1 mole C = 3.407 mol C 12.01 g C Moles of Hydrogen = 4.58 g x 1 mole H = 4.54 mol H 1.008 g H Moles of Oxygen = 54.50 g x 1 mole O = 3.406 mol O 16.00 g O C: 3.407 =1 H: 4.54 = 1.33 O: 3.406 = 1 3.406 3.406 3.406 1.33 x 1 = 1.33 1.33 x 2 = 2.66 1.33 x 3 = 3.99 = 4 C3H4O3 = empirical formula

  11. Determining Molecular Formulas It makes sense that the formula we calculate from % Composition by mass will always be the empirical formula (lowest common denominator formula). If we want to calculate the molecular formula, we must know the approximate molar mass of the compound in addition to its empirical formula. Because we know that the molar mass of a compound is just a integral multiple of the molar mass of its empirical formula, the rest is simple!

  12. Example: A sample of a compound of nitrogen (N) and oxygen (O) contains 1.52 g of N and 3.47 g of O. The molar mass of this compound is known to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound. Moles of N = 1.52 g x 1 mol N = 0.108 mol N x 10 = 1 mol N 14.01 g N Moles of O = 3.47 g x 1 mol O = 0.217 mol O x 10 = 2 mol O 16.00 g O Empirical Formula = NO2 Empirical Molar Mass = 14.01 g + 2 ( 16.00 g) = 46.02 grams The molar mass is said to be 90 – 95 grams. Therefore, the integer multiple of the empirical molar mass and the actual molar mass is 2. Therefore the Molecular Formula must be 2 x the empirical formula. The accurate molar mass = 92.04 grams. Molecular Formula = 2 x NO2 = (NO2)2 or N2O4

  13. Writing a Chemical Reaction Reactant A + Reactant B Product C + Product D Writing the physical state of the reactant or product is very important! Therefore we represent solids with an (s) subscript. We write a liquid with a (l) and a gas with a (g) subscript. When a liquid or a solid solute will dissolve inside of the solvent water, we write the subscript (aq) for the term aqueous solution. This simply means that the solute is dissolved inside of the solvent water to produce ions in water. Example: 2CuCl(aq) + H2S(g) Cu2S(s) + 2HCl(aq) The numbers in front of the compound(s) or element(s) are called coefficientsand they represent the amount, in moles, of each of the elements present.

  14. Different Chemical Reactions Double Displacement or Single Displacement: Reactant A + Reactant B Product C + Product D Double Displacement (Metathesis) Reactions: Cation1Anion1 + Cation2Anion2Cation1Anion2+ Cation2Anion1 Single Displacement Reaction: Element1 + Cation2Anion2 Element1asCationAnion2 + Cation2(no charge) Element3 + Cation2Anion2 Cation2Element3asAnion+ Anion2(no charge)

  15. Synthesis Reaction Or Direct Combination Reaction: Reactant A + Reactant Bg Product C A + B gAB Decomposition Reaction: Reactant A g Product Z + Product Y Many Different Types, Some Examples Are: ABC gAB + BC ABC gA + BC ABC gAC + BC AB gA + B AB gAB + A AB gAB + B AxBy gAvBu Combustion Reaction: Hydrocarbon + Oxygen Carbon Dioxide + Water CH4 + 2O2 (g)gCO2(g) + 2H2O(l) Combustion can also have Hydrocarbons with Oxygen in them, such as the following: C6H12O6 + O2 (g)g CO2 (g) + H2O(l)

  16. Amounts of Reactants and Products Chemists usually ask the question, “How much product will be formed from specific amounts of starting materials (reactants)?” Stoichiometry is the quantitative study of reactants and products in a chemical reaction. The Mole Method is used to calculate the amount of product formed in a reaction from any type of units given for the reactants. The mole method means that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. This means that the following reaction would be read as, “2 moles of carbon monoxide gas combine with 1 mole of oxygen gas to form 2 moles of carbon dioxide gas” 2 CO(g) + O2(g) 2CO2(g)

  17. Steps to the Mole Method • Write the correct formulas for all reactants and products, and balance the resulting equations. • Convert the quantities of some or all given or known substances (usually reactants) into moles. • Use the coefficients in the balanced equation to calculate the number of moles of the sought or unknown quantities (usually products) in the problem. • Using the calculated numbers of moles and the molar masses, convert the unknown quantities to whatever units are required (typically grams). • Check that your answer is reasonable in physical terms.

  18. Using the Mole Method Let’s Figure out this Example: Ammonia reacts with diatomic oxygen to form nitrogen monoxide and water vapor. When 50.0 g of O2 are allowed to react with ammonia, how much (mass) of nitrogen monoxide will be formed? NH3 + O2NO + H2O 50.0 g O2 ? g NO moles O2 moles NO

  19. Mass-Energy Problems Given the thermochemical equation SO2(g) + ½ O2(g) SO3(g) + heat(99.1 kJ) Calculate the heat evolved when 74.6 g of SO2 (molar mass = 64.07 g/mol) is converted to SO3. 74.6 g SO2 1 mol SO2 -99.1 kJ = -115 kJ 64.07 g SO2 1 mol SO2 Another example: Calculate the heat evolved when 266 g of white phosphorus (P4) burn in air according to the equation: P4(s) + 5O2(g) P4O10(s) + heat(3013 kJ)

  20. Limiting Reagents When chemists carry out a reaction, the reaction usually does not have the exact proportions indicated by the balanced equation (stoichiometric amounts). Some of the reactants are therefore used up while others will have left over amounts at the end of the reaction. The reactant used up first in a reaction is called the limiting reagent. When the limiting reagent is all used up, no more products can be formed. The excess reagents are the reactants that are present in quantities greater than necessary to react with the quantity of the limiting reagent.

  21. Limiting Reagent Problems Ammonia reacts with diatomic oxygen to form nitrogen monoxide and water vapor. When 40.0 g of NH3 and 50.0 g of O2 are allowed to react, (a) Which reagent will be INXS and which reagent will be the Limiting Reagent (LR)? Remember, the LR will determine how much of the final product will be formed. • How much (mass) of nitrogen monoxide will be formed? • How much (mass) of the INXS reagent will be left over after the reaction? NH3 + O2NO + H2O 40.0 g NH3 50.0 g O2 ? g NO moles NH3 moles O2 moles NO (moles of O2 needed)(moles of NH3 needed)

  22. Calculating Combustion Reactions When a compound containing Carbon, Hydrogen (and possibly Oxygen) is completely combusted, all of the Carbon is converted to CO2 and all of the Hydrogen is converted to H2O. When can calculate the mass of Carbon, Hydrogen and Oxygen in the original hydrocarbon compound by knowing this: Example: A sample of Isopropyl Alcohol is known to contain only C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g H2O. Calculate the mass of H, C and O in the original sample. Step 1-Calculate the mass of C and H in the original sample by converting in the following way: g CO2g mol CO2g mol C g g C g H2Og mol H2O g mol H g g H 0.561 g CO2 x 1 mol CO2 x 1 mol C x 12.0 g C = 0.154 g C 1 44.0 g CO2 1 mol CO2 1 mol C 0.306 g H2O x 1 mol H2O x 2 mol H x 1.01 g H = 0.0343 g H 1 18.0 g H2O 1 mol H2O 1 mol H Step 2- Calculate the mass of O in the original sample by subtracting the above masses of H and C from the original mass of the sample: 0.255 g sample – (0.154 g C + 0.0343 g H) = 0.067 g O

  23. Reaction Yield and Percent Yield The amount of limiting reagent present at the start of a reaction determines the theoretical yield of the reaction, that is, the amount of the product that would result if all the limiting reagent reacted like shown in the balanced stoichiometric reaction. In experiments, the actual yield, or the amount of product actually obtained from a reaction, is almost always less than the theoretical yield. To determine how efficient a given reaction is, chemists often figure the percent yield, which is: % yield = actual yield x 100% theoretical yield

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