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1.) Given the % composition of each element in the following compound. Calculate and determine the empirical formula. Name the formula as well.
1.6% H 22.2% N 76.2% O.
Hint, assume a 100 g sample, then you will have 1.6 g H, 22.2 g N and 76.2 g O

2. A sample of a compound is broken down into its constituent elements and yields 13.0 g of lead, 1.77 g of nitrogen and 4.03 g of oxygen. What is the empirical formula for this compound?

3.) If 4.04g of N combine with 11.46g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?

Respuesta :

Eduard22sly

1. The empirical formula of the compound is HNO₃. The name of the compound is trioxonitrate (v)

2. The empirical formula of the compound is PbN₂O₄ or  Pb(NO₂)₂

3. The molecular formula of the compound is Nâ‚‚Oâ‚…

1. How to determine the empirical formula

  • Hydrogen (H) = 1.6 g
  • Nitrogen (N) = 22.2 g
  • Oxygen (O) = 76.2 g
  • Empirical formula =?

Divide by their molar mass

H = 1.6 / 1 = 1.6

N = 22.2 / 14 = 1.586

O = 76.2 / 16 = 4.7625

Divide by the smallest

H = 1.6 / 1.586 = 1

N = 1.586 / 1.586 = 1

O = 4.7625 / 1.586 = 3

Thus, the empirical formula of the compound is HNO₃ and the name of the compound is trioxonitrate (v)

2. How to determine the empirical formula

  • Lead (Pb) = 13 g
  • Nitrogen (N) = 1.77 g
  • Oxygen (O) = 4.03 g
  • Empirical formula =?

Divide by their molar mass

Pb =13 / 207 = 0.0628

N = 1.77 / 14 = 0.1264

O = 4.03 / 16 = 0.2519

Divide by the smallest

Pb = 0.0628 / 0.0628 = 1

N = 0.1264 / 0.0628 = 2

O = 0.2519 / 0.0628 = 4

Thus, the empirical formula of the compound is PbNâ‚‚Oâ‚„ or Pb(NOâ‚‚)â‚‚

3. How to determine the molecular formula

We'll begin by determining the empirical formula of the compound. This can be obtained as follow:

  • Nitrogen (N) = 4.04 g
  • Oxygen (O) = 11.46 g
  • Empirical formula =?

Divide by their molar mass

N = 4.04 / 14 = 0.2886

O = 11.46 / 16 = 0.71625

Divide by the smallest

N = 0.2886 / 0.2886 = 1

O = 0.71625 / 0.2886 = 2.5

Multiply by 2 to express in whole number

N = 1 × 2 = 2

O = 2.5 × 2 = 5

Thus, the empirical formula of the compound is Nâ‚‚Oâ‚…

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

  • Empirical formula = Nâ‚‚Oâ‚…
  • Molar mass of compound = 108 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = molar mass

[Nâ‚‚Oâ‚…]n = 108

[(2×14) + (16×5]n = 108

[28 + 80]n = 108

108n = 108

Divide both side by 108

n = 108 / 108

n = 1

Molecular formula = [Nâ‚‚Oâ‚…]n

Molecular formula = [N₂O₅] × 1

Molecular formula = Nâ‚‚Oâ‚…

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