Chapter 6: Chemical Composition – Study Notes

Chapter 6: Chemical Composition

Formula Mass and Molar Mass

The formula mass is the mass of a formula unit or an individual molecule, calculated as the sum of the atomic masses of all atoms in the chemical formula. The molar mass is the mass of one mole of a substance, numerically equal to the formula mass but expressed in grams per mole (g/mol).

• Formula Mass: The sum of the atomic masses of all atoms in a molecule or formula unit, usually expressed in atomic mass units (amu).

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).

• Calculation Example: For water ($H_2O$): $\text{Formula mass of } H_2O = 2 \times (1.008\ \text{amu H}) + 1 \times (16.00\ \text{amu O}) = 18.02\ \text{amu}$ $\text{Molar mass of } H_2O = 2 \times (1.008\ \text{g/mol}) + 1 \times (16.00\ \text{g/mol}) = 18.02\ \text{g/mol}$

• Units: Formula mass in amu; molar mass in g/mol.

Example: 1 mole of $H_2O$ contains 2 moles of H and 1 mole of O.

Molar Mass of Compounds: Practice Problem

To calculate the molar mass of a compound, sum the molar masses of all atoms present in the formula.

• Practice: Calculate the molar mass of $(NH_4)_3PO_4$ to two decimal places.

Avogadro’s Number

Avogadro’s number ($N_A$) is the number of particles (atoms, molecules, or ions) in one mole of a substance. It is a fundamental constant in chemistry.

• Value: $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

• Applications:

  • 1 mole of $O_2$ contains $6.022 \times 10^{23}$ $O_2$ molecules.

  • 1 mole of $NH_3$ contains $6.022 \times 10^{23}$ $NH_3$ molecules.

  • 1 mole of Fe contains $6.022 \times 10^{23}$ Fe atoms.

  • 1 mole of $Na^+$ contains $6.022 \times 10^{23}$ $Na^+$ ions.

Using Mol-to-Mol Conversion Factors

Stoichiometric relationships in balanced chemical equations allow conversion between moles of different substances.

• Definition: A mole-to-mole factor (stoichiometric ratio) relates the moles of any two substances in a balanced equation.

• Example Equation: $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)$

Reactants/Products	Mole Ratio
Fe to O2	4 mol Fe : 3 mol O2
Fe to Fe2O3	4 mol Fe : 2 mol Fe2O3
O2 to Fe2O3	3 mol O2 : 2 mol Fe2O3

Practice: For $3H_2(g) + N_2(g) \rightarrow 2NH_3(g)$, write the mol-to-mol factors for $H_2$ and $N_2$, and for $NH_3$ and $H_2$.

Using Mol-to-Mass Conversion Factors

Conversion between moles and mass uses the molar mass as a conversion factor.

• Example for $H_2$:

  • $1\ \text{mol}\ H_2 \leftrightarrow 2.016\ \text{g}\ H_2$

  • Conversion factors: $\frac{1\ \text{mol}\ H_2}{2.016\ \text{g}\ H_2}$ and $\frac{2.016\ \text{g}\ H_2}{1\ \text{mol}\ H_2}$

Using Mol-to-Molecule/Atom/Ion Conversion Factors

Avogadro’s number allows conversion between moles and number of particles.

• For $H_2$:

  • $1\ \text{mol}\ H_2 \leftrightarrow 6.022 \times 10^{23}\ H_2$ molecules

  • Conversion factors: $\frac{1\ \text{mol}\ H_2}{6.022 \times 10^{23}\ H_2\ \text{molecules}}$ and $\frac{6.022 \times 10^{23}\ H_2\ \text{molecules}}{1\ \text{mol}\ H_2}$

• For Fe:

  • $1\ \text{mol}\ Fe \leftrightarrow 6.022 \times 10^{23}\ Fe$ atoms

  • Conversion factors: $\frac{1\ \text{mol}\ Fe}{6.022 \times 10^{23}\ Fe\ \text{atoms}}$ and $\frac{6.022 \times 10^{23}\ Fe\ \text{atoms}}{1\ \text{mol}\ Fe}$

Stoichiometry: Summary

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction, based on balanced equations.

• Example Reaction: $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)$

• Conversions:

  • Moles of Fe to moles of O2 and Fe2O3

  • Mass calculations using molar masses: $4 \times 55.845\ \text{g/mol} = 223.38\ \text{g Fe}$ $3 \times 32.00\ \text{g/mol} = 96.00\ \text{g O}_2$ $2 \times 159.69\ \text{g/mol} = 319.38\ \text{g Fe}_2O_3$

Converting Between Moles and Number of Atoms

To find the number of atoms in a given number of moles, multiply by Avogadro’s number.

• Example: Number of atoms in 10.0 moles of Mg: $10.0\ \text{mol}\ \times\ 6.022 \times 10^{23}\ \text{atoms/mol} = 6.022 \times 10^{24}\ \text{atoms}$

Converting Between Grams and Moles of an Element

To convert grams to moles, divide the mass by the molar mass of the element.

• Example: Number of moles of silicon in 43.67 g of Si: $\text{Moles} = \frac{43.67\ \text{g}}{28.09\ \text{g/mol}} = 1.56\ \text{mol}$ Additional info: Molar mass of Si is 28.09 g/mol.

Converting Between Grams and Number of Atoms

To find the number of atoms in a given mass, first convert grams to moles, then multiply by Avogadro’s number.

• Example: Number of lead atoms in 207 g of Pb: $\text{Moles} = \frac{207\ \text{g}}{207.2\ \text{g/mol}} = 1.00\ \text{mol}$ $\text{Atoms} = 1.00\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol} = 6.022 \times 10^{23}\ \text{atoms}$ Additional info: Molar mass of Pb is 207.2 g/mol.

Converting Between Grams and Moles of Compounds

To find the mass of a compound from moles, multiply the number of moles by the molar mass.

• Example: Mass of 0.328 moles of ammonia ($NH_3$): $0.328\ \text{mol} \times 17.03\ \text{g/mol} = 5.58\ \text{g}$ Additional info: Molar mass of NH3 is 17.03 g/mol.

Converting Between Mass of a Compound and Number of Molecules

To find the mass from the number of molecules, convert molecules to moles using Avogadro’s number, then multiply by molar mass.

• Example: Mass of $9.53 \times 10^{21}$ $CO_2$ molecules: $\text{Moles} = \frac{9.53 \times 10^{21}}{6.022 \times 10^{23}} = 0.0158\ \text{mol}$ $\text{Mass} = 0.0158\ \text{mol} \times 44.01\ \text{g/mol} = 0.695\ \text{g}$ Additional info: Molar mass of CO2 is 44.01 g/mol.

Converting Between Moles of a Compound and Moles of Constituent Element

To determine the moles of an element in a compound, multiply the moles of compound by the number of moles of the element per mole of compound.

• Example: Number of moles of Al in 9.00 moles of $Al_2O_3$: $9.00\ \text{mol}\ Al_2O_3 \times \frac{2\ \text{mol}\ Al}{1\ \text{mol}\ Al_2O_3} = 18.00\ \text{mol}\ Al$

Converting Between Mass of a Compound and Mass of Constituent Element

To find the mass of an element in a compound, use the mass percent or mole ratio.

• Example: Mass of C in 5.98 g of $C_3H_8$: $\text{Moles of } C_3H_8 = \frac{5.98\ \text{g}}{44.10\ \text{g/mol}} = 0.136\ \text{mol}$ $\text{Moles of C} = 0.136\ \text{mol} \times 3 = 0.408\ \text{mol}$ $\text{Mass of C} = 0.408\ \text{mol} \times 12.01\ \text{g/mol} = 4.90\ \text{g}$ Additional info: Molar mass of $C_3H_8$ is 44.10 g/mol; each molecule contains 3 C atoms.

Composition of Compounds

The percentage of each element in a compound can be calculated using the formula mass and the mass of each element in one mole of the compound.

• Formula: $\text{Percent of element X} = \frac{\text{Mass of element X in 1 mole of compound}}{\text{Mass of 1 mole of compound}} \times 100\%$

• Note: Due to rounding, the sum of percentages may not always equal exactly 100%.

Element	Mass in 1 mole	Molar Mass of Compound	Percent Composition
C	12.01 g	49.48 g	24.3%
H	2.02 g	49.48 g	4.1%
Cl	35.45 g	49.48 g	71.6%
Additional info: Example for CH2Cl2 (methylene chloride).

Determination of an Empirical Formula

The empirical formula is the simplest whole-number ratio of atoms in a compound. It can be determined from mass or percent composition data.

• Steps:

  1. If given percentages, assume 100 g of compound (percentages become grams).

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

  3. Write a formula using the mole values as subscripts.

  4. Divide all subscripts by the smallest number of moles.

  5. If subscripts are not whole numbers, multiply all by a factor to obtain whole numbers (e.g., 2 for 0.5, 3 for 0.33 or 0.67, 4 for 0.25 or 0.75).

• Example: A compound contains 24.5 g N and 70.0 g O. Find the empirical formula. Additional info: Convert grams to moles, find ratio, and write formula.

Determination of a Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula.

• Formula: $\text{Molecular Formula Mass} = n \times (\text{Empirical Formula Mass})$ where $n$ is a positive integer.

• Steps:

  1. Calculate the empirical formula mass.

  2. Divide the compound’s molar mass by the empirical formula mass to find $n$.

  3. Multiply the subscripts in the empirical formula by $n$ to get the molecular formula.

• Example: A compound contains 81.68% C and 18.32% H. Find the empirical formula, then use the molar mass to determine the molecular formula. Additional info: Assume 100 g sample, convert to moles, find ratio, and apply steps above.