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Atomic Unit Calculations
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Calculating Atomic Mass Units (amu) Definition: A unit of mass used to express atomic and molecular weights.
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Mass of Fluorine Mass of one proton = 1.67 x 10 -27 Mass of one neutron = 1.67 x 10 -27 Mass of one electron = negligible Fluorine has: 9 protons, and 10 neutrons Total mass of Fluorine = (9 protons x 1.67 x 10 -27 g) + (10 neutrons x 1.67 x 10 -27 g) = 3.173 x 10 -26 g This is too much work!!!!
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Mass of Fluorine What is the mass of fluorine? Mass = Protons(+) + Neutrons(0) Mass of Fluorine = 9 protons + 10 neutrons = 19 amu
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Ions Cations = positive charge Anions = negative charge How many electrons does F - have? Fluorine has 9 electrons when neutral Minus one charge indicates there is one more electron then normal. So, F – has 10 electrons
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Gram Formula Mass Definition - The mass in grams of one mole of a molecular substance.massmole – The mole is a unit of measurement used in chemistry to express amounts of a chemical substance
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What’s a mole? MOLE CALCULATIONS 6.02 x 10 23 particles
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1 mole of H 2 = 6.02 x 10 23 molecules of H 2 1 mole of H 2 = 2 x (1.0 g) = 2.0 g 1 mole of He = 6.02 x 10 23 atoms of He 1 mole of He = 4.0 g
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1 mole of H 2 O = 6.02 x 10 23 molecules of H 2 O 1 mole of H 2 O: H = 1.0 O = 16.0 18.0 g x 2 = 2.0 g x 1 = 16.0 g
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1 mole of MgCl 2 = 6.02 x 10 23 formula units of MgCl 2 1 mole of MgCl 2 : Mg = 24.3 Cl = 35.5 95.3 g x 1 = 24.3 g x 2 = 71.0 g
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1.KMnO 4 K = 39.1 Mn = 54.9 O = 16.0 x 1 = 39.1 x 1 = 54.9 x 4 = 64.0 158.0 g 2.KCl K = 39.1 Cl = 35.5 x 1 = 39.1 x 1 = 35.5 74.6 g
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3.Na 2 SO 4 Na = 23.0 S = 32.1 O = 16.0 x 2 = 46.0 x 1 = 32.1 x 4 = 64.0 142.1 g 4.Ca(NO 3 ) 2 Ca = 40.1 N = 14.0 O = 16.0 x 1 = 40.1 x 2 = 28.0 x 6 = 96.0 164.1 g
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Law of Conservation of Matter Matter can not be created or destroyed In a chemical reaction, the mass of the reactants must equal the mass of the products. Everything must equal Mass of Reactants = Mass of Products
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Law of Conservation of Matter Example: Hydrogen + Oxygen Water H O H 2 0 1 H ≠ 1 H 2 1 O = 1 O How do we balance the equation?
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L. O. C. cont. Hydrogen + Oxygen Water H O H 2 0 Use numbers to help balance equations 2 H + 1 O 1 H 2 O
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Practice Balance the following equations H 2 + O 2 H 2 O Na + Cl 2 NaCl
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Conversions This Again???
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Conversions We will now be converting units into and from moles. The number of particles (atoms/molecules) in one mole is … 6.02 x 10 23 This is known as Avogadro’s number (use this as a conversion factor)
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Conversions Example: Convert 1 gram of water into atoms Water = H 2 0 1 st step: Set up dimensional analysis moles atoms 1 g H 2 0 _________ _________ = # of atoms in 1 g H 2 O grams moles (Identify conversion factors)
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Conversions Step 2 – Plug in numbers Water = H 2 O How many moles is in one gram of water? First, determine how many grams are in one mole of water. 1 mole of hydrogen = 1 gram 1 mole of oxygen = 16 grams So, 1 mole of H 2 O = ( 1 gram H x 2) + (16 grams O) = 18 grams
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Conversions One mole of H 2 O = 6.02 x 10 23 atoms 6.02 x 10 23 atoms ______________ 1 mole
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Conversion Practice Convert 1 gram of glucose (C 6 H 12 O 6 ) into the number of glucose molecules….
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Lewis Dot Structure Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol Lines are drawn between atoms that are bonded to one another (pairs of dots can be used instead of lines)
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Lewis Dot Structure cont.
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Percent Composition Definition: The mass percent of each element in a compound. To calculate the percent composition of a component in compound: 1.)Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator. periodic tablemolecular mass calculator Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms. 2.)Divide the mass due to the component by the total molar mass of the compound and multiply by 100.
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Percent Composition
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Percent Composition Practice Calculate the % composition by mass of each of the following. 1.) H 2 O 2.) NaCl
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Empirical Formula A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms
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Determining an Empirical Formula Start with the number of grams of each element, given in the problem. – If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given.
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Determining an Empirical Formula Convert the mass of each element to moles using the molar mass from the periodic table. periodic table Divide each mole value by the smallest number of moles calculated.
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Determining an Empirical Formula Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. – If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple. e.g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3. e.g. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.
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