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.
For Further Reading
CHEMISTRY - DrBob222, Monday, January 28, 2008 at 8:03pm
Let X = mass Ag
Let Y = mass Ba
========================
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.
thank you so much for your help.
this is what i have thus far:
X(331.74g/2*107.87g) + Y(253.33g/137.33g)=0.944g
which is:
X(1.54g)+Y(1.84g)= 0.944
is that correct? how would i solve for the nitrates? we didn't measure each one separately--we measured the original sample. the original sample with nitrates measured .953 g
The other equation I posted with my first post.
It is
X*(molar mass AgNO3/atomic mass Ag) + Y*(molar mass Ba(NO3)2/atomic mass Ba) = 0.953.
Solve the two equations simultaneously. I see you have rounded the 1.54 and 1.84 to the correct number of significant figures. From a personal standpoint, especially with problems like this where the DIFFERENCES in the calculation may be small, I like to carry the extra places and round at the end. With small differences it is quite easy to make rounding errors. After you find X and Y (these are the grams Ag and grams Ba, then percent is the %Ag = (mass Ag/mass sample)*100 and %Ba = (mass Ba/mass sample)*100.
Let me know if you don't understand.
thank you,
here is what i have.
X(1.53768425g)+Y(1.844680696g)=0.944g
X(1.574765922g)+Y(1.902934537g)=0.953g
to solve this i know i need to use a system of equations, but i'm having trouble setting the coefficients to equal each other.
I see you took my advice literally about carrying extra places. It isn't necessary to use ALL of those numbers that show up on your calculator. I think 1.5377X + 1.8447Y = 0.944 will be quite sufficient and I suspect even one number fewer still will work ok. As for making the coefficients equal, I think that is the hard way of doing it although it can be done. I would solve equation 1 for either X or Y and substitute into the other one. For example,
1.5377X + 1.8447Y = 0.944
X = (0.944-1.8447Y)/1.5377
Now substitute this value of X into equation 2 and solve. That will get a Y value which can be substituted into equation 1 to obtain X.
I worked though the algebra and found a + 0.532 g Y but a negative number for X. So what's the problem. I suspect it's the numbers you obtained in performing the experiment. I also note that the differences of two small numbers in the algebra makes a large diffference in the answer. For example, just a small error I made (very small) changed the Ba value from 0.605 to 0.532.
thank you so much for all of your help!
To find the percentage of silver and barium in the original sample, you will need to set up a system of equations using the information provided.
Let's start by defining the variables:
Let X be the mass of silver (Ag)
Let Y be the mass of barium (Ba)
The first equation we can set up is based on the mass of the collected precipitate:
X(molar mass of silver chromate / 2 * atomic mass of silver) + Y(molar mass of barium chromate / atomic mass of barium) = mass of precipitate (0.944 g)
Next, we need the mass of the sample of nitrates. Let's call it M.
The second equation can be set up based on the mass of the sample:
X(molar mass of silver nitrate / atomic mass of silver) + Y(molar mass of barium nitrate / atomic mass of barium) = mass of sample (M)
Now, you have two equations with two unknowns (X and Y) that you can solve simultaneously. Once you find the values of X and Y, you can calculate the percentage of silver and barium in the original sample.
Keep in mind that you will need the molar masses and atomic masses of silver, barium, and the various compounds involved to perform the calculations.
If you still have difficulties, you may need to provide additional information, such as the molar masses of the compounds involved or the mass of the sample of nitrates used.