Stoichiometry: Mole, Gram Conversions

Stoichiometry Introduction

• Stoichiometry involves three main types of conversions for chemical reactions.
• These conversions relate to the amounts of reactants and products in a balanced equation.

Mole-to-Mole Conversion

• This is the simplest type of stoichiometric conversion.
• It involves converting moles of substance A to moles of substance B.
• Identification of the mole ratio is crucial for this conversion.
• The mole ratio is derived from the coefficients in the balanced chemical equation.

Mole-to-Gram and Gram-to-Mole Conversion

• Involves converting moles of substance A to grams of substance B.
• The reverse is also possible: converting grams of substance A to moles of substance B.
• This type of conversion generally requires two steps.

Gram-to-Gram Conversion

• This is the most complex of the basic stoichiometric conversions.
• Grams of substance A are converted to grams of substance B.
• This type of conversion usually involves three steps.

Example 1: Sulfur Dioxide and Oxygen Gas

• Sulfur dioxide ($SO_2$) reacts with oxygen gas ($O_2$) to form sulfur trioxide ($SO_3$).
• Balanced chemical equation: $2SO_2 + O_2 \rightarrow 2SO_3$

Example 1A: Moles of Sulfur Dioxide to Moles of Sulfur Trioxide

• Problem: If 3.4 moles of $SO_2$ reacts with excess $O_2$, find moles of $SO_3$ formed.
• Mole ratio between $SO_2$ and $SO_3$ is 2:2 (or 1:1).
• Calculation: 3.4 moles $SO_2$ will produce 3.4 moles of $SO_3$.

Example 1B: Moles of Sulfur Dioxide to Moles of Oxygen Gas

• Problem: How many moles of $O_2$ react completely with 4.7 moles of $SO_2$?
• Mole ratio between $SO_2$ and $O_2$ is 2:1.
• Calculation: 4.7 moles $SO_2 * (1 \text{ mol } O_2 / 2 \text{ mol } SO_2) = 2.35 \text{ moles } O_2$

Example 2: Propane and Oxygen Gas

• Propane ($C_3H_8$) reacts with oxygen gas ($O_2$) to form carbon dioxide ($CO_2$) and water ($H_2O$).
• Balanced chemical equation: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Example 2A: Moles of Propane to Grams of Carbon Dioxide

• Problem: If 2.8 moles of $C_3H_8$ reacts with excess $O_2$, find grams of $CO_2$ formed.
• Steps: Convert moles of $C_3H_8$ to moles of $CO_2$ using the mole ratio, then to grams of $CO_2$ using its molar mass.
• Molar mass of $CO_2$: 44.01 g/mol.
• Mole ratio between $C_3H_8$ and $CO_2$ is 1:3.
• Calculation: $2.8 \text{ mol } C_3H_8 * (3 \text{ mol } CO_2 / 1 \text{ mol } C_3H_8) * (44.01 \text{ g } CO_2 / 1 \text{ mol } CO_2) = 369.68 \text{ g } CO_2 $

Example 2B: Moles of Propane to Grams of Oxygen Gas

• Problem: How many grams of $O_2$ react completely with 3.8 moles of $C_3H_8$?
• Steps: Convert moles of $C_3H_8$ to moles of $O_2$ using the mole ratio, then to grams of $O_2$ using its molar mass.
• Molar mass of $O_2$: 32 g/mol.
• Mole ratio between $C_3H_8$ and $O_2$ is 1:5.
• Calculation: $3.8 \text{ mol } C_3H_8 * (5 \text{ mol } O_2 / 1 \text{ mol } C_3H_8) * (32 \text{ g } O_2 / 1 \text{ mol } O_2) = 608 \text{ g } O_2$

Example 2C: Grams of Propane to Moles of Water

• Problem: If 25 g of $C_3H_8$ reacts with excess oxygen, how many moles of $H_2O$ will form?
• Steps: Convert grams of $C_3H_8$ to moles of $C_3H_8$ using its molar mass, then to moles of $H_2O$ using the mole ratio.
• Molar mass of $C_3H_8$: 44.094 g/mol.
• Mole ratio between $C_3H_8$ and $H_2O$ is 1:4.
• Calculation: $25 \text{ g } C_3H_8 * (1 \text{ mol } C_3H_8 / 44.094 \text{ g } C_3H_8) * (4 \text{ mol } H_2O / 1 \text{ mol } C_3H_8) = 2.27 \text{ mol } H_2O$

Example 2D: Grams of Water to Moles of Carbon Dioxide

• Problem: If 38 g of $H_2O$ are produced, how many moles of $CO_2$ were produced?
• Steps: Convert grams of $H_2O$ to moles of $H_2O$ using its molar mass, then to moles of $CO_2$ using the mole ratio.
• Molar mass of $H_2O$: 18.016 g/mol.
• Mole ratio between $H_2O$ and $CO_2$ is 4:3.
• Calculation: $38 \text{ g } H_2O * (1 \text{ mol } H_2O / 18.016 \text{ g } H_2O) * (3 \text{ mol } CO_2 / 4 \text{ mol } H_2O) = 1.58 \text{ mol } CO_2$

Example 3: Aluminum and Chlorine Gas

• Aluminum (Al) reacts with chlorine gas ($Cl_2$) to form aluminum chloride ($AlCl_3$).
• Balanced chemical equation: $2Al + 3Cl_2 \rightarrow 2AlCl_3$
• Aluminum has a +3 charge, while chloride has a -1 charge.

Example 3A: Grams of Aluminum to Grams of Aluminum Chloride

• Problem: If 35 g of Al reacts with excess chlorine, how many grams of $AlCl_3$ will form?
• Steps: Convert grams of Al to moles of Al using its molar mass, then to moles of $AlCl_3$ using the mole ratio, and finally to grams of $AlCl_3$ using its molar mass.
• Molar mass of Al: 26.98 g/mol.
• Molar mass of $AlCl_3$: 133.33 g/mol.
• Mole ratio between Al and $AlCl_3$ is 2:2 (or 1:1).
• Calculation: $35 \text{ g } Al * (1 \text{ mol } Al / 26.98 \text{ g } Al) * (2 \text{ mol } AlCl_3 / 2 \text{ mol } Al) * (133.33 \text{ g } AlCl_3 / 1 \text{ mol } AlCl_3) = 172.96 \text{ g } AlCl_3$

Example 3B: Grams of Aluminum to Grams of Chlorine Gas

• Problem: How many grams of $Cl_2$ will react completely with 42.8 g of Al?
• Steps: Convert grams of Al to moles of Al using its molar mass, then to moles of $Cl_2$ using the mole ratio, and finally to grams of $Cl_2$ using its molar mass.
• Molar mass of Al: 26.98 g/mol.
• Molar mass of $Cl_2$: 70.9 g/mol.
• Mole ratio between Al and $Cl_2$ is 2:3.
• Calculation: $42.8 \text{ g } Al * (1 \text{ mol } Al / 26.98 \text{ g } Al) * (3 \text{ mol } Cl_2 / 2 \text{ mol } Al) * (70.9 \text{ g } Cl_2 / 1 \text{ mol } Cl_2) = 168.75 \text{ g } Cl_2$