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Mole Relationships in Chemistry Atomic masses O Atoms are composed of protons, neutrons and electrons O Almost all of the mass of an atom comes from the protons and neutrons O All atoms of the same element will have the same number of protons Atomic Masses O As a reference point, we use the atomic mass unit (u), which is equal to 1/12th of the mass of a 12C atom O One atomic mass unit (u) = 1.66 x 10-24 gram O Using this relative system, the mass of all other atoms can be assigned O Examples 7Li 14N 29Si = 7.016 004 u = 14.003 074 01 u = 28.976 4947 u Calculating Average Atomic Mass O Most elements occur in nature as a mixture of isotopes O Thus, atomic masses on periodic table are usually average values Average atomic mass % natural abundance = ∙ atomic mass1 100 1 % natural abundance + ∙ atomic mass2 … 100 2 Average Atomic Masses O One can calculate the average atomic weight of an element if the abundance of each isotope for that element is known O Example O Silicon exists as a mixture of three isotopes. Determine it’s average atomic mass based on the following data. Isotope Mass (u) Abundance 28Si 29Si 30Si 27.9769265 28.9764947 29.9737702 92.23 % 4.67 % 3.10 % Average Atomic Masses 28Si 92.23 100 29Si 4.67 100 (28.9764947 u) = 1.35 u 30Si 3.10 100 (29.9737702 u) = 0.929 u (27.9769265 u) = 25.80 u Average atomic mass for silicon = 28.08 u Atomic Masses and the Mole O The number of atoms in 12.000 grams of 12C can be calculated: One atom 12C = 12.000 u = 12 x (1.661 x 10-24 g) = 1.993 x 10-23 g / atom # atoms = 12.000 g (1 atom / 1.993 x 10-23 g) = 6.021 x 1023 atoms O The number of atoms of any element needed to equal its atomic mass in grams will always be 6.022 x 1023 atoms O Called the mole The Mole O 1 mole of any element = 6.022 x 1023 atoms O Atoms, ions and molecules are too small to directly measure in atomic mass units O Using moles gives us a practical unit! O We can then relate atoms, ions and molecules, using an easy to measure unit O The gram The Mole Molar Mass O Atoms come in different sizes and masses O A mole of atoms of one type would have a different mass than a mole of atoms of another type H 1.008 grams / mol O 16.00 grams / mol Mo 95.94 grams / mol Pb 207.2 grams / mol O We rely on a straight forward system to relate mass and moles Masses of Atoms and Molecules O Law of Definite Composition states that compounds always have a definite proportion of the elements that make it up O These proportions can be expressed as ratios of atoms, equivalent mass values, percentage by mass or volumes of gaseous elements O Example O Water always contains 2 H atoms for every O atom, which is 2 g H for every 16 g O or 11.1% H and 88.9% O by mass Mass Percent O How to Calculate Mass % O Obtained by comparing the MASS OF EACH ELEMENT present in 1 mole of the compound to the TOTAL MASS of 1 mole of the compound O A pure compound should show the same percent mass of each element consistently O So given a formula , you should be able to figure out the percent mass of each element 𝑀𝑎𝑠𝑠 % 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 × 100 = Elemental Analysis to Determine Mass Percent of a Compound O A sample is ‘burned,’ completely converting it to CO2 and H2O O Each is collected and measured as a weight gain O By adding other traps elements like oxygen, nitrogen, sulfur and halogens can also be determined O2 CO2 trap furnace sample H2O trap Elemental Analysis Example O A compound known to contain only carbon, hydrogen and nitrogen is examined by elemental analysis. The following information is obtained. Original sample mass Mass of CO2 collected Mass of H2O collected = 0.1156 g = 0.1638 g = 0.1676 g O Determine the % of each element in the compound Elemental Analysis O Mass of carbon 12.01 g C = 0.04470 g C 0.1638 g CO2 44.01 g CO2 O Mass of hydrogen 2.016 g H = 0.01875 g H 0.1675 g H2O 18.01 g H2O O Mass of nitrogen 0.1156 g sample - 0.04470 g C - 0.01875 g H = 0.05215 g N Elemental Analysis O Since we know the total mass of the original sample, we can calculate the % of each element 0.04470 g %C= x 100% = 38.67 % 0.1156 g % H = 0.01875 g x 100% = 16.22 % 0.1156 g % N = 0.05215 g x 100% = 45.11 % 0.1156 g Determining Chemical Formulas O Empirical formula O The lowest whole number ratio of elements in a compound O CH2 O Molecular formula O The actual ratio of elements in a compound O C2H4 O C3H6 O Empirical and molecular formula can be the same! O H2O Calculating Empirical Formulas from Mass % Pretend that you have a 100 gram sample of the compound O O Change the % to grams Convert the grams to moles for each element Write the number of each element as a subscript in a chemical formula O O O Keep each number as a decimal at this point! Divide each subscript by the smallest number Multiply the result by some integer to get rid of any fractions O O O May not be necessary How to Convert between Empirical Formulas and Molecular Formulas O Since the empirical formula is the lowest ratio, the actual molecule would weigh more O Molecular formula can always be obtained by multiplying by some whole number O To do so, divide the actual molecular molar mass (usually given in the problem) by the mass of 1 mole of the empirical formula O Gives whole number you MUST multiply the empirical formula by to get the molecular formula molar mass x empirical formula mass Using the Mole to Calculate Concentration of Solutions O Concentration is commonly expressed in terms of molarity O Defined mathematically as: moles of 𝒔𝒐𝒍𝒖𝒕𝒆 nsolute M= = volume of 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 (𝐋) Lsolution O “M” is read as “molar” O Molarity recognizes that compounds have different molar masses O A 1-molar solution of sucrose contains the same number of molecules as 1-molar solution of ethanol Other Methods of Expressing Concentration O When making different solutions with a specific molarity, the number of milliliters of solvent needed to prepare 1 liter of solution will vary O Sometimes it is necessary to know the exact proportions of solute to solvent that are in a particular solution O Various methods have been devised to express these proportions Molality Molality (m) = moles solute kilograms of solvent mol = kg •Recognizes that the ratio between moles of solute and kg of solvent can vary •A 1-molal solution of sucrose contains the same number of molecules as 1-molal ethanol Stoichiometry Chemical reactions O A chemical change involves the reorganization of the atoms in one of more substances O Chemical equation represents this process with the reactants on the left side of the arrow and the products on the right side of the arrow O Chemical reactions follow the Law of Conservation of Mass so the equation must be balanced! O Chemical equation for a reaction gives: O The nature of the reactants and products O Physical states! O The relative number of each Information conveyed by the balanced equation for the combustion of methane Stoichiometric calculations how to calculate amounts of reactants and products Stoichiometric calculations with limiting reactant Theoretical yield and Percent yield O Theoretical yield O The amount of a product formed when the limiting reactant is completely consumed O Determined using stoichiometry! O Percent yield O Percentage of theoretical yield that is actually produced in the laboratory 𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑥 100 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑