Understanding Molar Mass and % Composition

Understanding Molar Mass and Mole Calculations

What is Molar Mass?

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in the molecule.

Example:
For calcium carbonate (CaCO₃):

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol (and there are 3 oxygen atoms)

So, the molar mass of CaCO₃ = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol.

Converting Grams to Moles

To convert a given mass of a substance to moles, use the molar mass as a conversion factor.

Example:
Convert 13.5 g of CaCO₃ to moles.

$$\text{Moles}=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{13.5\text{g}}{100.09\text{g/mol}}\approx 0.135\text{mol}$$

This tells us that 13.5 grams of calcium carbonate contains approximately 0.135 moles of the substance.

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1. Example Calculation: Molar Mass of TBAF

• Compound Formula: $\text{C}_{16}\text{H}_{36}\text{FN}$ (Tetra-butylammonium fluoride)
• Steps:
  • Identify the number of each atom in the compound.
  • Multiply the atomic mass of each element by the number of atoms present.
  • Sum these values to get the total molar mass.
• Calculation:
  • $16 \times 12.01 \, \text{g/mol} + 36 \times 1.01 \, \text{g/mol} + 1 \times 19.00 \, \text{g/mol} + 1 \times 14.01 \, \text{g/mol} = 261.53 \, \text{g/mol}$
    
• Result: Molar mass of TBAF is approximately 261.53 g/mol.

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2. Determining the Mass of a Given Amount of Moles

• Explanation of using molar mass to calculate the mass of a specific number of moles.
• Formula: $\text{mass} = \text{moles} \times \text{molar mass}$.
• Example with TBAF:
• Problem: Calculate the mass of 2.5 moles of TBAF.
• Solution: $$2.5\text{moles}\times 261.53\text{g/mol}=653.825\text{g}\approx 650\text{g (rounded to significant figures)}$$
• Final Answer: $6.5 \times 10^2 \, \text{g}$
  

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3. What is Percent Composition?

• Definition of percent composition as the percentage by mass of each element in a compound.
• Importance in understanding the makeup of a compound and its chemical properties.

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4. Example Calculation: Percent Composition of Cinnamaldehyde

• Compound Formula: $\text{C}_9\text{H}_8\text{O}$
• Steps:
  • Calculate molar mass of Cinnamaldehyde. $$9\times 12.01+8\times 1.01+1\times 16.00=132.17\text{g/mol}$$
  • Find mass contribution of each element and use the formula for percent composition.
• Calculations:
  • Carbon (C): $$\frac{108.09\text{g}}{132.17\text{g/mol}}\times 100=81.78\mathrm{\%}$$
  • Hydrogen (H): $$\frac{8.08\text{g}}{132.17\text{g/mol}}\times 100=6.11\mathrm{\%}$$
  • Oxygen (O): $$\frac{16.00\text{g}}{132.17\text{g/mol}}\times 100=12.11\mathrm{\%}$$
• Result: Percent composition of Cinnamaldehyde is 81.78% C, 6.11% H, and 12.11% O.

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5. Why Percent Composition is Useful

• Applications in identifying compounds based on their composition.
• Importance in pharmaceutical and chemical industries for quality control and product formulation.

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Empirical vs. Molecular Formulas

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of elements in that compound. It does not necessarily show the exact number of atoms but gives the simplest proportions.

Example:
For glucose (C₆H₁₂O₆), the empirical formula is CH₂O because each ratio of C:H is simplified to 1:2:1.

Molecular Formula

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

Example: The molecular formula of glucose is C₆H₁₂O₆, which is a multiple of its empirical formula CH₂O.

Examples of Molecular Formulas Based on the Empirical Formula CH₂O

If the empirical formula is CH₂O, possible molecular formulas that are multiples of this ratio could include:

1. CH₂O – Simplest form (empirical formula can be molecular formula too).
2. C₂H₄O₂ – Molecular formula with a 2:4:2 ratio.
3. C₆H₁₂O₆ – Molecular formula for glucose, with a 6:12:6 ratio.