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A mixture of cyclopropane gas and oxygen is used as an anesthetic. Cyclopropane contains 85.7% C And 14.3% hydrogen by mass. At 50.0 degrees celcius and .984 atm pressure, 1.56 g cyclopropane has a volume of 1.00L.

Required:
What is the molecular formula of cyclopropane?

Respuesta :

Eduard22sly

Answer:

C₃H₆

Explanation:

We'll begin by calculating the empirical formula for cyclopropane. This is illustrated below:

Carbon (C) = 85.7%

Hydrogen (H) = 14.3%

Divide by their molar mass

C = 85.7/12 = 7.14

H = 14.3/1 = 14.3

Divide by the smallest:

C = 7.14/7.14 = 1

H = 14.3/7.14 = 2

Therefore, the empirical formula for the cyclopropane is CH₂

Next, we shall determine the number of mole of cyclopropane.

This can be obtained as follow:

Temperature (T) = 50 °C = 50 °C + 273 = 323 K

Pressure (P) = 0.984 atm

Volume (V) = 1 L

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) of cyclopropane =.?

PV = nRT

0.984 × 1 = n × 0.0821 × 323

Divide both side by 0.0821 × 323

n = 0.984 / (0.0821 × 323)

n = 0.0371 mole

Next, we shall determine the molar mass of cyclopropane. This can be obtained as follow:

Number of mole of cyclopropane = 0.0371 mole

Mass of cyclopropane = 1.56 g

Molar mass of cyclopropane =.?

Mole = mass /Molar mass

0.0371 = 1.56/Molar mass

Cross multiply

0.0371 × Molar mass = 1.56

Divide both side by 0.0371

Molar mass = 1.56 /0.0371

Molar mass of cyclopropane = 42.05 g/mol.

Finally, we shall determine the molecular formula for cyclopropane. This can be obtained as follow:

[CH₂]ₙ = 42.05

[12 + (2×1)]ₙ = 42.05

[12 + 2]ₙ = 42.05

14n = 42.05

Divide both side by 14

n = 42.05/ 14

n = 3

Thus,

[CH₂]ₙ => [CH₂]₃ => C₃H₆

Therefore, the molecular formula for cyclopropane is C₃H₆

Lanuel

The molecular formula of cyclopropane ([tex]CH_2[/tex]) is equal to [tex]C_3H_6[/tex]

Given the following data:

  • Percent mass of cyclopropane = 85.7%
  • Percent mass of hydrogen = 14.3%
  • Temperature = 50.0°C to Kelvin = [tex]273+50=323\;K[/tex]
  • Pressure = 0.984 atm.
  • Mass of cyclopropane = 1.56 grams.
  • Volume of cyclopropane = 1.00 Liter.

Scientific data:

  • Ideal gas constant, R = 0.0821L⋅atm/mol⋅K
  • The atomic weight of carbon (C) = 12 g/mol
  • The atomic weight of hydrogen (H) = 1 g/mol

To determine the molecular formula of cyclopropane:

First of all, we would determine the molar mass and number of moles of cyclopropane by using the ideal gas law equation;

[tex]PV=\frac{M}{MM} RT[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • M is the mass of gas.
  • MM is the molar mass of gas.
  • R is the ideal gas constant.
  • T is the temperature.

Making MM the subject of formula, we have:

[tex]MM = \frac{MRT}{PV} \\\\MM = \frac{1.56\;\times \;0.0821\times \;323}{0.984\;\times \;1}\\\\MM = \frac{41.3686}{0.984}[/tex]

Molar mass, MM = 42.04 g/mol.

For number of moles:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{1.56}{42.04}[/tex]

Number of moles = 0.0371 moles

Next, we would determine the empirical formula of cyclopropane:

For carbon (C):

[tex]C = \frac{85.7}{12}[/tex]

Carbon (C) = 7.1417

For hydrogen (H):

[tex]H = \frac{14.3}{1}[/tex]

Hydrogen (H) = 14.3

Simplest whole number ratio:

[tex]C = \frac{7.1417 }{7.1417 } =1[/tex]

[tex]H= \frac{14.3 }{7.1417 } =2[/tex]

Empirical formula of cyclopropane = [tex]CH_2[/tex]

Now, we can determine the molecular formula of cyclopropane:

[tex](CH_2)n = 42.04\\\\(12 + 1\times2)n=42.04\\\\14n=42.04\\\\n=\frac{42.04}{14}[/tex]

n = 3

Molecular formula of cyclopropane ([tex]CH_2[/tex]) = [tex](CH_2)_3 =C_3H_6[/tex]

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