1. The molecular formula of the compound is C₆H₃₆O₁₂
2. The molecular formula of the compound is N₂H₄O₂
3 The empirical formula and molecular formula of the compound are: C₂H₃ and C₄H₆
4. The empirical formula of the compound is C₃H₆O
5. The empirical formula of the compound is ZrSiO₄
<h3>1. How to determine the molecular formula </h3>
- Empirical formula = CH₆O₂
- Molar mass = (285 + 315) / 2 = 600 / 2 = 300 g/mole
- Molecular formula =?
Molecular formula = empirical × n = molar mass
[CH₆O₂]ₙ = 300
[12 + (6×1) + (16×2)]ₙ = 300
50n = 300
Divide both side by 74
n = 300 / 50
n = 6
Molecular formula = [CH₆O₂]ₙ
Molecular formula = [CH₆O₂]₆
Molecular formula = C₆H₃₆O₁₂
<h3>2. How to determine the molecular formula </h3>
- Empirical formula = NH₂O
- Molar mass = (55 + 65) / 2 = 120 / 2 = 60 g/mole
- Molecular formula =?
Molecular formula = empirical × n = molar mass
[NH₂O]ₙ = 60
[14 + (2×1) + 16]ₙ = 60
32n = 60
Divide both side by 32
n = 60 / 32
n = 2
Molecular formula = [NH₂O]ₙ
Molecular formula = [NH₂O]₂
Molecular formula = N₂H₄O₂
<h3>3. How to determine the empirical formula and molecular formula</h3>
We'll begin by obtaining the empirical formula. This is illustrated below:
- Carbon (C) = 88.9%
- Hydrogen (H) = 11.1%
- Empirical formula =?
Divide by their molar mass
C = 88.9 / 12 = 7.4
H = 11.1 / 1 = 11.1
Divide by the smallest
C = 7.4 / 7.4 = 1
H = 11.1 / 7.4 = 3/2
Multiply by 2 to express in whole number
C = 1 × 2 = 2
H = 3/2 × 2 = 3
Thus, the empirical formula of the compound is C₂H₃
Finally, we shall determine the molecular formula of the compound. This is illustrated below:
- Molar mass of compound = 54 g/mol
- Empirical formula = C₂H₃
- Molecular formula =?
Molecular formula = empirical × n = molar mass
[C₂H₃]ₙ = 75.16
[(12×2) + (3×1)]ₙ = 54
27n = 54
Divide both side by 27
n = 54 / 27
n = 2
Molecular formula = [C₂H₃]ₙ
Molecular formula = [C₂H₃]₂
Molecular formula = C₄H₆
<h3>4. How to determine the empirical formula</h3>
- Carbon (C) = 62.07%
- Hydrogen (H) = 10.34%
- Oxygen (O) = 27.59%
- Empirical formula =?
Divide by their molar mass
C = 62.07 / 12 = 5.1725
H = 10.34 / 1 = 10.34
O = 27.59 / 16 = 1.724
Divide by the smallest
C = 5.1725 / 1.724 = 3
H = 10.34 / 1.724 = 6
O = 1.724 / 1.724 = 1
Thus, the empirical formula of the compound is C₃H₆O
<h3>5. How to determine the empirical formula</h3>
- Zr = 49.76%
- Si = 15.32%
- O = 34.91%
- Empirical formula =?
Divide by their molar mass
Zr = 49.76 / 91 = 0.547
Si = 15.32 / 28 = 0.547
O = 34.91 / 16 = 2.182
Divide by the smallest
Zr = 0.547 / 0.547 = 1
Si = 0.547 / 0.547 = 1
O = 2.182 / 0.547 = 4
Thus, the empirical formula of the compound is ZrSiO₄
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