Calculate the mass of Fe(NH4)2(SO4)26H2O in your unknown sample. Show you work.
Average mass of unknown sample:1.0714g
PLEASE EXPLAIN CALCULATION PROCESS
I have idea what you're doing? What's the procedure. What's the starting material? You need to explain what's going on.
IF the sample is 100% ferrous ammonium sulfate hydrate then you will have 1.0714 of that material, of course.
IRON ANALYSIS BY REDOX TITRATION
Percent Fe in unknown sample is 70.5%.
Theoretical % Fe in Fe(NH4)2(SO4)26H2O is 14.24%
THE SAMPLE IS UNKOWN
@DrBob222
Please check you numbers and repost everything. The value I'm getting for the grams of ferrous ammonium sulfate hexahydrate in a 1.0714 g sample and 70.5% Fe just doesn't seem possible.
To calculate the mass of Fe(NH4)2(SO4)2•6H2O in your unknown sample, you need to follow these steps:
Step 1: Determine the molar mass of Fe(NH4)2(SO4)2•6H2O:
To do this, you need to calculate the sum of the atomic masses of all the elements in the compound.
Fe: 1 atom x 55.85 g/mol = 55.85 g/mol
N: 2 atoms x 14.01 g/mol = 28.02 g/mol
H: 16 atoms x 1.01 g/mol = 16.16 g/mol
S: 2 atoms x 32.07 g/mol = 64.14 g/mol
O: 26 atoms x 16.00 g/mol = 416.00 g/mol
H2O: 6 molecules x 18.02 g/mol = 108.12 g/mol
Adding all the atomic masses together gives:
55.85 g/mol + 28.02 g/mol + 16.16 g/mol + 64.14 g/mol + 416.00 g/mol + 108.12 g/mol = 688.29 g/mol
Step 2: Calculate the number of moles of Fe(NH4)2(SO4)2•6H2O in the unknown sample:
To do this, divide the mass of the unknown sample (1.0714 g) by the molar mass of Fe(NH4)2(SO4)2•6H2O (688.29 g/mol):
1.0714 g / 688.29 g/mol = 0.001556 mol
Step 3: Calculate the mass of Fe(NH4)2(SO4)2•6H2O in the unknown sample:
To do this, multiply the number of moles of Fe(NH4)2(SO4)2•6H2O (0.001556 mol) by the molar mass of Fe(NH4)2(SO4)2•6H2O (688.29 g/mol):
0.001556 mol x 688.29 g/mol = 1.0725 g
Therefore, the mass of Fe(NH4)2(SO4)2•6H2O in the unknown sample is approximately 1.0725 g.