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Y_Kistochka [10]
3 years ago
6

What is the molecular formula for a compound that is 26.37 carbon 5.541 hydrogen 52.70 oxygen and 15.38 nitrogen and has a molar

mass of 182.16g
Chemistry
1 answer:
lukranit [14]3 years ago
6 0

<u>Answer: </u>The molecular formula of the compound is C_2H_5O_3N

<u>Explanation:</u>

To know the molecular formula of the compound, we will follow some steps:

<u>Step 1:</u> Converting all these percentages into mass.

We take the total mass of the compound to be 100 grams, so, the percentages given for each element becomes its mass.

Mass of Carbon = \frac{26.37}{100}\times 100g=26.37g

Mass of Hydrogen = \frac{5.541}{100}\times 100g=5.541g

Mass of Oxygen = \frac{52.7}{100}\times 100g=52.7g

Mass of Nitrogen = \frac{15.38}{100}\times 100g=15.38g

Molar mass of Carbon = 12 g/mole

Molar mass of Hydrogen = 1 g/mole

Molar mass of Oxygen = 16 g/mole

Molar mass of Nitrogen = 14 g/mol

<u>Step 2:</u> Converting the given masses into their respective moles.

Moles of Carbon = \frac{\text{Given mass of C}}{\text{Molar mass of C}}= \frac{26.37g}{12g/mole}=2.197moles

Moles of Hydrogen = \frac{\text{Given mass of H}}{\text{Molar mass of H}}= \frac{5.541g}{1g/mole}=5.541moles

Moles of Oxygen = \frac{\text{Given mass of O}}{\text{Molar mass of O}}= \frac{52.7g}{16g/mole}=3.29moles

Moles of Nitrogen = \frac{\text{Given mass of N}}{\text{Molar mass of N}}= \frac{15.38g}{14g/mole}=1.09moles

<u>Step 3: </u>Now, calculating mole ratio, we divide each number of moles by the smallest number of moles calculated.

For Carbon = \frac{2.197}{1.09}=2.01\approx 2  

For Hydrogen = \frac{5.541}{1.09}=5.08\approx 5  

For Oxygen = \frac{3.29}{1.09}=3.01\approx 3

For Nitrogen = \frac{1.09}{1.09}=1

The ratio of C : H : O : N = 2 : 5 : 3 : 1

<u>Step 4:</u> Writing the mole ratio of the element as the subscripts in molecular formula.

Molecular formula of the compound : C_2H_5O_3N

Hence, the molecular formula of the compound is C_2H_5O_3N

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mina [271]

Answer:

The first row of elements fits in period <u>6</u>, after the element <u>lanthanum (La)</u>. The second row of elements fits in period <u>7</u>, after the element <u>actinium (Ac). </u>

I hope this helps!

4 0
3 years ago
consider the titration of hclo4 with koh. what is the ph after 17.0 ml of 0.15 m koh has been added to 15 ml of 0.20 m hclo4?
bezimeni [28]

The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is  <u>3.347</u>.

Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.

<u>Calculation:-</u>

Normality of acid                                               Normality of base

= nMV                                                                        nMV

= 1 × 0. 15 × 0.017                                              1 ×  0. 20 ×0.015 L

= 2.55 × 10⁻³                                                             = 3 × 10⁻³

The overall base will be high

net concentration = 3× 10⁻³ - 2.55 × 10⁻³

                             = 0.45 × 10⁻³

                             = 4.5× 10⁻⁴

pH = -log[4.5 × 10⁻⁴]

    = 4 - log4.4

     = <u>3.347</u>

A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.

Learn more about titration here:-brainly.com/question/186765

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7 0
1 year ago
7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s
Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both NH_{3} and CO_{2} are gaseous. Hence equilibrium constant depends upon partial pressures of NH_{3} and CO_{2}.

Initially no NH_{3} and CO_{2} were present.

Hence mole fraction of NH_{3} and CO_{2} at equilibrium can be calculated from coefficient of NH_{3} and CO_{2} in balanced equation.

Mole fraction of NH_{3} = (number of moles of NH_{3})/(total number of moles of NH_{3} and CO_{2}) = \frac{2moles}{(2+1)moles}=\frac{2}{3}

Mole fraction of CO_{2} = (number of moles of CO_{2})/(total number of moles of NH_{3} and CO_{2}) = \frac{1moles}{(2+1)moles}=\frac{1}{3}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if
katovenus [111]

The integrated rate law for a second-order reaction is given by:

\frac{1}{[A]t} =   \frac{1}{[A]0} + kt

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction


<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min


Hence, \frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)

<span>                                        = 4.858</span>

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>
</span>

<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

7 0
3 years ago
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The 54.9-g sample of the compound X2O7 contains 33.6 g of oxygen atoms. What is the molar mass of element X?
vazorg [7]
Mass of X₂O₇ = 54,9g

2x + 33,6g = 54,9g
2x = 54,9g - 33,6g
2x = 21,3g  | :2
x = 10,65g/mol
7 0
3 years ago
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