# CHAPTER 3 STUDY GUIDE

darkeyez's version from 2016-10-26 16:24

## Section

Determine the percent composition of a compound given it's molecular formula. C2H5OHC= 12(2)=24. H=1(6)=6. O=16(1)= 16. Total=46. Mass of C= (2moles\1)(12g\1mole)= 24gC. Mass of H= (6moles\1)(1g\1mole)=6gH. Mass of O= (1mole\1)(16g\1mole)= 46g. C=(24\46)(100)=52%C. H=(6\46)(100)=13%H. O=(16\46)(100)=35%O
Derive the Empirical Formula of a substance given it's percent composition. 72%Cl, 24%C, 4%H. 72%Cl= 35.5g Cl. 24%C=12gC. 4%H= 1.0079gH. (72gCl\1)(1moleCl\35.5gCl)= 2 mol Cl. (24gC\1)(1moleC\12gC)=2molC. (4gH\1)(1moleH\1.0079gH)=4 moleH. Cl=(2\2)=1 mole Cl. C=(2\2)=1moleC. H=(4\2)=1 mole H. Empirical Formula= CH4C
Determine the molecular formula from the empirical formula given it's molar mass: 57.14gC, 6.16gH, 9.52gN, 27.18gO. 57.14gC=(1moleC\12gC)=(4.76\0.68)moleC→7(2)=14C. 6.16gH=(1moleH\1.01gH)=(6.10\0.68)moleH→9(2)=18H. 9.52gN=(1molN\14.0gN)=(0.68\0.68)moleN→1(2)=2N. 27.18gO=(1moleO\16.0gO)=(1.70\0.68)moleO→2.5(2)=5moleO. EMPIRICAL FORMULA= C14H18N2O5→C7H9N2O5