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AP Chemistry Chapter 3

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Section 1

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molecular formula vs empirical formula

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Chemistry Science

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Mar 1, 2020

Cards (23)

Section 1

(23 cards)

molecular formula vs empirical formula

Front

the subscripts in the molecular formula of substance are always whole-number multiples of the subscripts in its empirical formulas , whole number ratio= molecular weight/ empirical formula weight

Back

formula weight (ionic)/molecular weight(molecular)

Front

of a substance is the sum of the atomic weights of the atoms in the chemical formula of substance, coefficients times the weight on the periodic table

Back

decomposition reaction

Front

one substance undergoes a reaction to produce two or more other substances

Back

theoretical yields

Front

the quantity of the product calculated to form when all of a limiting reactant is consumed ,

Back

molar mass

Front

the mass in grams of one mole of a substance

Back

reactants

Front

the chemical formula to the left of the arrow represent the starting substances called reactants

Back

moles in balanced equations

Front

the coefficients in a balanced chemical equation indicate both the relative number of molecules and moles present in that reaction

Back

combination reactions

Front

two or more substances react to form one product, a combination reaction between a metal and a nonmetal produce an ionic solid

Back

coefficients

Front

the numbers in front of formulas

Back

actual yield

Front

the amount of product actually obtained

Back

empirical formulas to moles

Front

the ratio of the numbers of the numbers of moles of all elements in a compound gives the subscripts in the compounds empirical formula, thus the mole concept provides a way of calculating empirical formulas

Back

combustion analysis

Front

one technique for determining empirical formulas in the laboratory, used for compounds with primary H and C, steps: 1. find the grams of carbon from the mass of CO2 and hydrogen from the mass of H2O 2. once you find the mass of H and C you find the weight of the third element (aka O) by , Mass of O= mass of sample - (mass of C + mass of H) 3. then you find the empirical formula of that compound

Back

percentage yield

Front

of a reaction relates actual and theoretical yields = (actual yield/ theoretical yield) x 100

Back

chemical equations

Front

what we use to represent chemical reactions

Back

limiting reactants

Front

a reaction stops as soon as one reactant is totally consumed, aka the limiting reactant,

Back

mass to mole rule

Front

the atomic weight of an element in atomic mass units is numerically equal to the mass in grams of 1 mole of that element

Back

mole/ avogadro's number

Front

1 mole equals the amount of matter that contains as many objects (atoms, molecules, or whatever other objects we are considering) as the number of atoms in exactly 12 g of isotonically pure 12^C 1 mole (of any atom)= 6.022x10^23 atoms

Back

balanced equation

Front

when both sides of the equation has equal amounts of atoms on each side of it

Back

products

Front

the chemical formulas to the right arrow represent substances produced in the reaction called products

Back

combustion reactions

Front

are rapid reactions that produce a flame, most combustion reactions we observe involve O2 form air as a reactant

Back

excess reactants

Front

the reactants other than the limiting reactant

Back

percentage composition

Front

the percentage by mass contributed by each element in the substance, = ((number of atoms of element) (atomic weight of element)/ formula weight of substance ) x100

Back

stoichiometry

Front

is the area of study that examines the quantities of substances consumed and produced in chemical reactions

Back