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Andreyy89
3 years ago
9

What is the empirical formula of a hydrocarbon if complete combustion or 2.900 mg of the hydrocarbon produced 9.803 mg of CO2 an

d 2.006 mg of H2O? Be sure to write C first in the formula.
Chemistry
1 answer:
kow [346]3 years ago
5 0

Answer:

The empirical formula of the hydrocarbon is CH.

Explanation:

The following data were obtained from the question:

Mass of hydrocarbon = 2.9 mg

Mass of CO2 = 9.803 mg

Mass of H2O = 2.006 mg

Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.

This is illustrated below:

Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol

Mass of CO2 = 9.803 mg

Mass of C in the compound =?

Mass of C in the compound =

12/44 x 9.803

= 2.674 mg

Molar mass of H2O = (2x1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O = 2.006 mg

Mass of H in the compound =

2/18 x 2.006

= 0.223 mg

Finally, we shall determine the empirical formula of the hydrocarbon as follow:

Carbon (C) = 2.674 mg

Hydrogen (H) = 0.223 mg

Divide by their molar mass

C = 2.674 /12 = 0.223

H = 0.223 / 1 = 0.223

Divide both side by the the smallest

C = 0.223/0.223 = 1

H = 0.223/0.223 = 1

Therefore, the empirical formula of the hydrocarbon is CH.

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3 years ago
A chemist wants to extract a solute from 100 mL of water using only 300 mL of ether. The partition coefficient between ether and
ExtremeBDS [4]

Answer:

a. 0,903

b. 0,985

c. 0,996

Explanation:

Partition coefficient between ether and water is:

K = \frac{CONCether}{CONCwater}

a. As k = 3,10, a single extraction with 300mL ether will give:

3,10 = \frac{(q/300mL)}{(1-q)/100mL}

9,30 - 9,30q = q

9,30 = 10,30q

<em>q = 0,903</em>

b. Three extractions of 100 mL will extract:

First extraction:

3,10 = \frac{(q/100mL)}{(1-q)/100mL}

3,10 - 3,10q = q

3,10 = 4,10q

<em>q = 0,756</em>

The second extraction will 0,756 times of the solute that is in water. This solute is 1-0,756=0,244

Second extraction will extract 0,244*0,756 = <em>0,184</em>

In the same way, third extraction will extraxt: (0,244-0,184)*0,756 = <em>0,045</em>

Total solute extracted:

<em>q = 0,756 + 0,184 + 0,045 = 0,985</em>

<em />

c. Six extractions with 50mL of ether extract:

First extraction:

3,10 = \frac{(q/50mL)}{(1-q)/100mL}

3,10 - 3,10q = 2q

3,10 = 5,10q

<em>0,608 = q</em>

Second extraction = (1-0,608)*0,608 = 0,234

Third extraction =(1-0,608-0,234)*0,608 = 0,096

Fourth extraction =(1-0,608-0,234-0,096)*0,608 = 0,038

Fifth extraction =(1-0,608-0,234-0,096-0,038)*0,608 = 0,015

Sixth extraction =(1-0,608-0,234-0,096-0,038-0,015)*0,608 = 0,005

q = 0,608+0,234+0,096+0,038+0,015+0,005 = <em>0,996</em>

<em />

I hope it helps!

3 0
3 years ago
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