Question

1) A certain compound has an empirical formula of CH_6O_ 2. Its molar mass is between 285 and 315 g/mol. What is its molecular formula?
CH_6O_2
C_4H_24O_8
C_6H_36O_12
C_5H_30O_10

2) certain compound has an empirical formula of NH_2O. Its molar mass is between 55 and 65 g/mol. What is its molecular formula?
NH_2O
N_2H_2O_2
N_2H_4O_2
N_4H_8O_4

3) A certain compound is found to have the percent composition (by mass) of 88.9 percent C and 11.1 percent H. The molar mass of the compound was found to be 54.0 g/mol. Calculate the empirical formula and the molecular formula.

C_2H_3 and C_4H_6
C_2H_6 and C_3H_9
CH_2 and C_3H_6
CH_3 and C_2H_6

4) The characteristic odor of pineapple is due to ethyl butyrate. It contains 62.07 percent carbon, 10.34 percent hydrogen, and 27.59 percent oxygen. Determine the empirical formula.

C_6H_6O_2
C_6H_12O_2
C_3H_3O
C_3H_6O

5) Zircon is a diamond-like material that has the composition 49.76 percent Zr, 15.32 percent Si, and 34.91 percent O. What is the empirical formula of zircon?

ZrSiO
ZrSiO_2
ZrSiO_3
ZrSiO_4

Answer 1

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

1. How to determine the molecular formula

• 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

2. How to determine the molecular formula

• 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

3. How to determine the empirical formula and molecular formula

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

4. How to determine the empirical formula

• 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

5. How to determine the empirical formula

• 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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