Percent Composition, Empirical and Molecular Formulas in General Chemistry

Percent Composition

Definition and Calculation

Percent composition describes the percentage by mass of each element in a chemical compound. It is a fundamental concept in analytical chemistry, used to determine the relative amounts of elements in compounds.

• Percent by mass of an element is calculated using the formula:

$\text{Percent by mass} = \frac{\text{mass of one element in formula unit}}{\text{mass of one formula unit}} \times 100\%$

• Example: For water (H2O):

• Molar mass of H2O = 2(1.01) + 16.00 = 18.02 g/mol

• Percent H: $\frac{2 \times 1.01}{18.02} \times 100 = 11.2\%$

• Percent O: $\frac{16.00}{18.02} \times 100 = 88.8\%$

Determining the Formula of a Compound

Using Percent Composition

Given the relative percentages of each element in a compound, you can determine the empirical formula, which represents the simplest whole-number ratio of the elements.

• Example: If a compound contains 10% A, 20% B, and 30% C, you can find the ratio of A:B:C to determine the empirical formula.

• Empirical formula is the simplest ratio of elements in a compound.

Empirical Formula

Calculation Steps

The empirical formula is determined from percent composition data by converting mass percentages to moles and finding the simplest whole-number ratio.

1. Assume 100 g of the compound to convert percentages directly to grams.

2. Convert grams to moles for each element using their molar masses.

3. Divide each mole value by the smallest mole value to get the simplest ratio.

4. Round to nearest whole number or multiply by a factor to obtain whole numbers if necessary.

• Example: Compound with 64.82% C, 21.59% O, 13.59% H

• Assume 100 g: 64.82 g C, 21.59 g O, 13.59 g H

• Convert to moles:

  • $\frac{64.82\,g}{12.011\,g/mol} = 5.397\,mol\,C$

  • $\frac{21.59\,g}{16.00\,g/mol} = 1.349\,mol\,O$

  • $\frac{13.59\,g}{1.0079\,g/mol} = 13.48\,mol\,H$

• Divide by smallest (1.349):

  • C: $\frac{5.397}{1.349} = 4.001$

  • H: $\frac{13.48}{1.349} = 9.992$

  • O: $\frac{1.349}{1.349} = 1$

• Empirical formula: C4H10O

Handling Non-Whole Number Ratios

If the ratio is not a whole number, multiply all subscripts by the smallest integer that converts all to whole numbers.

Fractional Subscript	Multiply by
0.20	5
0.25	4
0.33	3
0.40	5
0.50	2
0.66	3
0.75	4
0.80	5

Example: For a nitrogen oxide with 36.85% N, use the above method to determine the empirical formula.

Molecular Formula

Definition and Calculation

The molecular formula shows the actual number of atoms of each element in a molecule. It may be a multiple of the empirical formula.

• Empirical formula gives the simplest ratio; molecular formula gives the actual composition.

• Example: NO2 and N2O4 have the same empirical formula (NO2) but different molecular formulas.

To determine the molecular formula:

1. Find the empirical formula mass.

2. Divide the compound's molar mass by the empirical formula mass to get a whole number.

3. Multiply all subscripts in the empirical formula by this number.

$\text{Molecular formula} = (\text{Empirical formula}) \times n$

$\text{n} = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

• Example: Empirical formula C4H10O, molar mass = 222.1 g/mol

• Empirical formula mass = $4 \times 12.01 + 10 \times 1.008 + 16.00 = 74.14$ g/mol

• $n = \frac{222.1}{74.14} = 3$

• Molecular formula: C12H30O3

Compound Formulas from Experimental Data

Determining Formulas from Mass Data

Experimental data can be used to determine the formula of a compound formed from two elements.

• Example: Tin (Sn) reacts with iodine (I2) to form a compound.

• Initial mass of Sn: 1.056 g

• Mass of Sn recovered: 0.601 g

• Mass of Sn reacted: 1.056 g - 0.601 g = 0.455 g

• Convert masses to moles:

  • Sn: $\frac{0.455\,g}{118.7\,g/mol} = 3.83 \times 10^{-3}\,mol$

  • I: $\frac{1.947\,g}{253.81\,g/mol} = 7.67 \times 10^{-3}\,mol$

• Divide by smallest mole value:

  • Sn: $\frac{3.83 \times 10^{-3}}{3.83 \times 10^{-3}} = 1$

  • I: $\frac{7.67 \times 10^{-3}}{3.83 \times 10^{-3}} = 2$

• Empirical formula: SnI2

Summary Table: Steps for Empirical and Molecular Formula Determination

Step	Description
1	Obtain mass or percent composition of each element
2	Convert mass to moles using molar mass
3	Divide by smallest mole value to get ratio
4	Adjust ratio to whole numbers if necessary
5	Write empirical formula
6	Calculate empirical formula mass
7	Divide molar mass by empirical formula mass
8	Multiply subscripts by result to get molecular formula

Additional info: These notes cover foundational skills in chemical analysis, including percent composition, empirical and molecular formula determination, and application to real experimental data. Mastery of these topics is essential for success in General Chemistry.