You need to know the molar ratio for Antimony Oxide! Since Sb has an oxidation of +3 and Oxygen has a oxidation of -2, you need two Sb atoms and three O atoms.
Sb2O3
Figure out how many moles you have of antimony and oxygen..
51.5 g Sb * 1 mol/ 121.76 g= 0.43 moles Sb
16.9 g O * 1 mol/ 16 g= 1.06 moles O
I don't think even in the simplest form, the masses of the elements given are sufficient for a stable compound of Sb2O3.
A sample of an oxide of antimony (Sb) contained 51.5 g of antimony combined with 16.9 g of oxygen. What is the simplest formula for the oxide?
6 answers
51.5 g of Sb*(1 mole/121.76 g)= moles of Sb=0.418
16.9 g of O * (1 mole/16.0 g)= moles of O=1.06
divide lowest moles by the other
Sb=0.418/0.418=1
O=1.06/0.418=2.52
2.52 *2 to get a whole number
Sb2O5
16.9 g of O * (1 mole/16.0 g)= moles of O=1.06
divide lowest moles by the other
Sb=0.418/0.418=1
O=1.06/0.418=2.52
2.52 *2 to get a whole number
Sb2O5
I could be wrong though.
Thank you Mr. Devron! I learnt something from you as well. The compound is Antimony Pentoxide, and I did indeed not realize that after dividing by the lowest moles you can multiply!
hey i got Sb2O5
for the oxide is O5 right?
for the oxide is O5 right?
Sb2O5 is correct. As for multiplying by whole numbers; remember that you can round to whole numbers as long as you don't round too much. That is, 8.1 can usually and safely be rounded to 8.0 or 8.9 to 9.0 but rounding 2.5 to 3 or 2 isn't done for 2.5 can be multiplied by whole numbers to get a whole number. Numbers like 2.25, 2.33, 2.5, 2.67, 2.75 usually aren't rounded but multiplied by whole numbers to obtain whole numbers; i.e., 2.33 x 3 = 6.99 which then rounds to 7.0