A sample is to be analyzed for its chloride content by precipitating and weighing silver chloride. What weight (g) of sample would have to be taken so that the weight of precipitate is equal to 1/2 the percent chloride in the sample?
am = atomic mass
mm = molar mass
%Cl = [mass ppt AgCl x ( am Cl/mm AgCl)/mass sample]*100 = ?
The way I read it you want mass ppt = 1/2 the %Cl or 2*ppt = % so
just put in numbers to do that like this.
mass sample = [mass ppt AgCl x (amcl/mm AgCl)/%Cl]*100
mass sample = (12.5 x (35.453/143.32)/25]*100
Since my interpretation that you want 12.5 g ppt x 2 = 25% I put the 12.5 for mass ppt and 25 for the %Cl;i.e. just two numbers so that 2*ppt gives %Cl.
mass sample then = 1/2*(35.453/143.32)*100 = ?
In this scenario you weigh an unknown sample of 12.368 g and for the calculation of %Cl you just multiply mass ppt x 2.
Note that we are talking about major weighings here (25 g is a lot of unknown sample) so you convert to mg and weigh the ppt in mg.
If your interpretation is different than mine just plug in as I did and solve the equation to meet your thoughts. This is the way it was done in the "old days" in order to save time with calculations; however, in today's techy world a computer is hooked up to the procedure and it takes the mass of the sample, remembers it, factors in the conversion from AgCl to Cl and the mass of the ppt and rolls out the answers faster than ever before. When I was in school it took me more than 30 minutes to make a calculation. No calculator was available. We used a book of logarithm tables to do that. Talk about tedious. But it worked. Apparently I'm no worse because of it. :-)
Oh chloride, you're so tricky! Let's see if I can rustle up some calculations for you, with a side of humor of course.
To find the weight of the sample needed, we'll need to consider the percent chloride in the sample. But don't worry, I won't "sodium" you with any confusing formulas!
Let's assume the percent chloride in the sample is x%. So if we take 1/2 of that, we get (1/2)x%. Simple enough, right?
To find the weight of the sample needed, we'll need to find the weight of the precipitate first. The weight of the precipitate will be equal to the weight of the sample times the percent chloride in the sample. But remember, we only need 1/2 of that percentage. So, the weight of the precipitate is given by:
Weight of precipitate = (1/2)x% * weight of sample
Now we just need to find the weight of the sample. Since the weight of the precipitate is equal to the weight of the sample times (1/2)x%, we can set up an equation:
Weight of precipitate = (1/2)x% * weight of sample
Now we can solve for the weight of the sample. Divide both sides of the equation by (1/2)x%:
Weight of sample = Weight of precipitate / [(1/2)x%]
Voila! We have the weight of the sample. Just be sure to insert the correct values for the weight of the precipitate and the percent chloride in the sample.
I hope that explanation didn't "chloride" your brain with too much information! Let me know if you need any more help or if you need a little break from all this chemistry talk.
To determine the weight of the sample needed, we need to follow these steps:
Step 1: Calculate the weight of precipitate required:
- Since we want the weight of the precipitate to be equal to half the percent chloride in the sample, we need to find the percent chloride first.
- Let's assume the percent chloride in the sample is x. The weight of the sample (W) needed to obtain the desired weight of precipitate is equal to (1/2)*(x/100) as a decimal.
- The weight of precipitate required is also equal to (1/2)*(x/100) as a decimal.
Step 2: Calculate the formula mass of silver chloride:
- Silver chloride has a formula mass of 143.32 g/mol because the atomic mass of silver is 107.87 g/mol, and the atomic mass of chlorine is 35.45 g/mol.
Step 3: Convert the weight of precipitate to moles:
- Divide the weight of precipitate required by the formula mass of silver chloride to obtain the moles of silver chloride.
Step 4: Calculate the moles of chloride:
- Since silver chloride has a 1:1 stoichiometric ratio with chloride, the moles of silver chloride are equivalent to the moles of chloride.
Step 5: Convert the moles of chloride to grams:
- Multiply the moles of chloride by the atomic mass of chlorine (35.45 g/mol) to obtain the weight of the sample needed.
By following these steps, you can determine the weight (g) of the sample needed to obtain the desired weight of precipitate.
To determine the weight of the sample required, we need to understand the process of precipitating and weighing silver chloride and the given information.
Let's break down the problem step by step:
1. Start with the percentage of chloride in the sample. Let's assume it is "X%."
2. We are told that the weight of the precipitate (silver chloride) should be equal to half the percentage of chloride in the sample. So, we need to find 1/2 of X%.
3. To calculate this, we convert the percentage to a decimal value by dividing X by 100: X/100.
4. Next, calculate half of X% by multiplying X/100 with 1/2: (X/100) * (1/2). This gives you the fraction of the chloride content that needs to be converted into silver chloride.
5. Now, we need to find the weight of the sample that corresponds to the weight of the precipitate. This can be determined using the equation:
Weight of Sample = Weight of Precipitate / Fraction of Chloride
6. Since we know that the weight of the precipitate should be equal to half the percentage of chloride in the sample, we substitute the value we found in step 4 as the weight of precipitate.
Weight of Sample = (X/100) * (1/2) / (X/100)
7. Simplifying the equation, we get:
Weight of Sample = 1/2
Therefore, the weight of the sample required is 1/2 gram.
To summarize, to find the weight of the sample needed so that the weight of the precipitate is equal to 1/2 the percent chloride in the sample, you need to take a sample that weighs 1/2 gram.