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spayn [35]
4 years ago
11

Using the example in the above information determine the empirical formula of a compound if a sample contains 0.130 g of nitroge

n and 0.370 g of oxygen?
Chemistry
1 answer:
olga2289 [7]4 years ago
4 0

Answer: The empirical formula for the given compound is NO_2

Explanation : Given,

Mass of O = 0.370 g

Mass of N = 0.130 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.370g}{16g/mole}=0.0231moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.130g}{14g/mole}=0.00928moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.

For Oxygen  = \frac{0.0231}{0.00928}=2.4\approx 2

For Nitrogen = \frac{0.00928}{0.00928}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of O : N = 2 : 1

Hence, the empirical formula for the given compound is NO_2

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The question is incomplete, here is the complete question:

Consider the following multistep reaction:

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<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

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The intermediate reaction of the mechanism follows:

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<u>Step 3:</u>  CD_2+D\rightarrow CD_3;\text{(fast)}

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Rate law for the reaction follows:

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As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

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\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2  

Hence, the rate law for the reaction is \text{Rate}=k'[C][D]^2

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