Respuesta :
empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound
C H O
mass 64.27 g 7.19 g 28.54 g
number of moles 64.27 g / 12 g/mol 7.19 g/1 g/mol 28.54 g / 16 g/mol
= 5.356 mol = 7.19 mol = 1.784 mol
divide by least number of moles
5.356 / 1.784 7.19 / 1.784 1.784 / 1.784
= 3.002 4.03 = 1.000
rounded off to nearest whole number
C - 3
H - 4
O - 1
empirical formula - C₃H₄O
mass of empirical formula = 12 g/mol x 3 + 1 g/mol x 4 + 16 g/mol x 1 = 56 g
molecular mass = 168.19 g/mol
molecular formula is the actual ratio of elements making up the compound
number of empirical units = molar mass of molecule / empirical mass
empirical units = 168.19 g/mol / 56 g = 3.00
there are 3 empirical units making up the molecular formula
molecular formula = 3 x C₃H₄O
molecular formula = C₉H₁₂O₃
calculating for 100 g of compound
C H O
mass 64.27 g 7.19 g 28.54 g
number of moles 64.27 g / 12 g/mol 7.19 g/1 g/mol 28.54 g / 16 g/mol
= 5.356 mol = 7.19 mol = 1.784 mol
divide by least number of moles
5.356 / 1.784 7.19 / 1.784 1.784 / 1.784
= 3.002 4.03 = 1.000
rounded off to nearest whole number
C - 3
H - 4
O - 1
empirical formula - C₃H₄O
mass of empirical formula = 12 g/mol x 3 + 1 g/mol x 4 + 16 g/mol x 1 = 56 g
molecular mass = 168.19 g/mol
molecular formula is the actual ratio of elements making up the compound
number of empirical units = molar mass of molecule / empirical mass
empirical units = 168.19 g/mol / 56 g = 3.00
there are 3 empirical units making up the molecular formula
molecular formula = 3 x C₃H₄O
molecular formula = C₉H₁₂O₃
