Answer to Question #101659 in Organic Chemistry for Heidi

Question #101659
The combustion of 8.23 mg sample of unknown substance gave 9.62 mg of carbon dioxide and 3.94 mg of water. Another sample, weighing 5.32 mg, gave 13.49 mg of silver chloride in a happen analysis. Determine the percent composition for this organic compound.
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Expert's answer
2020-01-24T05:00:17-0500

1) Find mass and moles of carbon

"n(C)= \\frac{9.62\\times 10^{-3}}{44.01} =0.000219 moles"

"m(C) =n\\times M = 0.000219\\times 12.01 = 0.00263 g =2.63 mg"


2) Find mass and moles of hydrogen

"n(H) = n(H_2O)\\times 2 =\\frac{3.94\\times10^{-3}}{18.02} \\times 2 = 0.000437 moles"

"m(H) = n\\times M = 0.000437\\times 1 = 0.437mg"


3) Find mass percent of chlorine in a sample

a) find mass of chlorine if silver nitrate:

"n(AgCl )= \\frac{m}{M} = \\frac{13.49\\times 10^{-3}}{143.32} = 0.0000941 mol"

"n(Cl) = n(AgCl) = 0.0000941 mol"

"m(Cl) = 0.0000941\\times35.45 = 0.00334 g = 3.34 mg"


b) find mass percent of chlorine in silver nitrate

"mass\\% (Cl)= \\frac{3.34}{5.32} =0.627"


4) Find mass and moles of chlorine in the first sample

"m(Cl) = m_{sample}\\times mass\\%(Cl) = 8.23\\times 0.627 = 5.16 mg"

"n(Cl) = \\frac{m}{M} = \\frac{5.16\\times10^{-3}}{35.45} = 0.000146 mol"


5) check up if there is an oxygen in a sample:

"8.23 = m(O) + m(C) + m(H) + m(Cl)"

"8.23 = m(O) +2.63 + 0.437 + 5.16"

"m(O) = 0.003 mg" (negligible)


6) find mole ratio

"n(C):n(H):n(Cl) = 0.000219:0.000438:0.000146 \n= 1.5:3:1 = 3:6:2"

The formula is "C_3H_6Cl_2"


7) Percent composition

mass percent "\\% (c) = \\frac{2.63}{8.23}\\times 100\\% = 32.0\\%"

mass percent "\\% (H)= \\frac{0.437}{8.23} \\times 100\\% = 5.31 \\%"

mass percent "\\% =\\frac{5.16}{8.23} \\times 100\\% = 62.7\\%"


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Comments

Assignment Expert
23.01.20, 14:26

Dear Heidi, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Heidi
23.01.20, 05:27

Sorry, I meant halogen analysis.

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