Study Notes for CHEM 1411: Stoichiometry of Formulas and Equations

The mass in atomic mass units (a mu) of one atom of an element or one molecule or formula unit of a compound is numerically equal to the mass in grams of 1 mole of the substance.
The mole (mol) is defined as the amount of substance containing the same number of entities as there are atoms in exactly 12 g of carbon-12. The term "entities" encompasses atoms, ions, molecules, formula units, or electrons.
One mole (1 mol) consists of approximately $6.022 \times 10^{23}$ entities, commonly referred to as Avogadro's number.

(Refer also to provided materials for visual density representation)

The molar mass (M) of a substance is the mass of one mole of its entities (atoms, molecules, or formula units).
For monatomic elements, the molar mass is the same as the atomic mass expressed in grams per mole. For example, the molar mass of Ne = 20.18 g/mol.
The molar mass of compounds can be found by summing the molar masses of the constituent atoms in the formula:
- Molar Mass of SO₂ = (1 × M of S) + (2 × M of O)
- Molar Mass of O₂ = 2 × M of O = $2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}$

The following table outlines the chemical formula of glucose (C₆H₁₂O₆):

Chemical Formula	Atoms/molecule	Moles of atoms/mole	Mass/molecule	Mass/mole
C₆H₁₂O₆	6 (C), 12 (H), 6 (O)	6 mol (C), 12 mol (H), 6 mol (O)	$6(12.01) = 72.06 ext{ amu}$ , $12(1.008) = 12.10 ext{ amu}$ , $6(16.00) = 96.00 ext{ amu}$	72.06 g, 12.10 g, 96.00 g

Interconverting Moles, Mass, and Number of Chemical Entities

The following equations express how to interconvert between moles, mass, and number of entities:
- $\text{mass (g)} = \text{amount (mol)} \times M (g/mol)$
- $\text{amount (mol)} = \frac{\text{mass (g)}}{M (g/mol)}$
- $\text{no. of entities} = \text{amount (mol)} \times 6.022 \times 10^{23}$

Problem: How many grams of Ag are in 0.0342 mol of Ag?
- Plan: Multiply the number of moles by the molar mass of Ag.
- Solution: $0.0342 ext{ mol Ag} \times 107.9 \text{ g/mol} = 3.69 ext{ g Ag}$

Problem: How many Ga atoms are in mol of gallium?
- Plan: Multiply amount (mol) by Avogadro’s number.
- Solution: $2.85 \times 10^{-3} \text{ mol Ga} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.72 \times 10^{21} \text{ Ga atoms}$

Problem: How many Fe atoms are in 95.8 g of Fe?
- Plan: Convert mass to moles, then multiply by Avogadro’s number.
- Solution:
  1. $\frac{95.8 ext{ g Fe}}{55.85 ext{ g/mol}} = 1.72 ext{ mol Fe}$
  2. $1.72 ext{ mol Fe} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.04 \times 10^{24} \text{ atoms Fe}$

Molecular vs. Empirical Formulas: The empirical formula indicates the simplest whole number ratio, while the molecular formula shows the actual number of atoms. For example: - Empirical formula for H₂O is HO, Molecular formula is H₂O.
Limiting Reactants: Determine which reactant will run out first, thus limiting the amount of product formed. Analyze stoichiometric relationships to assess reactant consumption.
Reaction Yields: Calculate theoretical and actual yields. Use the following formula for percent yield: $$\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
Understanding stoichiometric calculations, as they form the basis for determining quantitative relationships in chemical reactions is essential.

Molar Mass (M) = mass of substance / amount (in moles)
Mass % of element = (mass of element in formula / total mass of compound) × 100
Empirical formula calculation involves converting grams to moles and finding the simplest ratio.