Respuesta :
when determining empirical formulas the percentages are first divided by their respective molar masses
Na (23 g/mol) Fe (56 g/mol) Si (28 g/mol) O (16 g/mol)
10.0 24.2% 24.3% 41.5 %
10.0/23 24.2/56 24.3/28 41.5/16
=0.43 =0.43 = 0.86 =2.59
divide all these values by the lowest fraction which is 0.43
0.43/0.43 0.43/0.43 0.86/0.43 2.59/0.43
=1 =1 =2 =6.02
When we round off these values the ratio of Na:Fe:Si:O = 1:2:6
therefore empirical formula would be NaFeSi2O6
Na (23 g/mol) Fe (56 g/mol) Si (28 g/mol) O (16 g/mol)
10.0 24.2% 24.3% 41.5 %
10.0/23 24.2/56 24.3/28 41.5/16
=0.43 =0.43 = 0.86 =2.59
divide all these values by the lowest fraction which is 0.43
0.43/0.43 0.43/0.43 0.86/0.43 2.59/0.43
=1 =1 =2 =6.02
When we round off these values the ratio of Na:Fe:Si:O = 1:2:6
therefore empirical formula would be NaFeSi2O6
To make it easier, assume that we have a total of 100 g of aegirine. Hence, we have 10g of Na, 24.2g of Fe, 24.3g of Si and 41.5g of O. Know we will convert each of these masses to moles by using the atomic masses of Na, Fe, Si, and O:
[tex] \frac{10g Na}{23g/mole} = 0.43 mole Na[/tex]
[tex]( \frac{24.2g Fe}{55.9g/mole} ) = 0.43 mole Fe[/tex]
[tex] ( \frac{24.3g Si}{28.1g/mole} ) = 0.86 mole Si[/tex]
[tex] ( \frac{41.5g O}{16g/mole} ) = 2.59 mole O[/tex]
Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the mineral:
Na = 0.43/0.43 = 1
Fe = 0.43/0.43 = 1
Si = 0.86/0.43 = 2
O = 2.59/0.43 = 6
So, the empirical formula of the compound NaFeSi2O6
