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Cerrena [4.2K]
2 years ago
8

WILL GIVE BRAINLIEST!!!

Chemistry
1 answer:
Afina-wow [57]2 years ago
4 0

Answer:

0.373 moles of ammonium carbonate

Explanation:

To solve this question we must find the molar mass of ammonium carbonate. With the molar mass and the mass we can find its moles, as follows:

(NH₄)₂CO₃ contains: 2 moles N, 8 moles H, 1 mole C, 3 moles O. Molar mass:

2N = 14.0g/mol*2 = 28.0

8H = 1.0g/mol*8 = 8.0

1C = 12.0g/mol*1 = 12.0

3O = 16.0g/mol*3 = 48.0

Molar mass: 28.0 + 8.0 + 12.0 + 48.0 = 96.0g/mol

The moles of ammonium carbonate in 35.8g are:

35.8g * (1mol / 96.0g) =

<h3>0.373 moles of ammonium carbonate</h3>
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Using the given formula, the density of the material is 2.015 g/mL

<h3>Calculating Density </h3>

From the question, we are to determine the density of the material

From the given formula

Density = Mass / Volume

And from the given information,

Mass = 65.5 g

and volume = 32.5 mL

Putting the parameters into the equation,

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<u>Answer:</u> The molecular formula for the compound is C_6H_6

<u>Explanation:</u>

We are given:

Percentage of C = 92.26 %

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Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles

  • <u>Step 2:</u> 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 7.68 moles.

For Carbon = \frac{7.68}{7.68}=1

For Hydrogen = \frac{7.74}{7.68}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

  • <u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{m}{M}RT

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

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

T = temperature of the gas = 100^oC=(100+273)K=373K

Putting values in above equation, we get:

820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

n=\frac{78.10g/mol}{13g/mol}=6

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 6)}H_{(1\times 6)}=C_6H_6

Hence, the molecular formula for the compound is C_6H_6

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