THE MOLE
THE MOLE Many common quantities have names that are used to describe them: Two of something are called a pair. Three of something are called a trio. Six of something are a half-dozen, and twelve are a dozen. In chemistry, scientists use the term mole to represent a specific quantity. A mole is a very large number, and is describes usable quantities of very small things- atoms, molecules, and compounds
AVAGADRO S NUMBER Named after the Italian physicist who first calculated it s value. Avagadro calculated the mole to be equal to 602,213,670,000,000,000,000,000. Rounded to 4 significant figures, Avagadro s number is 6.022 x 10 23. Because it is such a large number, it is only practical to use Avagadro s number when counting extremely small items. Marbles are much too large to count in moles A mole of marbles would cover the Earth to a depth of over six kilometers ( a layer of marbles over 3 1/2 miles deep)! However, because atoms are so small, Avagadro s number is a very good quantity to use for counting them.
DIMENSIONAL ANALYSIS You can have a pair of gloves, a pair of shoes, or a pair of earrings- a pair is two of anything. A mole works exactly the same way- it is 6.022 x 10 23 of anything. 1 mole = 6.022 x 10 23 things How many atoms of silver are found in 2.25 moles of silver? 2.25 moles Ag x 6.022 x 10 23 atoms = 1.35495 x 10 24 atoms Ag 1 mole
DIMENSIONAL ANALYSIS Twelve roses is a dozen roses; eighteen roses is 1.5 dozen roses. Using dimensional analysis, a known number of molecules or atoms can be expressed as multiples of moles. 6.022 x 10 23 things = 1 mole Express 7.382 x 10 24 atoms Na as moles of Na: 7.382 x 10 24 atoms Na x 1 mole = 12.26 moles Na 6.022 x 10 23 atoms
COUNTING BY MASS If the mass of an individual object is known, then the mass of a quantity of the object can be used to determine how many objects there are. For example... If one candy corn has a mass of 3.1 grams, then 31 grams of candy corn equals 10 pieces: 31 g x 1 piece = 10 pieces 3.1 g
COUNTING BY MASS The relationship that Avagadro discovered was that the atomic mass of an atom, expressed in grams, was always equal to 6.022 x 10 23 atoms of that element: 10.81 g of boron = 6.022 x 10 23 atoms of boron 63.55 g of copper = 6.022 x 10 23 atoms of copper 207.2 g of lead = 6.022 x 10 23 atoms of lead The atomic mass of an atom or moleculeexpressed in grams- is called the molar mass of that substance.
COUNTING BY MASS How many atoms of gold are in a ring that has a mass of 124.6 grams? First, create an equivalence statement using the molar mass of gold (atomic mass expressed in grams) and Avagadro s number: 196.97 g Au = 6.022 x 10 23 atoms Au Dimensional analysis can be used to make the conversion: 124.6 g Au x 6.022 x 10 23 atoms Au = 3.809 x 10 23 atoms Au 196.97 g Au
WORKING WITH MOLECULES The concepts described so far can be applied to molecules as well as elements. A mole of water molecules is equal to 6.022 x 10 23 water molecules. However, when converting to/from mass, the equivalence statement is based on the molar mass of water. Since a water molecule is made of 2 hydrogen and 1 oxygen atoms, the individual atomic masses of all of the atoms need to be added together to get the molecular mass: 1 mole H2O = 18.016 g H2O 1.008 + 1.008 + 16.00 = 18. 016, so 6.022 x 10 23 molecules H2O = 18.016 g H2O
WORKING WITH MOLECULES - PRACTICE NaCl CH4 MgBr2 O2 C6H12O6 22.99 + 35.45 = 58.44 g 12.01 + (4 x 1.008) = 16.04 g 24.31 + (2 x 79.90) = 184.11 g 16.00 + 16.00 = 32.00 g (6 x 12.01) + (12 x 1.008) + (6 x 16.00) = 180.16 g
EQUIVALENCE STATEMENTS - A REVIEW 1 mole of a substance = 6.022 x 10 23 atoms (or molecules) of the substance 1 mole = atomic (or molecular) mass in grams atomic (or molecular) mass in grams = 6.022 x 10 23 atoms (or molecules)
PERCENT COMPOSITION The percent by mass of each element in a compound is called the percent composition of that compound. For example, 100.00 grams of C6H12O6 contains 40.00 grams of carbon, 6.71 grams of hydrogen, and 53.29 grams of oxygen. Percent composition = mass of element/mass of compound x 100% % composition of carbon = 40.00/100.00 x 100% = 40.00% carbon % composition of hydrogen = 6.71/100.00 x 100% = 6.71% hydrogen % composition of oxygen = 53.29/100.00 x 100% = 53.29% oxygen 40.00% + 6.71% + 53.26% = 100.00%
PERCENT COMPOSITION BY FORMULA Percent compositions can be calculated by using just the formula of a compound. Assume that you have 1 mole of the substance, and use molar masses to calculate the percentages. For example, 1 mol of glucose (C6H12O6) has a molar mass of 180.16 grams. One mol of glucose contains 6 mol of carbon (6 x 12.01 g = 72.06 grams), 12 mol of hydrogen (12 x 1.008 g = 12.09), and 6 mol of oxygen (6 x 16.00 g = 96.00 grams): 72.06/ 180.16 x 100% = 40.00% carbon 12.09/180.16 x 100% = 6.71% hydrogen 96.00/180.16 x 100% = 53.29% oxygen
MOLECULAR AND EMPIRICAL FORMULAS Erythrose is a simple sugar with the formula C4H8O4. Ribose is a simple sugar with the formula C5H10O5. Glucose is a simple sugar with the formula C6H12O6. These are known as the molecular formulas for these substances, and are specific to one particular substance (C6H12O6 is always glucose, never ribose). However, in all three cases, the sugars have the same number of carbon atoms as oxygen atoms, and twice as many hydrogen atoms as either. The simplest representation of this pattern, known as the empirical formula, is CH2O. More than one substance can have the same empirical formula.
MOLECULAR AND EMPIRICAL FORMULAS If you know the percent composition of the elements in the compound, then the empirical formula can be calculated. For example, an unknown substance contains 40.05 % sulfur and 59.95% oxygen: 1- Assume that you have 100 g of your substance. This will convert the % values to masses (in g): 40.05% sulfur becomes 40.05 g S, and 59.95% oxygen becomes 59.95 g O. 2- Convert these masses to moles: 40.05 g S x 1 mol S = 1.249 mol S 32.07 g S 59.95 g O x 1 mol O = 3.747 mol O 16.00 g O
MOLECULAR AND EMPIRICAL FORMULAS 3- Divide both molar quantities by the smallest of the quantities: 40.05 g S x 1 mol 32.07 S = g S 1.249 1.249 mol S = 1 59.95 g O x 1 mol 16.00 O = g O 3.747 mol O 1.249 = 3 These numbers are the subscripts for the empirical formula; 1 sulfur atom and 3 oxygen atoms = SO3
MOLECULAR AND EMPIRICAL FORMULAS Calculating the molecular formula requires the empirical formula and one other piece of data- the molar mass of the substance in question (in this class, this data will be provided for you). If the empirical formula of a substance is CH, and the molar mass is 78.12 g, the molecular formula is calculated by: 1- calculate the molar mass of the empirical formula (one mole of CH has a mass of 13.02 g). 2- divide the experimental s molar mass by the empirical s molar mass (78.12 13.02 = 6 ). 3- multiply every subscript by this number = C6H6. This is the molecular formula of this substance.