Chapter 3.  Reactants are left of the arrow  Products are right of the arrow  The symbol  is placed above the arrow to indicate that the rxn is being.

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Presentation transcript:

Chapter 3

 Reactants are left of the arrow  Products are right of the arrow  The symbol  is placed above the arrow to indicate that the rxn is being heated  Must have an equal number of atoms of each element on each side of the arrow  Remember you can change coefficients but NEVER change the subscripts  State of substance sometimes shown  Solid (s), liquid (l), gas (g), aqueous (aq)

 Combination (synthesis) reactions  2+ substances react to form one product  2 Mg(s) + O 2 (g)  2 MgO(s)  Decomposition reactions  One substance reacts to product 2+ substances  CaCO 3 (s)  CaO(s) + CO 2 (g)  Combustion  Reactions that produce a flame  Most have O 2 as reactant  Combustion of hydrocarbons and oxygen containing hydrocarbon derivatives generates CO 2 and H 2 O  If insufficient O 2 present will form CO in addition to CO 2

 Formula weight- sum of the atomic weights of each atom in the chemical formula  Aka molecular weight of molecules  Percent Composition- calculate the % of each element in the total mass of the compound (# of atoms of element)(atomic mass of element)x 100 Total mass of the compound

 6.02 x units  An element’s atomic mass expressed in grams contains 1 mole of atoms- molar mass  Going to moles divide, away from moles multiply Molar mass 6.02 x L

 Determining empirical from % composition  Convert % to mass (assume 100g)  Calculate moles of each element  Determine mole ratios (divide all moles by smallest #)  Multiply if needed to ensure all whole #’s  Molecular Formula from empirical Whole- number multiple = molecular weight Empirical formula weight  Multiply empirical by multiple to get molecular formula

 Combustion analysis  Calculate the mass of C and H  Moles of C per mole CO 2  Moles of H per mole H 2 O  Determine mass of O  Mass of O = mass of sample – (mass of C + mass of H)  Calculate moles of C, H, O  Divide by lowest to determine ratio

Grams reactant Often starting point Moles of reactant Divide by molar mass Moles of product Coefficients from balanced equation Unknown / known Grams of product

 Substance that controls the amount of product that can form  Solving:  Convert grams to moles  Divide by coefficient in balanced equation  Smallest # is limiting reactant  Stoichiometry problem- use limiting reactant

 Theoretical yield  Quantity of product calculated to form when all of the limiting reactant reacts  Actual yield  Amount of product actually obtained  Less than theoretical  Percent yield Actual yield x 100 Theoretical yield  Goal = close to 100%