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Mole Relationships in Chemistry The Mole Concept and Atomic Masses O The mole concept and molar mass is historically based on two laws from Joseph-Louis Proust in 1797 O The Law of Definite Proportions O This 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 - 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 O The Law of Multiple Proportions O These laws are based on mass O John Dalton’s atomic theory used these laws to formulate his atomic theory O His atomic theory is based on counts of atoms O The relationship between mass and count is done through the mole Deriving The Mole from Masses 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 Why is the Mole So Important? O 1 mole of any element = 6.022 x 1023 atoms O Using moles gives us a practical AND measurable unit! O Atoms, ions and molecules are too small to directly measure in atomic mass units O So by using the mole, we can relate atoms, ions and molecules, using an easy to measure unit -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 which you learned in first year chemistry! Using Molar Mass to Calculate % Composition (Mass %) O We can use the molar mass of a compound to determine the relative masses of the elements in a compound 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 of element in 1 mole of compound Mass % = × 100 mass of 1 mole of compound 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 Using % Composition to Determine Chemical Formulas O Based on the relative masses of the elements in a compound, we can directly calculate an empirical chemical formula O The lowest whole number ratio of elements in a compound O CH2 O We need additional information to determine the molecular formula O The actual ratio of elements in a compound O C2H4 O C3H6 O There are instances where empirical and molecular formula can be the same! O H2O Calculating Empirical Formulas from % Composition 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 Molecular molar mass x Empirical molar 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