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A compound was analyzed and found to contain the following percent composition: 2.056% hydrogen, 32.69% S, and 65.26% oxygen. Calculate the empirical formula.

Sagot :

Answer:

H2SO4

Explanation:

Firstly,

2.056% hydrogen = 2.056g of H

32.69% Sulphur = 32.69g of S

65.26% oxygen = 65.26g of O

Next, we convert each mass into moles by dividing by their respective molar mass:

H = 1, S = 32, O = 16

H = 2.056/1 = 2.056mol

S = 32.69/32 = 1.021mol

O = 65.26/16 = 4.078mol

Next, we divide each mole value by the smallest mole value (1.021)

H = 2.056/1.021 = 2.01

S = 1.021/1.021 = 1

O = 4.078/1.021 = 3.99

Approximately the ratio of H to S to O is 2:1:4, hence, the empirical formula for H, S and O is H2SO4.

The empirical formula of a chemical compound is the simplest whole-number ratio of atoms present in a compound

The correct answer of the following is [tex]H_2SO_4[/tex]

The data is given as follows:-

  • 2.056% hydrogen = 2.056g of H
  • 32.69% Sulphur = 32.69g of S
  • 65.26% oxygen = 65.26g of O

We have to find the moles in the reaction.

The mas of the element is as follows:-

  • H = 1
  • S = 32
  • O = 16

After solving the equation we get

[tex]H = \frac{2.056}{1} = 2.056mol\\\\S = \frac{32.69}{32} = 1.021mol\\\\O = \frac{65.26}{16} = 4.078mol[/tex]

Next, we divide each mole value by the smallest mole value (1.021)

[tex]H = \frac{2.056}{1.021} = 2.01\\\\S = \frac{1.021}{1.021} = 1\\\\O = \frac{4.078}{1.021} = 3.99[/tex]

The ratio of H to S to O is 2:1:4,

Hence, the empirical formula for H, S, and O is [tex]H_2SO_4.[/tex]

For more information, refer to the link:-

https://brainly.com/question/15804584

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