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Natali5045456 [20]
3 years ago
11

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

What is the molecular formula of the compound
Chemistry
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

The molecular formula = C_{6}H_{6}

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,  

n=\frac{m}{M}

Using ideal gas equation as:

PV=\frac{m}{M}RT

where,  

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15

M=78.31\ g/mol

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,  

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = C_{6}H_{6}

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Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 730.5 mmHg  

V = Volume of the gas = 9.49 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol

According to the reaction shown below as:-

2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when \frac{2}{3}\times 0.3794 moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 735.5 mmHg  

V = Volume of the gas = 9.99 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

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