A 0.5000g sample of FeO is ignited to Fe2O3; what is the percentage gain in weight?
Assuming no losses, the sample of Fe2O3 will have a mass of
0.5000 x (molar mass Fe2O3/2*molar mass FeO) = ?
Weight gain is the difference between ? and 0.5000 = let's say y.
Then % gain = 100y/0.5000 = ?
You must be a fairly good typist; you typed four (4) questions in less than a minute. Also I note you didn't send a lot of time telling how you thought the problems might be solved.
To calculate the percentage gain in weight, we need to determine the difference between the final and initial weights, and then express it as a percentage.
Step 1: Determine the initial weight of FeO.
Given that the initial weight of the FeO sample is 0.5000g.
Step 2: Determine the final weight of Fe2O3.
The FeO is ignited to form Fe2O3, which means oxygen is added to the FeO.
The molar mass of FeO is 71.85 g/mol, while the molar mass of Fe2O3is 159.69 g/mol.
To calculate the final weight of Fe2O3, we need to consider the stoichiometry of the reaction. From the balanced chemical equation:
4 FeO + O2 → 2 Fe2O3
This equation tells us that two moles of Fe2O3 are formed for every four moles of FeO. Thus, the mass of Fe2O3 formed is proportional to the molar mass ratio of Fe2O3 to FeO.
Mass of Fe2O3 = (molar mass of Fe2O3 / molar mass of FeO) * initial weight of FeO
Mass of Fe2O3 = (159.69 g/mol / 71.85 g/mol) * 0.5000g
Step 3: Calculate the difference in weight.
The difference in weight is the final weight of Fe2O3 minus the initial weight of FeO.
Difference in weight = Final weight - Initial weight
Step 4: Calculate the percentage gain in weight.
The percentage gain in weight is the difference in weight expressed as a percentage of the initial weight.
Percentage gain in weight = (Difference in weight / Initial weight) * 100
Now, plug in the values from Steps 1, 2, and 3 to calculate the percentage gain in weight.