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3 years ago

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14.A 4.25 gram sample of an unknown gas is found to occupy a volume of 1.70 L at a pressure of 883 mm Hg and a temperature of 58

°C. The molar mass of the unknown gas is ______
g/mol.

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

58.5g/mol

Explanation:

Given parameters:

Mass of gas = 4.25g

Volume = 1.7L

Pressure = 883mmHg

760mmHg = 1 atm

883mmHg = $\frac{883}{760}$ = 1.16atm

Temperature = 58°C = 58 + 273 = 331K

Unknown:

Molar mass of sample = ?

Solution:

To solve this problem, we use the ideal gas equation to find the number of moles;

PV = nRT

P is the pressure

V is the volume

n is the number of moles

R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹

T is the temperature

n = $\frac{PV}{RT}$ = $\frac{1.16 x 1.7}{0.082 x 331}$ = 0.07mole

Since;

mass = number of moles x molar mass

4.25 = 0.07 x molar mass

Molar mass = 58.5g/mol

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So, lets probe those 3 cases.

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4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

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That does not correspond to any metal.

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where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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We know that:

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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