A 7.7 g sample of iron ore is treated as follows.
The iron in the sample is all converted by a
series of chemical reactions to Fe2O3. The
mass of Fe2O3 is measured to be 25.4 g. What
was the mass of iron in the sample of ore?
Answer in units of g.
You end up with 25.4 g Fe2O3.
25.4g Fe2O3 x (2*atomic mass Fe/molar mass Fe2O3) = mass Fe.
Piecemeal it's done this way.
mols Fe = 25.4 g/molar mass Fe2O3.
Since there are 2 mols Fe for every mole Fe2O3, then mols Fe = twice mols Fe2O3.
Then g Fe = mols Fe x atomic mass Fe.
To find the mass of iron in the sample of ore, we need to subtract the mass of Fe2O3 from the initial mass of the ore.
Given:
Mass of Fe2O3 = 25.4 g
Initial mass of the ore = 7.7 g
We know that the formula for iron(III) oxide (Fe2O3) consists of two iron atoms (Fe) and three oxygen atoms (O). The molar mass of iron(III) oxide can be calculated as follows:
(2 * Atomic mass of iron) + (3 * Atomic mass of oxygen) = Molar mass of Fe2O3
Using the atomic masses:
(2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
Now, we can set up a ratio based on the molar mass to find the mass of iron:
7.7 g (Initial mass of the ore) / 159.69 g/mol (Molar mass of Fe2O3) = x g (Mass of iron)
Solving for x:
x = (7.7 g * 159.69 g/mol) / 1
Calculating the mass of iron in the sample of ore:
x = 1230.413 g/mol
Therefore, the mass of iron in the sample of ore is approximately 1230.413 g.