A 43.5 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 7.7 grams. What was the percent iron in the sample of ore?
Answer in units of %

Question

A 43.5 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 7.7 grams. What was the percent iron in the sample of ore?
Answer in units of %

julian.pitalua · Answer

Molecular mass of Fe2O3= 158,91g/mol. 

Calculating the percent of iron in the compound

100/158,91g= %Fe/110,94g(Fe)

Then with the percent that we got up here. we now can get the grams of Fe that are in the compound.

 100/7,7g= %Fe/ grams(Fe).

Now, the mass of Fe that is in the compound.

that's the mass that we had before...so..

100/43.5g= %Fe/ grams(Fe)