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Lyrx [107]
2 years ago
15

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

Chemistry
1 answer:
vovikov84 [41]2 years ago
6 0

Answer:

\rm C_2H_8N_2O_3.

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

  • \rm C: approximately 12.
  • \rm H: approximately 1.
  • \rm N: approximately 14.
  • \rm O: approximately 16.

The relative atomic mass of an element is numerically equal to the mass (in grams, \rm g,) of one mole of atoms of this element.

For example, the relative atomic mass of \rm C is approximately 12. Therefore, each mole of \rm C\! atoms would have a mass of 12\; \rm g.

This sample contains 24\; \rm g of carbon. That would correspond to approximately \displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol of \rm C atoms.

Similarly, for the other three elements:

\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol.

\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol.

\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol.

Hence, the ratio between these elements in this compound would be:

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3.

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3 is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, \rm C, is present in this compound, then:

  • \rm C (carbon) and then \rm H (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
  • The other elements in this compound will be listed in alphabetical order.

If there is no carbon \rm C in this compound, then list all the elements in this compound in alphabetical order.

Both \rm C (carbon) and \rm H (hydrogen) are found in this compound. Therefore, the first element in the list would be \rm C\!. The second would be \rm H\!, followed by \rm N\! and then \rm O\!.

Hence, the empirical formula of this compound would be \rm C_2H_8N_2O_3.

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7 0
1 year ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

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1 year ago
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2 years ago
A race car is driven by a professional driver at 99 . What is this speed in and ? 1 mile = 1.61 kilometers 1 hour = 60 minutes E
slavikrds [6]

Answer: The speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>

Explanation:

Given, The speed of a race car = 99 miles/ hour

To convert the speed into  kilometers per hour and kilometers per minute

Since   1 mile = 1.61 kilometers

So, Speed of car = (99 ) x (1.61 )

= 159.39 kilometers per hour.

Also, 1 hour = 60 minutes

Then, Speed of car = (159.39) ÷60

= 2.6565≈2.65 kilometer per minute.

Hence, the speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>

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