Hey guys, i have a question about how to figure out how much mass of a compound i need to find 130ppm here is the question.
What mass of Fe(NH4)2(SO4)2*6H2O is required to prepare 2.25 L of a 130 ppm iron solution?
Now i know that ppm is (mg of desired compoun)/(kg of solvent used) But in this case im not quite sure what the solvent would be?
Would the solvent be the previously explained Fe..... compound just minus the Fe? or would the compound just bethe 6H2O? im confused, if anyone could help that would be great.
Isn't that definition of yours 1 ppm = mg desired compound/kg SOLUTION. USUALLY when a solvent is not specified it is understood to be H2O. And I remember for solutions that 1 ppm = 1 mg/L.
So 130 ppm = 130 mg/L solution or
130*2.25 = about (estimate) 292 mg of Fe. To convert that to what must be weighed to obtain that 292, just convert by
292 estimate x (molar mass Fe(NH4)2(SO4)2.6H2O/atomic mass Fe) = ?
Oh okay, I understand. Wow thanks alot.
To calculate the mass of Fe(NH4)2(SO4)2*6H2O required to prepare the solution, you need to determine the solvent in this case. In this question, the solvent refers to the liquid in which the compound is dissolved, which is typically water.
Here's how you can solve the problem step-by-step:
1. Start by converting the given concentration of 130 ppm (parts per million) to a decimal fraction:
ppm = mg/L --> 130 ppm = 130 mg/L = 0.130 g/L
2. Next, convert the volume of the solution from liters to milliliters (since ppm and milliliters often go together):
2.25 L = 2.25 x 1000 mL = 2250 mL
3. Now, you know that the concentration of iron (Fe) in the solution is 0.130 g/L, and the volume is 2250 mL (or 2250 g since 1 mL of water is approximately equal to 1 g). You can use these values to calculate the mass of Fe(NH4)2(SO4)2*6H2O required.
To do this, set up a proportion using the molar mass of Fe(NH4)2(SO4)2*6H2O:
(0.130 g Fe / 1 L solution) = (X g Fe(NH4)2(SO4)2*6H2O / 2250 g solution)
The molar mass of Fe(NH4)2(SO4)2*6H2O can be calculated as follows:
Fe = 1 atom * (55.85 g/mol) = 55.85 g/mol
N = 2 atoms * (14.01 g/mol) = 28.02 g/mol
H = 16 atoms * (1.008 g/mol) = 16.13 g/mol
S = 2 atoms * (32.07 g/mol) = 64.14 g/mol
O = 16 atoms * (16.00 g/mol) = 256.00 g/mol
H2O = 12 atoms * (2.02 g/mol) = 24.24 g/mol
Molar mass of Fe(NH4)2(SO4)2*6H2O = 55.85 + 28.02 + 16.13 + (2 * 64.14) + (16 * 16.00) + (12 * 2.02) + (6 * 24.24) = 561.53 g/mol
4. Now, solve the proportion for X (mass of Fe(NH4)2(SO4)2*6H2O):
(0.130 g / 1) = (X g / 2250 g)
X = (0.130 g / 1) * (2250 g / 1)
X ≈ 292.5 g
Therefore, approximately 292.5 grams of Fe(NH4)2(SO4)2*6H2O is required to prepare 2.25 L of a 130 ppm iron solution.