An organic compound was synthesized and found to contain only C, H, N, O, and Cl. It was observed that when .150g sample of the compound was burned, it produced .138g of CO2 and .0566g of H2O. All the Nitrogen in a different .200g sample of the compound was converted to NH3, which was found to have a mass of .0238g. Finally, the chlorine in a .125g sample of the compound was converted to AgCl. The AgCl, when dried, was found to weigh .251g.

a. Calculate the percent by mass of each element in the compound.
b. Determine the empirical formula for the compound.

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I did B first so I could could do A. As much as I would like to post my work, it is A LOT. However, I do have answers so that you can check if it's right (although I doubt this), or if I'm in the right direction.

Thus, for the empirical formula I got C3HN8O8CL10. For the molar mass to calculate percent composition, I got 631 grams.

As for percent composition, here's the list of the following elements and my answer:
9.97% C
.27% H
17.74% N
20.28% O
56.10% Cl2

The problem with this is that I get about 104% when I add the percentages all together, which means the empirical formula is probably a little off.

I know it's a little hard to help me with this without much work, but any suggestions/advice for this? Any help would be great. Thank you very much!

Let me try to get you started in the right direction. If you had shown your work I could have found the error. No work means I don't know where you went wrong. For example,

%Cl = (mass Cl/mass sample) *100
mass Cl = 0.251 g AgCl x (1 mole Cl/1 mole AgCl)*100 = 0.251 x (35.453/143.321)= 0.06209 g Cl
%Cl = 0.06209/0.125 = 49.67%
Similarly, the N must be
%N = [0.0238 x (14/17)]/0.2 = 9.8%

etc. You need to check out the molar masses and atomic masses and I used those in my memory and they may be off a little. As for the percents adding up to more than 100, that can't be for % oxygen is determined by adding C, H, N, and Cl and subtracting from 100 so they MUST add to 100.
I'll let you work with this for awhile. Post your work if you get stuck.

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Well, it seems like you're on the right track, but there might be a small mistake somewhere in your calculations. Let's take a closer look at it together:

First, let's calculate the moles of each element in the given masses:

- Moles of carbon (C) = 0.138g CO2 * (1 mol CO2/44.01g CO2) * (1 mol C/1 mol CO2)
- Moles of hydrogen (H) = 0.0566g H2O * (1 mol H2O/18.02g H2O) * (2 mol H/1 mol H2O)
- Moles of nitrogen (N) = 0.0238g NH3 * (1 mol NH3/17.03g NH3) * (1 mol N/1 mol NH3)
- Moles of oxygen (O) = [mass of the sample] - [mass of C] - [mass of H] - [mass of Cl]
- Moles of chlorine (Cl) = 0.251g AgCl * (1 mol AgCl/143.32g AgCl) * (1 mol Cl/1 mol AgCl)

Next, we need to find the simplest whole number ratio:

- Divide the moles of each element by the smallest number of moles calculated.
- Round the ratios to the nearest whole number.

Finally, we can use the ratios to write the empirical formula:

C(x)H(y)N(z)O(w)Cl(v)

Now, let's calculate the percent composition based on the empirical formula:

- %C = (mass of C in empirical formula / molar mass of empirical formula) * 100%
- %H = (mass of H in empirical formula / molar mass of empirical formula) * 100%
- %N = (mass of N in empirical formula / molar mass of empirical formula) * 100%
- %O = (mass of O in empirical formula / molar mass of empirical formula) * 100%
- %Cl = (mass of Cl in empirical formula / molar mass of empirical formula) * 100%

Make sure to double-check your calculations and check for any rounding errors. If you still have any specific questions or need further help, feel free to ask!

To solve this problem, we need to break it down step by step. Let's start with finding the empirical formula for the compound.

Step 1: Calculate the moles of CO2 produced.
We know that 0.138g of CO2 is produced from the combustion of the compound. To convert grams to moles, we use the molar mass of CO2 (44.01 g/mol):

moles of CO2 = 0.138g / 44.01 g/mol ≈ 0.00313 mol CO2

Step 2: Calculate the moles of H2O produced.
We know that 0.0566g of H2O is produced from the combustion of the compound.
To convert grams to moles, we use the molar mass of H2O (18.02 g/mol):

moles of H2O = 0.0566g / 18.02 g/mol ≈ 0.00314 mol H2O

Step 3: Calculate the moles of N in NH3.
We know that 0.0238g of NH3 is produced from the conversion of Nitrogen in the compound to NH3.
To convert grams to moles, we use the molar mass of NH3 (17.03 g/mol):

moles of NH3 = 0.0238g / 17.03 g/mol ≈ 0.00140 mol NH3

Step 4: Calculate the moles of Cl in AgCl.
We know that 0.251g of AgCl is produced from the conversion of Chlorine in the compound to AgCl.
To convert grams to moles, we use the molar mass of AgCl (143.32 g/mol):

moles of AgCl = 0.251g / 143.32 g/mol ≈ 0.00175 mol AgCl

Step 5: Calculate the moles of each element present in the compound.
Now that we have the moles of each compound, we can calculate the moles of each element present.

For CO2, there is 1 mole of C per mole of CO2, so the moles of C = 0.00313 mol CO2.
For H2O, there are 2 moles of H per mole of H2O, so the moles of H = 2 * 0.00314 mol H2O = 0.00628 mol H.
For NH3, there is 1 mole of N per mole of NH3, so the moles of N = 0.00140 mol NH3.
For AgCl, there is 1 mole of Cl per mole of AgCl, so the moles of Cl = 0.00175 mol AgCl.

Step 6: Calculate the empirical formula.
To find the empirical formula, we need to find the ratio of moles of each element present in the compound. We divide each mole value by the smallest one (0.0014) to get the simplest whole-number ratio.

C: 0.00313 mol CO2 / 0.0014 mol N ≈ 2.24
H: 0.00628 mol H / 0.0014 mol N ≈ 4.49
N: 0.0014 mol N / 0.0014 mol N = 1
Cl: 0.00175 mol AgCl / 0.0014 mol N ≈ 1.25

By multiplying these ratios by 4 (to get whole numbers), we obtain the empirical formula: C9H18N4Cl.

Now let's move on to calculating the percent by mass of each element.

Step 7: Calculate the molar mass of the empirical formula.
The molar mass of C9H18N4Cl can be calculated by adding the molar masses of each element:

molar mass = (9 * Atomic mass of C) + (18 * Atomic mass of H) + (4 * Atomic mass of N) + (1 * Atomic mass of Cl)

Step 8: Calculate the percent composition of each element.
To calculate the percent by mass of each element, divide the mass of each element by the molar mass of the empirical formula and multiply by 100.

% C = (9 * Atomic mass of C) / molar mass * 100
% H = (18 * Atomic mass of H) / molar mass * 100
% N = (4 * Atomic mass of N) / molar mass * 100
% Cl = (1 * Atomic mass of Cl) / molar mass * 100

Comparing the calculated percentages to your results, it seems there might be a calculation error or a rounding error. Double-check your calculations and make sure you are using accurate atomic masses.

Hope this helps! Let me know if you need further assistance.