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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 N₂O₅

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 PbN₂O₄ or Pb(NO₂)₂

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 N₂O₅

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

  • Empirical formula = N₂O₅
  • Molar mass of compound = 108 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = molar mass

[N₂O₅]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 = [N₂O₅]n

Molecular formula = [N₂O₅] × 1

Molecular formula = N₂O₅

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