Chemistry, 21.10.2020 16:01 charisaramsey

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?

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Answer from: marvinsductant6710

The molecular formula of cyclopropane ( CH_2 ) is equal to C_3H_6

Given the following data:

Percent mass of cyclopropane = 85.7%Percent mass of hydrogen = 14.3%Temperature = 50.0°C to Kelvin = $273+50=323\;K$ 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⋅KThe atomic weight of carbon (C) = 12 g/molThe 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;

$PV=\frac{M}{MM} RT$

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:

$MM = \frac{MRT}{PV} \\\\MM = \frac{1.56\;\times \;0.0821\times \;323}{0.984\;\times \;1}\\\\MM = \frac{41.3686}{0.984}$

Molar mass, MM = 42.04 g/mol.

For number of moles:

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{1.56}{42.04}$

Number of moles = 0.0371 moles

Next, we would determine the empirical formula of cyclopropane:

For carbon (C):

$C = \frac{85.7}{12}$

Carbon (C) = 7.1417

For hydrogen (H):

$H = \frac{14.3}{1}$

Hydrogen (H) = 14.3

Simplest whole number ratio:

$C = \frac{7.1417 }{7.1417 } =1$

$H= \frac{14.3 }{7.1417 } =2$

Empirical formula of cyclopropane = CH_2

Now, we can determine the molecular formula of cyclopropane:

$(CH_2)n = 42.04\\\\(12 + 1\times2)n=42.04\\\\14n=42.04\\\\n=\frac{42.04}{14}$

n = 3

Molecular formula of cyclopropane ( CH_2 ) = (CH_2)_3 =C_3H_6

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Answer from: jose477

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

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A mixture of cyclopropane gas and oxygen is used as an anesthetic. Cyclopropane contains 85.7% C And...

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