Quiz - Empirical Formula | Fundamentals of Chemical Reactions

General Chemistry 2 - Quiz - Empirical Formula

1

Analysis of a compound known to contain only Mg, P, and O gives this analysis: 21.8% Mg, 27.7% P, 50.3% O. What is its empirical formula?

1. Convert mass percentages to moles:

  • Assume 100 g of the compound.
  • Mass of Mg = 21.8 g, Mass of P = 27.7 g, Mass of O = 50.3 g

2. Calculate the number of moles of each element:

  • Moles of Mg: 21.8 g24.31 g.mol-1 ​≈ 0.897 mol
  • Moles of P: 27.7 g30.97 g.mol-1 ≈ 0.895 mol
  • Moles of O: 50.3 g16.00 g.mol-1 ​≈ 3.144 mol

3. Determine the simplest whole-number ratio:

  • Divide each by the smallest number of moles calculated (0.895).
  • Mg: 0.8970.895 ​≈ 1
  • P: 0.8950.895 ≈ 1
  • O: 3.1440.895 ≈ 3.51

4. Simplify the ratio:

  • The ratio of Mg:P:O is approximately 1:1:3.51.
  • To get whole numbers, multiply each by 2 to approximate 2:2:7.

5. Empirical formula:

  • The simplest whole-number ratio is approximately Mg2​P2​O7​.
2

A chloride salt of rhenium contains 63.6% Re by mass. What is its empirical formula?

1. Assume you have 100 g of the compound.

  • Mass of Re = 63.6 g
  • Mass of Cl = 100 g - 63.6 g = 36.4 g

2. Convert these masses to moles using the atomic masses of Re and Cl.

  • Moles of Re: 63.6 g186.21 g.mol-1 ​≈ 0.341 mol
  • Moles of Cl: 36.4 g35.45 g.mol-1 ≈ 1.027 mol

3. Determine the simplest whole-number ratio of the elements.

  • Ratio of Re to Cl: 0.3410.341 ≈ 1
  • Ratio of Cl to Re: 1.0270.341 ≈ 3

The simplest whole-number ratio of Re to Cl is approximately 1:3, indicating the empirical formula is ReCl3

3

An oxide of manganese contains 2.29 g of manganese per gram of oxygen. What is the empirical formula of this compound?

1. Given Data:

  • Mass of manganese per gram of oxygen: 2.29 g Mn / 1 g O

2. Convert masses to moles:

  • Moles of Mn: 63.6 g186.21 g.mol-1 ≈ 0.341 mol
  • Moles of Cl: 36.4 g35.45 g.mol-1 ≈ 1.027 mol

3. Determine the simplest whole-number ratio of the elements.

  • Ratio of Re to Cl: 0.3410.341 ≈ 1
  • Ratio of Cl to Re: 1.0270.341 ≈ 3

The simplest whole-number ratio of Re to Cl is approximately 1:3, indicating the empirical formula is ReCl3

4

The percentages by mass of C, H, and Cl in a compound are C 52.2%, H 3.7%, and Cl 44.1%. How many carbon atoms are in the simplest formula of the compound?

1. Assume 100 g of the compound:

  • Mass of C = 52.2 g
  • Mass of H = 3.7 g
  • Mass of Cl = 44.1 g

2. Convert these masses to moles:

  • Moles of C: 52.2 g12.01 g.mol-1 ​≈ 4.35 mol
  • Moles of H: 3.7 g1.008 g.mol-1 ≈ 3.67 mol
  • Moles of Cl: 44.1 g35.45 g.mol-1 ​≈ 1.24 mol

3. Determine the simplest whole-number ratio:

  • Divide each by the smallest number of moles calculated (1.24).
  • C: 4.351.24 ​≈ 3.51
  • H: 3.671.24 ≈ 2.96
  • Cl: 1.241.24 = 1

4. Simplify the ratio:

  • The ratio of C : H : Cl is approximately 3.5 : 3 : 1.
  • To get whole numbers, multiply each by 2 to approximate 7 : 6 : 2.

5. Empirical formula:

  • The simplest whole-number ratio is approximately C7H6Cl2​.
5

The mass percentages in a compound are carbon 57.48%, hydrogen 4.22% and oxygen 38.29%. What is its empirical formula?

1. Assume 100 g of the compound:

  • Mass of C = 57.48 g
  • Mass of H = 4.22 g
  • Mass of O = 38.29 g

2. Convert these masses to moles:

  • Moles of C: 57.48 g12.01 g.mol-1 ​≈ 4.79 mol
  • Moles of H: 4.22 g1.008 g.mol-1 ≈ 4.19 mol
  • Moles of O: 38.29 g16.00 g.mol-1 ​≈ 2.39 mol

3. Determine the simplest whole-number ratio:

  • Divide each by the smallest number of moles calculated (2.39).
  • C: 4.792.39 ​≈ 2
  • H: 4.192.39 ≈ 1.75
  • O: 2.392.39 = 1

4. Simplify the ratio:

  • The ratio of C : H : O is approximately 2 : 1.75 : 1.
  • To get whole numbers, multiply each by 4 to approximate 8 : 7 : 4.

5. Empirical formula:

  • The simplest whole-number ratio is approximately C8H7O4​.