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Levart [38]
3 years ago
15

A percent composition analysis yields 32.37% sodium, 22.57 sulfur and 45.05% oxygen what is the empirical formula for the compou

nd
Chemistry
1 answer:
Dovator [93]3 years ago
8 0

Answer:  

Na₂SO₄

Explanation:  

Assume that you have 100 g of the compound.  

Then you have 32.37 g Na, 22.57 g S, and 45.05 g O.  

1. Calculate the moles of each element.

Moles of Na = 32.37 × 1/22.99  =   1.408   mol Na

Moles of S    = 22.57 × 1/32.06  = 0.7040 mol S

Moles of O   = 45.05 × 1/16.00   = 2.816    mol O

2. Calculate the <em>molar ratio</em>s.  

Divide all values by the smallest number.

Na: 1.408/0.7040   = 2.000

S:  0.7040/0.7040 = 1

O: 2.816/0.7040    = 4.000

3. Determine the empirical formula

Round off all numbers to the closest integer.  

Na: 2

S:   1

O:  4

The empirical formula is Na₂SO₄.

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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 %

Percentage of H = 7.74 %

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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