Mole Relationships in Chemistry

Transcription

1 Mole Relationships in Chemistry

2 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

3 Atomic Masses O As a reference point, we use the atomic mass unit (u), which is equal to 1/12 th of the mass of a 12 C atom O One atomic mass unit (u) = 1.66 x gram O Using this relative system, the mass of all other atoms can be assigned O Examples 7 Li = u 14 N = u 29 Si = u

4 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 = % natural abundance atomic mass 1 atomic mass 2

5 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 28 Si % 29 Si % 30 Si %

6 Average Atomic Masses 28 Si ( u) = u 29 Si ( u) = 1.35 u 30 Si ( u) = u Average atomic mass for silicon = u

7 Atomic Masses and the Mole O The number of atoms in grams of 12 C can be calculated: One atom 12 C = u = 12 x (1.661 x g) = x g / atom # atoms = g (1 atom / x g) = x atoms O The number of atoms of any element needed to equal its atomic mass in grams will always be x atoms O Called the mole

8 The Mole O 1 mole of any element = x 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

9 The Mole

10 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 grams / mol O grams / mol Mo grams / mol Pb grams / mol O We rely on a straight forward system to relate mass and moles

11 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

12 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 100 mass of 1 mole of compound

13 Elemental Analysis to Determine Mass Percent of a Compound O A sample is burned, completely converting it to CO 2 and H 2 O 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 O 2 furnace CO 2 trap H 2 O trap sample

14 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 CO 2 collected Mass of H 2 O collected = g = g = g O Determine the % of each element in the compound

15 Elemental Analysis O Mass of carbon g C g CO g CO 2 O Mass of hydrogen g H g H 2 O O Mass of nitrogen g H 2 O = g C = g H g sample g C g H = g N

16 Elemental Analysis O Since we know the total mass of the original sample, we can calculate the % of each element g % C = x 100% = % g % H = g x 100% = % g % N = g x 100% = % g

17 Determining Chemical Formulas O Empirical formula O The lowest whole number ratio of elements in a compound O CH 2 O Molecular formula O The actual ratio of elements in a compound O C 2 H 4 O C 3 H 6 O Empirical and molecular formula can be the same! O H 2 O

18 O O O O O O O O Calculating Empirical Formulas from Mass % Pretend that you have a 100 gram sample of the compound 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 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 May not be necessary

19 How to Convert between Empirical Formulas and Molecular Formulas O 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 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 x molar mass empirical formula mass

20 Using the Mole to Calculate O Concentration of Solutions Concentration is commonly expressed in terms of molarity O Defined mathematically as: M = moles of solute volume of solution (L) = n solute L solution O O O M is read as molar Molarity recognizes that compounds have different molar masses A 1-molar solution of sucrose contains the same number of molecules as 1-molar solution of ethanol

21 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

22 Molality moles solute kilograms of solvent Molality (m) = = 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

23 Stoichiometry

24 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

25 Information conveyed by the balanced equation for the combustion of methane

26 Stoichiometric calculations how to calculate amounts of reactants and products

27 Stoichiometric calculations with limiting reactant

28 Theoretical yield and O Theoretical yield Percent 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