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ivanzaharov [21]
3 years ago
8

A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for

this compound?
Chemistry
1 answer:
iren [92.7K]3 years ago
5 0

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

 

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

 

Divide the three by the lowest ratio which is 3.22:

 

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

 

So the empirical formula is:

CHN

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Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
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Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

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aA+bB\rightarrow cC+dD

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The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

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The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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