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just olya [345]
3 years ago
8

A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental

Chemistry
1 answer:
OleMash [197]3 years ago
3 0

Answer:

the empirical (lowest raios) is

C2H4Cl    

Explanation:

A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.

What is the empirical formula of this compound?

the compound has ONLY C, H, and Cl

the % Cl  = 100% - 24.27% -4.03% = 71.7%

in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H

the number of moles are Cl=71.7/70.91 =1.01= ~ 1

                                          C = 24.27/12.0 = 2.02 =~ 2

                                           H = 403/1.01 = 3.97 =~   4

so   the empirical (lowest raios) is

C2H4Cl      

                                 

                               

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<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

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So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

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Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

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Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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