Percent Composition, Empirical Formulas, and Molecular Formulas

Percent Composition Fundamentals

• Percent composition is defined as the percentage by mass of each individual element present within a chemical compound.

• The general formula used to calculate the percent composition of an element is: $\% \text{ of an element in a compound} = \frac{\text{Mass of element in } 1\,\text{mol of compound}}{\text{Molar mass of compound}} \times 100$

• To calculate percent composition, the initial necessary step is always to determine the molar mass of the entire compound.

Calculating Percent Composition: Carbon Dioxide Example

• Objective: Determine the percent composition of carbon dioxide ($CO_2$).

• Step 1: Calculate the molar mass of $CO_2$:

  • Number of carbon atoms: $1$
  • Atomic mass of carbon ($C$): $12.01\,\text{g/mol}$
  • Number of oxygen atoms: $2$
  • Atomic mass of oxygen ($O$): $16.00\,\text{g/mol}$
  • Molar mass calculation setup: $\text{Molar mass of } CO_2 = 1(12.01) + 2(16.00)$$\text{Molar mass of } CO_2 = 12.01 + 32.00$$\text{Molar mass of } CO_2 = 44.01\,\text{g/mol}$

• Step 2: Calculate the percent composition of Carbon ($C$): $\% \text{ composition of C} = \frac{12.01\,\text{g of C in } 1\,\text{mol of } CO_2}{44.01\,\text{g total in } 1\,\text{mol of } CO_2} \times 100$$\% \text{ composition of C} = 27.29\%\,\text{C in } CO_2$

• Step 3: Calculate the percent composition of Oxygen ($O$): $\% \text{ composition of O} = \frac{2(16.00\,\text{g of O in } 1\,\text{mol of } CO_2)}{44.01\,\text{g total in } 1\,\text{mol of } CO_2} \times 100$$\% \text{ composition of O} = 72.71\%\,\text{O in } CO_2$

Determining Elemental Mass from Percent Composition: Magnesium Hydroxide Example

• Problem scenario: Magnesium hydroxide is $54.87\%$ oxygen by mass. Calculate how many grams of oxygen are contained in a sample of $225\,\text{g}$ of the compound.

• Given quantities:

  • Percent of oxygen in $Mg(OH)_2$: $54.87\%$
  • Total mass of $Mg(OH)_2$: $225\,\text{g}$
  • Target quantity: Grams of $O$ ($\text{g of O} = ?$

• Calculation strategy:

  • Find what $54.87\%$ of $225\,\text{g}$ represents using a proportion.

• Proportional setup: $\frac{54.87}{100} = \frac{\text{g of O}}{225\,\text{g total}}$

• Solving by cross-multiplication: $12345.75 = 100(\text{g of O})$$\frac{12345.75}{100} = \frac{100(\text{g of O})}{100}$$\text{Mass of Oxygen} = 123.4\,\text{g of O}$

Practice Problems for Percent Composition

• Practice Problem 1: Find the percent composition of calcium nitrate, $Ca(NO_3)_2$.

• Practice Problem 2: Potassium chlorate is $31.91\%$ potassium by mass. How many grams of potassium would be expected in a $150\,\text{g}$ sample of the compound?

Determining Chemical Formulas and Empirical Formula Fundamentals

• Empirical formula definition: The symbols of the elements and subscripts written to represent the lowest whole-number mole ratio of atoms in a chemical compound.

  • Percent composition data is directly used to calculate the empirical formula.

• Molecular formula definition: The actual chemical formula representing the total number of atoms of each element in a molecule of the compound, which is not always reduced to the lowest whole-number ratio.

• Systematic steps to determine an empirical chemical formula from percent composition:

  • Step 1: Find out how many grams ($\text{g}$) are present of each constituent element.
  • Step 2: Convert the mass of each element into the number of moles ($\text{mol}$).
  • Step 3: Compare the ratio of moles to determine the ratio of atoms in the compound by dividing all calculated mole quantities by the smallest mole value obtained.

Empirical Formula Determination: Sodium-Chlorine-Oxygen Compound

• Problem scenario: A sample of $110.0\,\text{g}$ of an unknown compound contains $18.8\%\,\text{Na}$, $29.0\%\,\text{Cl}$, and $52.2\%\,\text{O}$. Determine its chemical formula based on this percent composition.

• Step 1: Determine the mass in grams for each individual element within the $110.0\,\text{g}$ total mass:

  • Sodium ($Na$): $18.8\% \text{ of } 110.0\,\text{g} = 20.7\,\text{g Na}$

• Chlorine ($Cl$): $29.0\% \text{ of } 110.0\,\text{g} = 31.9\,\text{g Cl}$

• Oxygen ($O$): $52.2\% \text{ of } 110.0\,\text{g} = 57.4\,\text{g O}$

• Step 2: Convert the mass of each element into moles using atomic masses ($Na = 22.99\,\text{g/mol}$, $Cl = 35.45\,\text{g/mol}$, $O = 16.00\,\text{g/mol}$):

  • Moles of Sodium: $20.7\,\text{g Na} \times \frac{1\,\text{mol}}{22.99\,\text{g Na}} = 0.900\,\text{mol Na}$

• Moles of Chlorine: $31.9\,\text{g Cl} \times \frac{1\,\text{mol}}{35.45\,\text{g Cl}} = 0.900\,\text{mol Cl}$

• Moles of Oxygen: $57.4\,\text{g O} \times \frac{1\,\text{mol}}{16.00\,\text{g O}} = 3.59\,\text{mol O}$

• Step 3: Divide each mole value by the smallest calculated value ($0.900\,\text{mol}$) to find the lowest whole-number mole ratio:

  • Mole ratio for Sodium: $\frac{0.900\,\text{mol Na}}{0.900} = 1\,\text{mol Na}$

• Mole ratio for Chlorine: $\frac{0.900\,\text{mol Cl}}{0.900} = 1\,\text{mol Cl}$

• Mole ratio for Oxygen: $\frac{3.59\,\text{mol O}}{0.900} = 4\,\text{mol O}$

• Final Empirical Formula:

  • Combining the integer ratios ($1\,\text{Na}$, $1\,\text{Cl}$, $4\,\text{O}$) yields the chemical formula: $NaClO_4$

Method for Determining Molecular Formulas

• The procedure to determine a molecular formula requires knowing the empirical formula and the experimentally measured molar mass of the compound:
  • Step 1: Calculate the molar mass of the compound's empirical formula.
  • Step 2: Compare the empirical formula mass to the actual (given) molar mass of the compound.
  • Case A: If the calculated empirical molar mass is identical to the actual molar mass: $\text{Molecular formula} = \text{Empirical formula}$
• Case B: If the empirical molar mass differs from the actual molar mass, divide the actual molar mass by the empirical molar mass (the smaller number) to determine the integer multiplication factor: $\text{Multiplication Factor} = \frac{\text{Actual Molar Mass}}{\text{Empirical Molar Mass}}$
• Step 3: Multiply all subscripts in the empirical formula by this integer multiplication factor to obtain the molecular formula.

Molecular Formula Worked Examples

• Example 1: Determining the molecular formula for empirical formula $CH_2$ with a given actual molar mass of $42\,\text{g/mol}$.

  • Elemental atomic masses:
  • Carbon ($C$): $12.01\,\text{g/mol}$
  • Hydrogen ($H$): $1.008\,\text{g/mol}$
  • Calculate empirical molar mass of $CH_2$: $\text{Molar mass of } CH_2 = 1(12.01) + 2(1.008)$$\text{Molar mass of } CH_2 = 12.01 + 2.016$$\text{Molar mass of } CH_2 = 14.026\,\text{g/mol}$

• Determine the multiplication factor: $\frac{42\,\text{g/mol}}{14.026\,\text{g/mol}} = 3$

• Apply the multiplication factor to the empirical formula subscripts: $3(CH_2) = C_3H_6$

  • The resulting molecular formula is $C_3H_6$.

• Example 2: Determining the molecular formula for empirical formula $CH_2$ with a given actual molar mass of $70\,\text{g/mol}$.

  • Elemental atomic masses:
  • Carbon ($C$): $12.01\,\text{g/mol}$
  • Hydrogen ($H$): $1.008\,\text{g/mol}$
  • Calculate empirical molar mass of $CH_2$: $\text{Molar mass of } CH_2 = 1(12.01) + 2(1.008)$$\text{Molar mass of } CH_2 = 12.01 + 2.016$$\text{Molar mass of } CH_2 = 14.026\,\text{g/mol}$

• Determine the multiplication factor: $\frac{70\,\text{g/mol}}{14.026\,\text{g/mol}} = 5$

• Apply the multiplication factor to the empirical formula subscripts: $5(CH_2) = C_5H_{10}$

• The resulting molecular formula is $C_5H_{10}$.