Module 8 Part 1 (Chapter 7)
Module 8 Part 1 (Chapter 7) CHM 1025
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This 3 page Class Notes was uploaded by Madison Woy on Friday October 30, 2015. The Class Notes belongs to CHM 1025 at University of Florida taught by Melanie veige in Fall 2015. Since its upload, it has received 14 views. For similar materials see Intro to Chem (Online) in Chemistry at University of Florida.
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Date Created: 10/30/15
Module 8 Pt1 ch 7 71 M A group of atoms molecules or formula units that contain 6022x1OA23 of these items 0 Avagadros Number The number of items in a mole One mole of any element contains Avogadro39s number of atoms For an ionic compound 1 mol contains Avogadro39s number of formula units 0 Formula units The group of ions represented by the formula of an ionic compound Avogadro39s number as a conversion factor 6022 x 1037 aunl 13Fc 4 400met39Fc X V 241 X 10quot 111011180ch l metf6 n chemical formulas the subscripts specify the number of atoms in one molecule and the number of moles of each element in the molecule 0 For example gt Carbon Dioxide CO2 has One carbon atom and 2 oxygen atoms This means that 002 has one mol of carbon and 2 mol of oxygen 72 Molar massthe mass in grams of one mole of an element equal numerically to its atomic mass The molar mass of a compound is equal to the sum of the masses of the elements in the formula We are counting 6022x1OA23 atoms of an element when we weigh the number of grams equal to its molar mass 0 For example carbon has an atomic mass of 1201 on on the periodic table This means one mol of Carbon Atoms has a mass of 1201 g To determine the molar mass of a compound multiply the molar mass of each element by its subscript in the formula and add the results 73 The molar mass of an element or compound is one of the most useful conversion factors in chemistry because it converts moles of a substance to grams For example conversion factors for H20 can be written as 1802gH20 d lmolHQO lmolHZO m 1802ng 1 mol of H20 2 1802 g of H20 Mass percent compositionthe percent by mass of the elements in a formUla mass orquot m clcmcnl Muss pcrccnl of an clcmcnl v ILIOH lnml muxx nl lhc compound The equation to find the mass percent composition of the compound is as follows muss otkumh clcmcrll l pcrccnl unmpusmun 39 IU39r molar muxx 1 the u39ntlpnd d Molecular formulathe actual formula that gives the number of atoms of each type of the elements in the compound gmpirical formuathe simplest or smallest wholenumber ratio of the atoms in a formula O For Name Molecularactualformula Empiricalsimplestlotmula example l u m liclircnc 39 39ll 39Y rlllI llTI H V H lru il 39H V Rm mm tilu Hutm l 39H l To calculate empirical formulas 0 step one calculate the moles of each element 0 step two divide by the smallest number of moles 0 step three use the lowest whole number ratio of moles as subscripts lf dividing by the smallest number of moles results in a decimal and the decimal is less than 1 or greater than 9 simply round to the nearest whole number 0 If the decimal is between 1 and 9 multiply by a integer until a whole number is obtained 75 Empirical formulas do not necessarily represent the actual number of atoms in a molecule A a molecular formula is related to the empirical formula by a small integer such as one two or three The relationship between these formulas can be seen below Molecular formula lolar mass Small integer Empirical formula Empirical formula mass 1min liluxx 0i cmnpt iunti Small integer cutip39riml formula mzm
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