Answer to Question #300941 in Organic Chemistry for Tina

Question #300941

 Combustion of an 8.23 mg sample of a compound gave 9.63 mg of carbon dioxide and 3.94 mg water. Analysis of another 5.32 mg sample of the same compound by the Carius method gave 13.49 mg of AgCl. 


1
Expert's answer
2022-02-23T01:04:05-0500

1) Find mass and moles of carbon

n(C)=9.62×10344.01=0.000219molesn(C)= \frac{9.62\times 10^{-3}}{44.01} =0.000219 moles

m(C)=n×M=0.000219×12.01=0.00263g=2.63mgm(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(H2O)×2=3.94×10318.02×2=0.000437molesn(H) = n(H_2O)\times 2 =\frac{3.94\times10^{-3}}{18.02} \times 2 = 0.000437 moles

m(H)=n×M=0.000437×1=0.437mgm(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)=mM=13.49×103143.32=0.0000941moln(AgCl )= \frac{m}{M} = \frac{13.49\times 10^{-3}}{143.32} = 0.0000941 mol

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

m(Cl)=0.0000941×35.45=0.00334g=3.34mgm(Cl) = 0.0000941\times35.45 = 0.00334 g = 3.34 mg


b) find mass percent of chlorine in silver nitrate

mass%(Cl)=3.345.32=0.627mass\% (Cl)= \frac{3.34}{5.32} =0.627


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

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

n(Cl)=mM=5.16×10335.45=0.000146moln(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) + m(C) + m(H) + m(Cl)

8.23=m(O)+2.63+0.437+5.168.23 = m(O) +2.63 + 0.437 + 5.16

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


6) find mole ratio

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

The formula is C3H6Cl2C_3H_6Cl_2


7) Percent composition

mass percent %(c)=2.638.23×100%=32.0%\% (c) = \frac{2.63}{8.23}\times 100\% = 32.0\%

mass percent %(H)=0.4378.23×100%=5.31%\% (H)= \frac{0.437}{8.23} \times 100\% = 5.31 \%

mass percent %=5.168.23×100%=62.7%\% =\frac{5.16}{8.23} \times 100\% = 62.7\%

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