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sp2606 [1]
3 years ago
9

Complete combustion of 2.90g of a hydrocarbon produced 9.32g of CO2 and 3.18g of H2O. What is the empirical formula for the hydr

ocarbon?
Chemistry
2 answers:
ValentinkaMS [17]3 years ago
8 0

Mass of CO₂ = 9.32g

Molar mass of CO₂ = 44 g/mol

Mass of H₂O = 3.18 g

Molar mass of H₂O = 18 g/mol

Moles = mass/ molar mass

9.32 g CO₂ x (1 mol CO₂ / 44 g CO₂) = 0.2118 mol CO₂  

Every CO₂ molecule has 1 Carbon atom, therefore 0.2118 mol of CO₂ will have 0.2118 moles of C  

3.18 g H₂O x (1 mol H₂O / 18 g H₂O) = 0.177 mol H₂O  

In every H₂O molecule there are 2 atoms of H therefore 0.177 mol of H₂O will have 2 x 0.177 or 0.354 moles of H

Now the ratio of C : H  = 0.2118 : 0.354

To get the whole number we divide both numbers in the ratio by the lowest number.    

C : H  

= (0.2118/0.2118) : (0.354 / 0.2118)  

= 1:1 .67

Since we cannot round, we multiply by 3 to clear the fraction:  

C= 1 x 3 =3

H = 1.67 x 3 = 5

Thus the empirical formula is C₃H₅.


Olenka [21]3 years ago
5 0

Answer;

= C3H5

Explanation and solution;

1 mole of CO2 contains 44 g, of which 12 g are carbon

Thus, mass of carbon in 9.32 g will be;

(12/44) × 9.32 g = 2.542 g

Mass of Hydrogen in 3.18 g of water;

= (2/18) × 3.18 g = 0.353 g

we then find the number of moles;

Moles of carbon ; 2.542 /12 = 0.2118 moles

Moles of Hydrogen = 0.353 moles

The ratios of C ; H ;

= 1 :  0.353 /0.2118

= 1 : 5/3

= 3: 5

Therefore; the empirical formula of the hydrogen carbon is; C3H5

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Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

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We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

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Consider the following system at equilibrium:
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Answer:

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8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

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1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

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I hope it helps you!

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