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How Scientists Determine Formulas You have learned how atoms form bonds how to write formulas how to name compounds how to count the atom in compounds. Don’t you wonder how scientists figure out the formulas in the first place?

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Atomic Masses: Counting Atoms by Weighing Let’s REVIEW a bit: Atoms have very tiny masses so scientists made a unit to avoid using very small numbers. 1 atomic mass unit (amu) = 1.66 10–24 g TOO TINY TO WORK WITH IN A LAB!

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The Mole brings measurements to a scale we can work with ! One mole of anything contains 6.022 × 10 23 units of that substance. Avogadro’s number is 6.022 × 10 23 A sample of an element with a mass equal to that element’s average atomic mass (expressed in g) contains one mole of atoms.

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The Mole

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The average atomic mass for an element is the weighted average of the masses of all the isotopes of an element. It is found on the periodic table. Use digits to the hundredths place for all problems in this class. EX] Oxygen 16.00 g/mol Hydrogen 1.01 g/mol *Sometimes, in demo problems, I will round the values. I do this so you see the concept more easily and don’t get bogged down in numbers.

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1.Understand the definition of molar mass 2.Calculate molar mass 3.Learn to convert between moles and mass 4.Learn to calculate the mass percent of an element in a compound Objectives

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Molar Mass A compound is a collection of atoms bound together. The molar mass of a compound is obtained by summing the masses of the component atoms.

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Molar Mass For compounds containing ions, the molar mass is obtained by summing the masses of the component ions.

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Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of N 2 = 28.02 g/mol (2 × 14.01 g/mol) Molar Mass of H 2 O = 18.02 g/mol (2 × 1.01 g/mol) + 16.00 g/mol Molar Mass of Ba(NO 3 ) 2 = 261.35 g/mol 137.33 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol) Remember how to “count atoms” 1 Ba, 2 N, 6 O…because the subscript 2 applies to everything in the parentheses that precedes it

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Percent Composition of Compounds Percent composition consists of the mass percent of each element in a compound: Mass percent =

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For iron in iron(III) oxide,(Fe 2 O 3 ): Mass % O? You can do it 2 ways: 3(16.00 g) = 48.00 g = 30.06% 2(55.85 g) + 3(16.00 g) g159.70 g OR 100%- 69.94%= 30.06%...percent should add up to 100% Note: You can check your work. Total percent should always be very close to 100%, although it may be slightly off due to rounding.

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Calculate % composition from masses of each element Scientists also do this calculation from actual masses obtained from experiments. This is actually a very important research tool! EX] A new compound is extracted from a plant root. It breaks down as shown below. What is its % composition? Total sample mass = 0.2370 g C = 0.9480 g/ 0.2370 g x 100% = 40.00% O = 0.1264 g/ 0.2370 g x 100% = 53.33% H = 0.0158 g/0.2370 g x 100% = 6.67% _______ 100.00%

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Exercise Consider separate 100.0 gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 Rank them from highest to lowest percent oxygen by mass. H 2 O, CO 2, C 3 H 6 O 2, N 2 O

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SO WHAT! Why do we care? Because, from this information, we can figure out the actual chemical formula! This is how scientists determine the chemical formulas for substances! Discovery, research…SCIENCE !

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1.Understand the meaning of empirical formula 2.Learn to calculate empirical formulas 3.Learn to calculate the molecular formula of a compound Objectives

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Empirical Formulas The word “empirical” means it is based on observations, experiments, or experience…not theory or logic only Scientists collect data in experiments and use it to determine empirical formulas The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. The empirical formula can be found from the percent composition of the compound. The empirical formula can be found from masses of each element also.

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Calculation of Empirical Formulas

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Refresher Course! You need to remember how to calculate moles of an element from grams of the element Divide the given mass by the molar mass EXAMPLE: How many moles in 36.03 g C|1mol C = 3.00 mol C |12.01 g C

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Grams or percent…It doesn’t matter! If you are given grams, calculate number of moles by dividing by the molar mass If you are given %composition, change units from percent to grams and calculate moles by dividing by the molar mass. (You assume a sample size of 100 g to do this.) Why do we do it 2 ways? – Because sometimes experiments yield results measured in grams, and sometimes the results are measured as percentages of the whole sample.

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B. Calculation of Empirical Formulas

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Subscripts must be WHOLE numbers Sometimes when you divide all mole values by the smallest, you don’t get whole numbers! Multiply by the smallest whole number that will convert them all to whole numbers. See hint below Look for common decimal equivalents of fractions.25=1/4.33=1/3.5=1/2.75=3/4.67=2/3.2=1/5.17=1/6 Multiply all values by the denominator to get whole numbers

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Calculation of Molecular Formulas The molecular formula is the exact formula of the molecules present in a substance. Remember, only IONIC compounds are simplified to the lowest whole number ratio This is because molecular compounds actually form molecules (!!groups of atoms held together by bonds formed from shared electrons!!) Molecules can be multiples of the simplest formula (empirical) The molecular formula is always an integer multiple of the empirical formula. Molecular formula = (empirical formula) n where n is a whole number

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Molecular formulas How do you know when you need to calculate a molecular formula? – You must be given the actual molar mass of the compound. What do you do to calculate the molecular formula? – Divide the given molar mass by the mass of the empirical formula – Multiply each SUBSCRIPT by this value

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EXAMPLE: Glucose Molar mass of empirical formula =30 g/mol Given molar mass of substance = 180 g/mol Given mass = 180 = 6 Emp mass 30 Multiply each subscript by the calculated integer MOLECULAR FORMULA: C 6 H 12 O 6

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Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2C3H5O2 What is the molecular formula? C 6 H 10 O 4

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p223

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p225

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