7.77 g of rush (an oxide of iron) is heated until all the O2 atoms combine with charcoal. When this is done, a pellet of pure iron with a mass of 5.44 g remains. What is the empirical forumla for rust?
5.44/55.847 = 0.0974 moles Fe.
g oxygen = 7.77-5.44 = 2.33
moles oxygen = 2.33/16 = 0.145 moles oxygen.
Now you want to find the small whole number ratio of Fe to O. The easiest way to do this is to divide the smaller number by itself (0.0974) and divide the other number by the same small number. that should give you a whole number ratio or one that can be converted to a small whole number ratio.
By the way, rust is not quite this formula. It is much more complicated than that.
Thank you!
I don't get one part though,
to find the mass of the oxygen, why did you minus 5.44 and 7.77?
7.77 is the mass of the iron oxide in the problem and it produces 5.44 g iron. Ergo, the mass oxygen must be 7.77 - 5.44. Said another way, the iron + oxygen has a mass of 7.77 grams. The oxygen is driven off and the amount of iron remaining is 5.44, so.......
OH, i understand now, 7.77 is the combination of iron and oxygen.. whereas 5.44 is just iron, so therefore to find the mass of the oxygen, you minus the two..
thank you so much!
You have it.
To determine the empirical formula of rust (an oxide of iron), you would need to compare the masses of iron and oxygen in the given compound. The empirical formula represents the simplest ratio of elements in a compound.
Let's start by calculating the mass of oxygen that combined with the charcoal. We know that the initial mass of rust is 7.77 g, and after heating, we are left with a pellet of pure iron weighing 5.44 g.
1. Calculate the mass of oxygen in the rust:
Mass of oxygen = Mass of rust - Mass of iron
Mass of oxygen = 7.77 g - 5.44 g
Mass of oxygen = 2.33 g
Next, we need to convert the mass of iron and oxygen to moles using their respective molar masses. The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
2. Calculate the number of moles of iron:
Moles of iron = Mass of iron / Molar mass of iron
Moles of iron = 5.44 g / 55.85 g/mol
Moles of iron = 0.0975 mol (approx.)
3. Calculate the number of moles of oxygen:
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Moles of oxygen = 2.33 g / 16.00 g/mol
Moles of oxygen = 0.1456 mol (approx.)
Now we need to find the simplest ratio between the moles of iron and oxygen. Divide both moles by the smaller number of moles to find the ratio.
4. Divide both moles by the smaller number of moles (0.0975 mol in this case):
Moles of iron = 0.0975 mol / 0.0975 mol = 1
Moles of oxygen = 0.1456 mol / 0.0975 mol ≈ 1.493
Finally, we need to round off the ratio to the nearest whole number. Therefore, the empirical formula for rust is:
FeO2 (approx.)
So, the empirical formula for rust is FeO2.