o Calculating Percent Compositions of Elements Within a Compound o Using Percent Compositions to Calculate Empirical Formulas Chapter 3 Outline Atomic Mass o Atomic Mass Units o Conversion amu to g The Mole and Avogadro s Number o The Mole o Avogadro s Number o Molar Mass o Calculating Moles from Mass o Calculating Mass from Moles Molecular Mass o Calculating Molecular Mass o Calculating Moles from Mass Percent Composition of Compounds Experimental Determination of Empirical Formulas o Combustion o Determining Molecular Formulas Chemical Reactions and Chemical Equations o Components of a Chemical Reaction Reactants Products States of Matter o Chemical Equations as Ratios of Moles o Balancing Chemical Equations Stoichiometry o Mass of Reactant to Mass of Product o Mass of Reactant to Mass of Reactant o Stoichiometry and Dimensional Analysis Excess and Limiting Reagents o Determining Limiting Reagents o Calculating Mass Remaining of Limiting Reagents o Stoichiometric Quantities as Perfect Values Reaction Yield o Calculating Reaction Yields o Stoichiometry and Reaction Yield o Limiting Reagents and Reaction Yield Chapter 3 Notes Atomic Mass o Defined as the average mass of all isotopes of an element o Atomic mass is in units of atomic mass unit This unit is representative protons and neutrons have a mass of 1 amu and electrons have a mass of 0 amu This is to say that all of the mass of an atom comes from the protons and neutrons located in the nucleus o 1 g 6 022 x 1023 amu o 1 amu 1 661 x10 24 g The Mole and Avogadro s Number o The mole is the SI unit for the amount of a substance It is the amount of a substance that contains as many basic units atoms molecules or particles as there are atoms in exactly 12 g of carbon 12 o The number of carbon 12 atoms that make up exactly 12 g is represented by Avogadro s number Avogadro s Number AN 6 022 x 1023 o Avogadro s number can be used to calculate The number of moles given the number of particles or mass of a substance The mass or number of particles give the number of moles of a substance o Because of the relationship between amu grams and Avogadro s number the atomic mass of a substance is interchangeable with the number of grams in one mole of the substance o o Molecular Mass o Most of the time there will be several elements combined These molecules and compounds each have their own molecular mass o To calculate the molecular mass of a substance find the sum of the molar masses of each element in the substance For example the molecular mass of glucose C6H12O6 is 6 12 01 12 1 008 6 16 00 180 156 g mol o To calculate the number of moles of a substance divide the mass of the substance by its molecular mass Percent Composition o Recall that percent simply means part over whole o In the context of percent composition it means the molecular mass of the all the atoms of an element over the molecular mass of the whole molecule The molecular mass of glucose is 180 156 g mol This is the whole To find the percent composition for carbon we find the mass of all the carbon atoms in glucose which is 6 12 01 72 06 g mol We then divide this by the total mass to get 72 06 180 156 0 39999 which we would round to 40 00 o We can use percent compositions to calculate empirical formulas Since percent composition converts it into an out of 100 representation we say that we have 100 grams of the substance This makes it so that the percentages for each element represent the masses of each element From here we convert each mass to moles by dividing by molar mass Recall that all formulas are ratios of moles Since the empirical formula is the smallest whole number ratio of moles we divide each value by the smallest to get the smallest ratio o It is important to note that if you calculate the percent composition of elements in an empirical formula it will be the same as if you had calculated it from the molecular formula This is because the molecular formulas are simply ratios of empirical formulas o Percent compositions to be very familiar with are below Carbon in carbon dioxide 12 01 44 01 Hydrogen in water 2 016 18 016 Experimental Determination of Empirical Formulas o The empirical formulas of hydrocarbons and other organic compounds specifically ones that only contain carbon hydrogen and oxygen are determined trough combustion o When these compounds are burned they produce carbon dioxide and water Both of these substances are collected and their masses are measured o By the law of conservation of mass we know that the mass of a particular element present at the beginning of a reaction must be present at the end of the reaction If the unknown compound is a hydrocarbon CxHy Multiply the mass of carbon dioxide formed by 12 01 44 01 o This gives the mass of carbon in the compound Multiply the mass of water formed by 2 016 18 016 o This gives the mass of hydrogen in the compound Convert each of these masses to moles Divide by the smallest number of moles to get the empirical formula of the hydrocarbon If the unknown compound is an organic compound CxHyOz Multiply the mass of carbon dioxide formed by 12 01 44 01 o This gives the mass of carbon in the compound Multiply the mass of water formed by 2 016 18 016 o This gives the mass of hydrogen in the compound Find the sum of the carbon and hydrogen o Subtract this mass from the mass of the unknown compound to determine the mass of oxygen stuck in the compound Convert each of these masses to moles Divide each by the smallest number of moles to get the ratio of moles that make up the empirical formula o To determine the molecular formula of a compound If you know the empirical formula Calculate the mass of the empirical formula Divide the molecular formula weight by the empirical formula weight to get the scalar factor Multiply subscripts of the empirical formula by this factor to get o To determine the molecular mass of a compound the molecular formula You must know the mass of the sample and the number of moles Divide the mass by the number of moles o This is the molecular weight of the substance in g mol Chemical Reactions and Chemical Equations o All chemical formulas and equations are ratios of moles o The components of a chemical reaction are The reactants located on the left side of the equation The products located on the right side of the equation States of matter s l g aq Stoichiometry o The elements that are on the left of the equation as parts of the reactants must appear on the right side of the equation
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