A 44.6 g 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 10.1 grams. What
was the percent iron in the sample of ore?
Answer in units of %

What fraction, F, of the mass of a Fe2O3 molecule is FE ?

total mass of molecule = 2(atomic mass Fe)+3(atomic mass O)

mass of Fe in that molecule = 2 (atomic mass Fe)
so
F = 2(atomic mass Fe)/[ the sum ]

then
Mass Fe = 10.1 * F

and percent = (Mass Fe/44.6)100

To find the percent of iron in the sample of ore, we need to calculate the mass percent. Here's how you can do it:

1. First, determine the mass of iron in the sample. In this case, the mass of Fe2O3 formed is given as 10.1 grams.

2. Next, convert the mass of Fe2O3 to the mass of iron (Fe). The molar mass of Fe2O3 is calculated as follows:
Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of Fe2O3 = (2 x molar mass of Fe) + (3 x molar mass of O)
= (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
= 159.69 g/mol

Now, we can set up a ratio using the equation:
mass of Fe2O3 / molar mass of Fe2O3 = mass of Fe / molar mass of Fe

Plugging in the values:
10.1 g / 159.69 g/mol = mass of Fe / 55.845 g/mol

3. Solve the ratio to find the mass of iron (Fe):
mass of Fe = (10.1 g / 159.69 g/mol) × 55.845 g/mol
= 3.51 g

4. Finally, calculate the percent iron in the sample:
Percent iron = (mass of Fe / mass of ore sample) × 100
= (3.51 g / 44.6 g) × 100
= 7.87%

Therefore, the percent iron in the sample of ore is approximately 7.87%.