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2 years ago

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Compound X has a molar mass of 135.13 and the following composition: element mass % carbon 44.44% hydrogen 3.73% nitrogen 51.83%

Write the molecular formula of X.

1 answer:

geniusboy [140]2 years ago

7 0

Hello!

First, let's see what a molar mass is. A molar mass is how much a mole weighs, or how much 6.02214076×10²³ of an element weighs.

Now, let's find the molar mass of each of these three elements.

Carbon: 12.0107 g/mol

Hydrogen: 1.00784 g/mol

Nitrogen: 14.0067 g/mol

Now, see how much of the molar mass of compound x each of these contributes.

Carbon: 44.44% * 135.13 = 60.051772 g/mol

Hydrogen: 3.73% * 135.13 = 5.040349 g/mol

Nitrogen: 51.83% * 135.13 = 70.037879 g/mol

Finally, calculate how many atoms there are in the molecule. Do this my dividing the "how much they contribute to the molar mass" by the molar mass.

Carbon: 60.051772 / 12.0107 = 5

Hydrogen: 5.040349 / 1.00784 = 5

Nitrogen: 70.037879 / 14.0067 = 5

Therefore, we know that there are 5 Carbons, 5 hydrogens, and 5 nitrogens.

Put this all together to get the molecular formula: C5H5N5 (note that the numbers should be written as subscript)

Hope this helps!

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What is the ratio of the concentration of potassium ions to the concentration of carbonate ions in a 0.015 m solution of potassi

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The ratio of the concentration of potassium ions (K⁺) to the concentration of carbonate ions (CO₃²⁻) = 2: 1

<h3>Further explanation</h3>

Stochiometry in Chemistry studies about chemical reactions mainly emphasizes quantitative, such as the calculation of volume, mass, amount, which is related to the number of actions, molecules, elements, etc.

Reaction equations are chemical formulations for reagents and product substances

Reaction coefficients are numbers in the chemical formula of substances involved in the reaction equation. Reaction coefficients are useful for balancing reagents and products.

The reaction coefficient shows the ratio of the number of moles or molecules of the reacting substance

The ionization reaction is the reaction of the decomposition of a substance into its ions when the substance is dissolved in water.

Molarity (M)

Molarity shows the number of moles of solute in every 1 liter of solution.

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Electrolytes will dissociate into ions when dissolved in water

While non-electrolyte solutions do not produce ions

Electrolyte solutions can be

1. Electrolytes are strong if in a solution can be fully ionized usually expressed by the degree of ionization α = 1

2. Electrolytes are weak if in ions the solution is only partially ionized. The degree of ionization ranges from 0 <α <1

Potassium carbonate (K₂CO₃) will dissociate into ions

K₂CO₃ ⇒ 2K⁺+ CO₃²⁻

Comparison of reaction coefficients = mole ratio = concentration ratio for the same volume, so

The mole ratio of the ions is

K⁺ : CO₃²⁻ = 2: 1

This comparison also shows the ratio of concentrations of ions: (K₂CO₃ concentration = 0.015 M)

K⁺: CO₃²⁻ = 2.0.015: 1.0.015

K⁺: CO₃²⁻ = 0.03: 0.015 = 2: 1

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3 0

2 years ago

Read 2 more answers

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

<u>Explanation:</u>

<u>For 1:</u> At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

<u>For 2:</u>

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

<u>For nitric acid:</u>

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

<u>For methylamine:</u>

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

To calculate the $pK_b$ of base, we use the equation:

$pK_b=-\log(K_b)$

where,

$K_b$ = base dissociation constant = $3.9\times 10^{-10}$

Putting values in above equation, we get:

$pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})$

We are given:

$pK_b=9.41$

$[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

$[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

Putting values in above equation, we get:

$pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-9.41=4.59$

Hence, the pH of the solution is 4.59

4 0

2 years ago

A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel

postnew [5]

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

$\rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}$

or, $n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}$

where, n' = no. of formula units per cell

$A_{c}$ = molecular weight of cation = 90.5 g/mol

$A_{a}$ = molecular weight of anion = 37.3 g/mol

$V_{c}$ = volume of cubic cell = 3.55 $g/cm^{3}$

a = edge length of unit cell = $3.9 \times 10^{-8} cm$

$N_{A}$ = Avogadro's number = $6.023 \times 10^{23}$

$\rho$ = density = 3.55 $g/cm^{3}$

Now, putting the given values into the above formula as follows.

$n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}$

= $\frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}$

= 0.9

= 1 (approx)

Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.

3 0

2 years ago