Determine the empirical and molecular formula given the percent compositions or masses.
8.81 g Carbon, 91.2 g Chlorine, Molar mass: 1362.5 g/mol
C"=12.011" Cl"=35.453"
"8.81g" Carbon "=\\frac{8.81}{12.011}=0.7335moles"
"91.2g" Chlorine "=\\frac{91.2}{35.453}=2.5724moles"
Mole ratio C: Cl "=0.7335:2.5724"
"=1:3.5"
"=2:7"
C2 Cl7 has a mass of "2(12.011)+7(35.453)"
"=272.193"
"\\frac{1362.5}{272.193}=5.00"
Molecular Formulae "= _5("C2 Cl7 ")"
"=" C"_{10}" Cl"_{35}"
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