The molar mass of the reagent, Fe(NH4)2(SO4)2.6H2O is 392.14 g/mol. Now suppose the measured mass of Fe(NH4)2(SO4)2.6H2O during the practical was 0.0218 g and this was dissolved in 50ml of deionized water that contains 1 to 2ml of concentrated sulphuric acid and transfer quantitatively to a 250ml volumetric flask using a funnel and a glass rod and dilute to the mark. Determine the iron concentration in the standard solution in ppm (mg L-1) units, showing full details of your working.
Convert 0.218 g of the compound to g Fe. That is done by 0.218 x (atomic [mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O]*100 = ?
Convert g Fe to mg = ?
Convert 250 mL to L.
Then mg Fe/L = your answer.
DRBob222 what do you mean when you say That is done by 0.218 x (atomic [mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O]*100 = ?
Could you please explain further?
OK, but first I made a typo. That is supposed to be 0.0218 g.
You are weighing Fe(NH4)2(SO4)2.6H2O but you don't want to calculate mg/L of that stuff, you want to calculate mg/L of Fe. Not all of that stuff is Fe.
What fraction is Fe? That is [atomic mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O] = fraction that is Fe.
So 0.0218 x fraction that is Fe = grams Fe in the mass of Fe(NH4)2(SO4)2.
The mass of the compound is 0.0218g and if I calculate the mass of Fe I get a mass of 0.31g which is more than the mass of the whole compound.
Thank you for all your help.
To determine the iron concentration in the standard solution in ppm (mg L-1) units, we need to follow the given steps and calculations. Here's how you can calculate it:
Step 1: Calculate the molarity of the Fe(NH4)2(SO4)2.6H2O standard solution.
First, we need to convert the mass of the reagent to moles using its molar mass.
Given molar mass of Fe(NH4)2(SO4)2.6H2O = 392.14 g/mol
Mass of Fe(NH4)2(SO4)2.6H2O = 0.0218 g
Using the formula: Moles = Mass / Molar mass
Moles of Fe(NH4)2(SO4)2.6H2O = 0.0218 g / 392.14 g/mol
Step 2: Calculate the volume of the standard solution.
The given volume of the standard solution is 250 mL.
Step 3: Calculate the molarity of the standard solution.
Given moles of Fe(NH4)2(SO4)2.6H2O = 0.0218 g / 392.14 g/mol
Given volume of standard solution = 250 mL = 0.25 L
Molarity (M) = Moles / Volume
Molarity of Fe(NH4)2(SO4)2.6H2O standard solution = (0.0218 g / 392.14 g/mol) / 0.25 L
Step 4: Calculate the iron concentration in ppm (mg L-1) units.
To convert molarity to ppm (mg L-1), we need to use the formula:
ppm = Molarity x Molar mass x 1000
Molar mass of iron (Fe) = 55.845 g/mol
Iron concentration in ppm (mg L-1) = (Molarity x Molar mass of Fe x 1000)
Now, perform the calculations using the given values and the formulas to obtain the final result.
I was gone for awhile. I think you just misplaced the decimal.
fraction = atomic mass Fe/molar mass Fe(NH4)2(SO4)2 = (55.85/392.14) = 0.142
Then mass Fe = mass Fe(NH4)2*(SO4)2 x fraction = 0.0218 x 0.142 = 0.00318 g Fe which is not larger than 0.0218.
Convert that to mg which is 3.18 mg and divide by L (0.250) which comes out to be 3.18/0.25 = ?