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likoan [24]
2 years ago
9

Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen, and 54.2 grams of oxygen.

Chemistry
1 answer:
Lilit [14]2 years ago
6 0

The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

<h3>What is Empirical Formula ?</h3>

Empirical formula is the simplest whole number ratio of atoms present in given compound.

Element   %   Atomic mass   Relative no. of atoms  Simplest whole ratio

C          40.6       12                   \frac{40.6}{12} = 3.3                        \frac{3.3}{3.3} = 1

H          5.1          1                      \frac{5.1}{1} = 5.1                          \frac{5.1}{3.1} = 2

O         54.2       16                    \frac{54.2}{16} = 3.3                        \frac{3.3}{3.3} = 1

The Empirical formula of compound is C₁H₂O₁ or CH₂O

<h3>How to find the Molecular formula of compound ?</h3>

Molecular formula = Empirical formula × n

n = \frac{\text{Molecular weight}}{\text{Empirical Formula weight}}

   = \frac{118.084}{30}

   = 4

Molecular formula = n × Empirical formula

                              = 4 (C₁H₂O₁)

Thus from the above conclusion we can say that The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

Learn more about the Empirical Formula here: brainly.com/question/1603500

#SPJ1

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Answer:

1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. 6 moles of Cl2

Explanation:

1. The balanced equation for the reaction. This is illustrated below:

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.

This is illustrated below:

From the balanced equation above,

4 moles of FeCl3 reacted with 3 moles of O2.

Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.

Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:

Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.

From the balanced equation above,

4 moles of FeCl3 will react to produced 6 moles of Cl2.

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The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
AlekseyPX

Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of H_2 = 0.729 M

The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

Now put all the given values in this expression, we get :

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

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concentration of NH_3 in the equilibrium mixture = 0.80-2\times 0.243=0.31

Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

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