Respuesta :
Empirical formula is the simplest ratio of whole numbers of components in a compound.
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
The empirical formula of enargite is calculated as below
find the moles of each element
that is moles =% composition/molar mass
=Cu = 48.41/63.5 =0.762 moles
As = 19.02/74.9 = 0.254 moles
S =32.57/32.1 = 1.015 moles
find the moles ratio by diving each mole with the smallest mole( 0.254 moles)
CU =0.762/0.254 = 3
As =0.254/0.254 =1
S= 1.015/0.254 = 4
therefore the empirical formula =Cu3AsS4
find the moles of each element
that is moles =% composition/molar mass
=Cu = 48.41/63.5 =0.762 moles
As = 19.02/74.9 = 0.254 moles
S =32.57/32.1 = 1.015 moles
find the moles ratio by diving each mole with the smallest mole( 0.254 moles)
CU =0.762/0.254 = 3
As =0.254/0.254 =1
S= 1.015/0.254 = 4
therefore the empirical formula =Cu3AsS4