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Transcript

Chapter 7: The Mole and Chemical Composition Notes Section 1: Avogadro’s Number and Mole Conversions Mole – a number - also the SI unit for amount of particles - 6.02 x 1023 things, like atoms or molecules - An Avogadro’s number of things (6.02 x 1023 is Avogadro’s Number) A mole is used in chemistry, like a dozen is used for donuts. Let’s say you buy 5 dozen donuts. How many donuts have you purchased? 12 donuts 5 dozen donuts X -------------------- = 60 donuts 1 dozen donuts Just like the above example, you can use Avogadro’s number to convert from moles to particles or from particles to moles. For example, if you have 0.50 moles of carbon atoms, how many atoms is that? 6.02 x 1023 atoms Carbon 0.50 mol Carbon X -------------------------------------- = 3.01 x 1023 atoms Carbon 1 mol Carbon You can also convert from atoms to moles. For example, if you have 1.2 x 1024 atoms of Helium, how many moles do you have? 1 mol He 1.2 x 1024 atoms He X ------------------------------- = 2.0 mol Helium 6.02 x 1023 atoms He Notice that the unit of mole is abbreviated (mol), we like to abbreviate whenever possible in chemistry, even if it is only 1 letter… You can also use this same technique to convert from moles of an element or compound to grams of that element or compound. See…we use a balance to measure out grams of substances, not moles. So it would be useful to be able to convert from grams to moles and vice versa. You use the molar mass (the number located below the symbol on the periodic table). Here is how it is done. Let’s say you have 0.50 moles of Carbon. How many grams is that? 12.01 grams Carbon 0.50 mol Carbon X --------------------------- = 6.0 grams of Carbon 1 mol Carbon This method only works if you label the units of the numbers you are working with. You must label your units with these kinds of problems! Chapter 7: The Mole and Chemical Composition Notes Section 2: Relative Atomic Mass and Chemical Formula Elements that are found in nature contain a mixture of different isotopes (an element with the same number of protons, but a different number of neutrons) of the element. The periodic table reflects this by showing the mass number as the average atomic mass. The average atomic mass is the weighted average of the atomic mass of all of the different isotopes of an element. Practice – If there are 2 isotopes of Copper, use their mass and relative abundance to calculate the average atomic mass of copper. Isotope Abundance Cu-62.94 69.17% Cu-64.93 30.83% Calculation: Average Atomic Mass = Sum of (Isotope x Decimal abundance) Average Atomic Mass = (62.94 amu x 0.6917) + (64.93 amu x 0.3083) Average Atomic Mass = (43.53 amu) + (20.02 amu) = 63.55 amu Second Practice – If there are 2 isotopes of Gallium, use their mass and relative abundance to calculate the average atomic mass of copper. Isotope Abundance Ga-68.926 60.00% Ga-70.925 40.00% Calculation: Average Atomic Mass = Sum of (Isotope x Decimal abundance) Average Atomic Mass = (68.926 amu x 0.6000) + (70.925 amu x 0.4000) Average Atomic Mass = (41.36 amu) + (28.37 amu) = 69.73 amu Formula Mass is defined as the molar mass of the elements in an element or compound’s formula. It is calculated by adding up the atomic mass of all of the atoms in a formula. Example: Find the formula mass of Zinc chloride Step 1 – Write the formula for zinc chloride = ZnCl2 Step 2 – Look up the atomic mass of elements = Zn (65.39 amu), Cl (35.45 amu) Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to guide you = 65.39 amu + 2 x 35.45 amu = 126.29 amu Second Example: Find the formula mass of Zinc sulfate Step 1 – Write the formula for zinc sulfate = ZnSO4 Step 2 – Look up the atomic mass of elements = Zn (65.39 amu), S (32.07 amu), O (16.00 amu) Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to guide you = 65.39 amu + 32.07 amu + 4 x 16.00 amu = 161.46 amu Third Example: Find the formula mass of Ammonium sulfate Step 1 – Write the formula for ammonium sulfate = (NH4)2SO4 Step 2 – Look up the atomic mass of elements = N (14.01 amu), H (1.01 amu), S (32.07 amu), O (16.00 amu) Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to guide you = 2 x 14.01 amu 8 x 1.01 amu + 32.07 amu + 4 x 16.00 amu = 132.17 amu Chapter 7: The Mole and Chemical Composition Notes Section 3: Formulas and Percentage Composition When chemists analyze the composition of a substance, the result is reported as a percentage of the elements that comprise the substance. This information is used to calculate the empirical formula for a compound. The empirical formula is the simplest whole number ratio of atoms in a molecule. Example: Chemical analysis of a liquid shows that it is 60.0% Carbon, 13.4% Hydrogen, and 26.6% Oxygen by mass. Calculate the empirical formula of the substance. Step 1 – Assume you have 100 grams of the sample Step 2 – Convert from grams of each element to moles 1 mol C 60.0 g C x ------------ = 5.00 mol C 12.01 g C 1 mol H 13.4 g H x ------------ = 13.3 mol H 1.01 g H 1 mol O 26.6 g O x ------------ = 1.66 mol O 16.00 g O Step 3 – In order to convert the ratios to whole numbers, divide the moles of each element by the lowest value of moles from step 2. 5.00 mol C --------------- = 3.01 mol C 1.66 1.3 mol H ------------- = 8.01 mol H 1.66 1.66 mol O -------------- = 1.00 mol O 1.66 Step 4 – Round the ratios to the closest whole number and write the formula for the compound C3H8O Molecular formulas are whole number multiples of the empirical formula. Sometimes the empirical and molecular formula for a compound are the same, like water for example (H2O). Molecular Formula = n (Empirical Formula) Where n equals some whole number Example Problem: The empirical formula for a compound is P2O5. The molar mass of the compound is 284 g/mol. Determine the molecular compound. Step 1 – Calculate the formula mass of the empirical formula = 141.94 g/mol Step 2 – Solve the Equation for n Experimental molar mass of compound n = ---------------------------------------------------------Molar mass of the empirical formula 284 g/mol n = --------------------- = 2.00 = 2 141.04 g/mol Step 3 – Multiply the subscripts of the empirical formula by n to find the molecular formula of the compound P2O5 P(2x2)O(5x2) = P2O10 Calculating percent composition from empirical formula Example: calculate the percent composition of Cu2S Step 1 – Calculate the formula mass of each component of the compound Cu – 2 x 63.55 g/mol = 127.10 g/mol S – 32.07 g/mol = 32.07 g/mol Step 2 – Add them up to find the molar mass 127.10 g/mol + 32.07 g/mol = 159.17 g/mol Step 3 – Calculate the fraction that each component contributes to the molecule 127.10 g/mol Mass % Cu = ------------------ x 100 = 79.85% Cu 159.17 g/mol 32.07 g/mol Mass % S = ------------------ x 100 = 20.15% S 159.17 g/mol