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1. Determine the molecular formula of an oxide of iron in which the mass of iron and oxygen are 69.9% and 30% respectively given that the molar mass of the oxide 159.898/mol, find the empirical and molecular formula.

2. a crystalline salt when heated becomes anhydrous and loses 51.2% of its weight the anhydrous salt analysis gave the percent composition as magnesium is equal to 20.0% and sulphur is equal to 26.66% and oxygen is equal to 53.33%.

3. In three moles of Ethane calculate the following
1. calculate number of carbon atoms.
2. number of moles of hydrogen atoms
3. number of molecules of Ethane.​

Respuesta :

1a. The empirical formula of the compound is Fe₂O₃

1b. The molecular formula of the compound is Fe₂O₃

2a. The molecular formula of the anhydrous salt is MgSO₄

2b. The formula of the crystalline salt is MgSO₄.7H₂O

3i. The number of mole of carbon atoms in the compound is 6 moles

3ii. The number of mole of hydrogen atoms in the compound is 18 moles

3iii. The number of molecules in 3 moles of ethane is 1.806×10²⁴ molecules

1a. How to determine the empirical formula

  • Fe = 69.9%
  • O = 30%
  • Empirical formula =?

Divide by their molar mass

Fe = 69.9 / 56 = 1.248

O = 30 / 16 = 1.875

Divide by the smallest

Fe = 1.248 / 1.248 = 1

O = 1.875 / 1.248 = 3/2

Multiply by 2 to express in whole number

Fe = 1 × 2 = 2

O = 3/2 × 2 = 3

Thus, the empirical formula of the compound is Fe₂O₃

1b. How to determine the molecular formula

  • Empirical formula = Fe₂O₃
  • Molar mass of compound = 159.89 g/mol
  • Molecular formula = ?

Molecular formula = empirical × n = molar mass

[Fe₂O₃]n = 159.89

[(56×2) + (16×3)]n = 159.89

160n = 159.89

n = 159.89 / 160

n = 1

Molecular formula = [Fe₂O₃]n

Molecular formula = [Fe₂O₃] × 1

Molecular formula = Fe₂O₃

2a. How to determine the molecual formula of the anhydrous salt

We'll begin by calculating the empirical formula

  • Mg = 20.0%
  • S = 26.66%
  • O = 53.33%
  • Empirical formula =?

Divide by their molar mass

Mg = 20.0 / 24 = 0.83

S = 26.66 / 32 = 0.83

O = 53.33 / 16 = 3.33

Divide by the smallest

Mg = 0.83 / 0.83 = 1

S = 0.83 / 0.83 = 1

O = 3.33 / 0.83 = 4

Thus, the empirical formula of the anhydrous salt is MgSO₄

The molecular formula of the anhydrous salt can be obtained as follow:

  • Empirical formula = MgSO₄
  • Molar mass of compound = 120 g/mol
  • Molecular formula = ?

Molecular formula = empirical × n = molar mass

[MgSO₄]n = 120

[24 + 32 + (16×4)]n = 159.89

120n = 120

n = 120 / 120

n = 1

Molecular formula = [MgSO₄]n

Molecular formula = [MgSO₄] × 1

Molecular formula = MgSO₄

2b. How to determine the formula of the crystalline salt

  • Water (H₂O) = 51.2%
  • Anhydrous salt (MgSO₄) = 100 - 51.2 = 48.8%
  • Formula of crystalline salt =?

Divide by their molar mass

MgSO₄ = 48.8 / 120 = 0.4

H₂O = 51.2 / 18 = 2.8

Divide by the smallest

MgSO₄ = 0.4 / 0.4 = 1

H₂O = 2.8 / 0.4 = 7

Thus, the formula of the crystalline salt is MgSO₄.7H₂O

3i. How to determine the mole of carbon atoms in 3 moles of C₂H₆

1 mole of C₂H₆ contains 2 moles of carbon atoms.

Therefore,

3 moles of C₂H₆ will contain = 3 × 2 = 6 moles of carbon atoms

3ii. How to determine the mole of hydrogen atoms in 3 moles of C₂H₆

1 mole of C₂H₆ contains 6 moles of hydrogen atoms.

Therefore,

3 moles of C₂H₆ will contain = 3 × 6 = 18 moles of hydrogen atoms

3iii. How to determine the number of molecules

From Avogadro's hypothesis,

1 mole of ethane = 6.02×10²³ molecules

Therefore,

3 moles of ethane = 3 × 6.02×10²³ molecules

3 moles of ethane = 1.806×10² molecules

Learn more about empirical formula:

https://brainly.com/question/24297883

Learn more about Avogadro's number:

https://brainly.com/question/26141731

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Complete question

2. A crystalline salt when heated becomes anhydrous and loses 51.2% of its weight the anhydrous salt analysis gave the percent composition as magnesium is equal to 20.0% and sulphur is equal to 26.66% and oxygen is equal to 53.33%. Ccalculate the molecular formula of the anhydrous and the crystalline salt. The molecular weight of the anhydrous salt is 120

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