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Chapter 3
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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)
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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
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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
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6.02 x 10 23 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 10 23 22.4 L
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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
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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
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Grams reactant Often starting point Moles of reactant Divide by molar mass Moles of product Coefficients from balanced equation Unknown / known Grams of product
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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
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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%
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