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nika2105 [10]
3 years ago
13

A 1.038 g sample of unknown containing C, H, and O yielded 2.48 g of CO2 and 0.510 g of H2O during combustion analysis. Determin

e the empirical formula of the compound.
A) C3H3O
B) C6H6O
C) CH3O
D) C6H6O2
E) C2H6O2
Chemistry
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

A. C3H3O

Explanation:

We have the following data:

m = mass of organic compound = 1.038g

Then mass of CO2 is 2.48g

Also mass of H2O is 0.510g

To calculate percentage of carbon = 12 x mass of CO2 x 100/44 x mass of organic compound

This gives us 65.1 percent

Formula to calculate percentage of H=

2 x mass of water x 100/18 x mass of organic compound.

= 2 x 0.510 x 100/18x 10038

This gives 5.4%

Percentage of Oxygen =100-(percentageof carbon +percentageof H)

= 100-(65.1+5.4)

= 100-70.5

= 29.5

We have to calculate emperical formula:

For C = 65.1

65.1/12

= 5.4

= 5.4/1.8

=3

For H=5.4

5.4/1

= 5.4

= 5.4/1.8

=3

For O = 29.5

= 29.5/16

= 1.8

= 1.8/1.8

=1

So the emperical formula of organic compound is given as

C3H3O1

= C3H3O

(The atomic weights of C, H, O are 12,1,16)

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Explanation:

Hello there!

In this case, since these problems about formulas, firstly require the determination of the empirical formula, assuming that the given percentages are masses, we can calculate the moles and mole ratio of oxygen to iron as shown below:

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In such a way, by rounding to the first whole number we multiply by 8 and divide by 5 to obtain:

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Regards!

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