My chem class just finished a lab in which the objective was to find the percentage of silver and barium in a nitrate salt mixture by precipitating the chromates, separating out the precipitate from the solvent and spectator ions via filtration, and analysis of the combined precipitate.

In the lab the salt was dissolved and the barium and silver ions were precipitated with excess potassium chromate solution. the precipitate was collected.

i am supposed to find the percentage of silver and barium in the original sample; however i am having a VERY difficult time calculating this.

my teacher suggested either using a system of equations (two unknown variables) or finding the moles and then working from there. but i am really lost.

here is the data:
30mL of K2CrO4 @ .25 M
mass of flask: 95.383g
mass of flask & sample: 96.336g

after the precipate was collected and dried it weighed 0.944 g

any help setting up the calculation is much appreciated. if i didn't explain it well enough i woudl be happy to elaborate. i'm just really struggling right now.

Let X = mass Ag

Let Y = mass Ba
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One equation you can use is:
X(molar mass Ag2CrO4/2*atomic mass Ag) + Y(molar mass BaCrO4/atomic mass Ba) = 0.944
Do you have the mass of the sample of nitrates you used? I hope so. Then the second equation would be as follows:
X(molar mass AgNO3/atomic mass Ag) + Y(molar mass Ba(NO3)2/atomic mass Ba) = mass sample taken.
When you find X and Y, then
% follows from that. Let me know if you still have problems.

To find the percentage of silver and barium in the original sample, you can use the concept of stoichiometry and use the information given in the lab.

1. Determine the number of moles of the precipitation reaction:
- First, calculate the moles of potassium chromate solution used:
Moles of K2CrO4 = volume (L) × concentration (mol/L)
= 0.030 L × 0.25 mol/L
- Next, determine the moles of the precipitate formed (silver chromate and barium chromate):
Moles of precipitate = moles of K2CrO4 (from the balanced equation)
- The balanced equation for this precipitation reaction is:
AgNO3 + Ba(NO3)2 + K2CrO4 → Ag2CrO4 + BaCrO4 + 2KNO3
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of K2CrO4 (1:1 ratio).
Hence, the moles of precipitate = moles of K2CrO4

2. Calculate the moles of silver and barium in the precipitate:
- Since the chemical formula of silver chromate is Ag2CrO4 and barium chromate is BaCrO4, we can see that there is a 1:1 mole ratio of Ag2CrO4 to Ag atoms, and a 1:1 mole ratio of BaCrO4 to Ba atoms.
- Moles of silver = moles of Ag2CrO4
- Moles of barium = moles of BaCrO4

3. Convert the moles of silver and barium to grams:
- To convert moles to grams, use the molar mass of the respective elements.
- The molar mass of silver (Ag) is approximately 107.87 g/mol.
- The molar mass of barium (Ba) is approximately 137.33 g/mol.
- Mass of silver = moles of silver × molar mass of silver
- Mass of barium = moles of barium × molar mass of barium

4. Calculate the percentage of silver and barium in the original sample:
- Percentage of silver = (mass of silver / mass of sample) × 100%
- Percentage of barium = (mass of barium / mass of sample) × 100%

Given the data:
- Mass of flask = 95.383 g
- Mass of flask and sample = 96.336 g
- Mass of precipitate (dried) = 0.944 g

By using this approach, you can find the percentage of silver and barium in the original sample.