Chemical Calculations | General Chemistry 2

 Mole number n (in mol): 

The number of moles in a sample is given by
 

• Mass to moles:

Calculate the number of mole in a 2.0 g sample of N₂:
 

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_2}}{{\mathrm{M}}_{{\mathrm{N}}_2}}$

m_N2 = 2.0 g
M_N2 = 2 x M_N = 2 x 14.0 = 28.0 g.mol^-1 

⇒ n = $\frac{2.0 \mathrm{g}}{28.0 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 7.1 x 10^-2 mol

• Moles to number of particles

Calculate the number of nitrogen atom N_N in a 2.0 g sample of N₂:
 

n = $\frac{{\mathrm{m}}_{{\mathrm{N}}_2}}{{\mathrm{M}}_{{\mathrm{N}}_2}}$ = $\frac{{\mathrm{N}}_{{\mathrm{N}}_2}}{{\mathrm{N}}_{\mathrm{A}}}$

m_N2 = 2.0 g
M_N2 = 28.0 g.mol^-1
N_N2 = 2 x N_N
N_A = 6.022 x 10^-23 mol^-1

⇒ N_N2 = N_A x $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$ = 6.022 x 10^-23 mol^-1 x $\frac{2.0 \mathrm{g}}{28.0 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 4.3 x 10²² molecules

⇒ N_N = 2 x N_N2 = 8.6 x 10²² atoms.

Stoichiometry:

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Stoichiometry is crucial for predicting the outcomes of reactions, optimizing reactant use, and determining reaction yields and limiting reactants. Balanced chemical equations serve as the foundation of stoichiometric calculations, ensuring that mass and atoms are conserved in chemical reactions.
 

Stoichiometric coefficients and mole ratios:

Stoichiometric coefficients are the numeric values written to the left of each species in a chemical equation to balance the equation. These coefficients can be interpreted as the number of molecules or the number of moles of a substance produced or consumed during the reaction. Mole ratios are derived from the coefficients of a balanced chemical equation.

3 H₂ + N₂ → 2 NH₃

• Molecular interpretation: 3 molecules of H₂ react with 1 molecule of N₂ to form 2 molecules of NH₃.
• Molar interpretation: 3 moles of H₂ react with 1 mole of N₂ to form 2 moles of NH₃.

Types of stoichiometry problems:

• How much of a product can be formed starting from a certain amount of reactants.
• How much of one reactant is necessary to react with a given amount of another.
• How much reactant is required to produce a desired amount of product.

How to solve stoichiometry problems:

1.  Balancing the chemical equation.
2. Determine the molar masses of reactants and products of interest.
3. Convert all given masses to moles using the molar mass of each substance.
4. Use the balanced equation to determine the stoichiometric ratios between reactants and products.
5. Use the stoichiometric ratios to calculate the number of moles of the desired products or reactants.
6. Convert the calculated moles of desired materials back to masses if necessary.

Limiting reactant:

The limiting reactant of a reaction is the reactant that is consumed completely in a chemical reaction. It determines the maximum amount of product that can be formed during a reaction.

Identifying the limiting reactant is crucial for predicting the maximum yield of a product and for optimizing the use of reactants in chemical reactions. Reactants are said to be in stoichiometric amounts when they are combined in the same relative amounts as those represented in the balanced chemical equation.
 

Steps to identify the limiting reactant:

• Start with a balanced chemical equation for the reaction.
• Convert the mass of each reactant to moles using their molar masses.
• Use the mole ratios from the balanced equation to compare the amount of reactants:

a A + b B → c C

1. Consider one of the starting reagents as the limiting reactant (for example A).
2. Calculate the mole number of the other reagent, B, required for a complete reaction of A.
   Be sure to use the stoichiometric ratios of the balanced equation: $\frac{{\mathrm{n}}_{\mathrm{A}}}{\mathrm{a}}$ = $\frac{{\mathrm{n}}_{\mathrm{B}}}{\mathrm{b}}$
3. Compare the amount of B needed for a complete reaction with the actual amount of B:
   If the amount of B needed > actual amount of B, B is the limiting reactant.
   If the actual amount of B > amount of B needed, A is the limiting reactant.

Theoretical vs. actual yield:

• Theoritical yield: The maximum amount of product that can be produced from a given amount of reactants, based on stoichiometric calculations.
• Actual yield: The amount of product actually obtained from a reaction.

How to calculate theoretical yield:

• Use the balanced equation to determine the mole ratios between reactants and products.
• Convert the mass of reactants to moles.
• Identify the limiting reactant.
• Use the mole ratio to calculate the moles of product formed.
• Convert moles of product to mass using the molar mass.

Calculate the theoretical yield of NH_3​ if 28 g of N₂ reacts with excess H₂​:

• Balanced equation: N₂ + 3 H₂ → 2 NH₃
• Molar mass of N₂​ = 28.02 g.mol^-1.
• Moles of N₂ = $\frac{28 \mathrm{g}}{28.02 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 1 mol.
• According to the balanced equation, the mole ratio N₂ to NH₃​ is 1:2 ⇒ Moles of NH₃ = 2 mol.
• Molar mass of NH₃ = 17.03 g.mol^-1.
• Theoretical yield of NH₃ = 2 mol x 17.03 g.mol^-1 = 34.06 g.

Percent yield:

Percent yield is a measure of the efficiency of a reaction, calculated as the ratio of the actual yield to the theoretical yield, expressed as a percentage. Percent yield is a critical parameter for evaluating the success of a reaction and for comparing the efficiency of different synthetic methods.
 

Percent composition by mass:

Percent composition by mass is the percentage by mass of each element in a compound. It is calculated as follows:
 

Calculation of Percent Composition:

• Determine the molar mass of the compound by summing the molar masses of all the atoms in the formula.
• Determine the mass contribution of each element by multiplying the number of atoms of the element by its atomic mass.
• Calculate the Percent Composition.

Percent composition of H₂O:

%H = 2 x $\frac{1.008}{18.016}$ x 100 = 11.19%

%O = $\frac{16.00}{18.016}$ x 100 = 88.81%

M_H = 1.008 amu
M_O = 16.00 amu
M_H2O = 18.016 amu

Calculate the percent composition of aluminium in Al₂(SO₄)₃:

• Find the molar mass of Al₂(SO₄)₃: M_Al2(SO4)3 = 2 M_Al + 3 M_S + 12 M_O = 342.14 g.mol^-1
• Calculate the percent composition:

%Al = n x $\frac{{\mathrm{M}}_{\mathrm{Al}}}{{\mathrm{M}}_{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}}$ x 100%

n = 2 [2 atoms of Al in one molecule of Al₂(SO₄)₃]
M_Al = 26.98 g.mol^-1
M_Al2(SO4)3 = 342.14 g.mol^-1

%Al = 2 x $\frac{26.98 \mathrm{g}.{\mathrm{mol}}^{-1}}{342.14 \mathrm{g}.{\mathrm{mol}}^{-1}}$ x 100 = 15.77 %

Molarity M:

Molarity is the number of moles of solute per liter of solution. It is commonly used in laboratories for reactions involving solutions, where volume measurements are convenient.
 

Molality m:

Molality is the number of moles of solute per kilogram of solvent. It is used in colligative properties calculations (such as boiling point elevation and freezing point depression) because it is temperature-independent.
 

Key differences:

• Molarity depends on the volume of the solution, which can change with temperature and pressure.
• Molality depends on the mass of the solvent, which is not affected by temperature and pressure.

Dilution:

Dilution is the process of preparing a less concentrated solution from a more concentrated one. This is achieved by adding more solvent to the solution, decreasing the concentration (molarity) of the solute without changing the total number of moles of solute present: moles of solute before dilution = moles of solute after dilution.
 

Mole-volume relationship:

The relationship between the number of moles, molarity, and volume is given by:
 

Dilution formula:

The relationship between the concentrations and volumes of the stock and diluted solutions is given by:
 

How much of a 2.0 M NaCl stock solution is needed to prepare 500 mL of a 0.5 M NaCl solution?

• C₁ = 2.0 M; C₂ = 0.5 M; V₂ = 500 mL
• C₁V₁ = C₂V₂ ⇒ (2.0 M) V₁ = 0.5 M x 500mL
• V₁ = $\frac{0.5 \mathrm{M} \mathrm{x} 500 \mathrm{mL}}{2.0 \mathrm{M}}$ = 125 mL

Steps for dilution:

1. Determine the final concentration and volume needed.
2. Calculate the volume of stock solution needed: use the dilution formula to calculate V₁.
3. Add enough solvent to the stock solution to reach the final volume V₂.

Volume of solvent required to dilute a solution 10 times:

• Principle: The total number of moles of solute present in the solution does not change during dilution.
• Diluting a solution 10 times means C₂ = C₁ / 10 ⇒ C₁ = 10 C₂
• Using the dilution formula: C₁V₁ = C₂V₂ ⇒ 10 C₂V₁ = C₂V₂ ⇒ 10 V₁ = V₂
• To dilute a solution 10 times, add 10 times the volume of solvent to the original solution.

Titration:

A laboratory technique used to determine the unknown concentration of an acid or base using a neutralization reaction. A neutralization reaction is a reaction between an acid and a base resulting in the formation of water and a salt.
 

Key components of a titration:

• Titrant: The solution of known concentration added to the analyte.
• Analyte: The solution of unknown concentration being analyzed.
• Indicator: A chemical that changes color at the endpoint of the titration.
• Endpoint: The point at which the indicator changes color, signifying that the reaction is complete and the moles of acid equal the moles of base.

Titration principle:

• A solution of known concentration (the titrant) is gradually added to a solution of unknown concentration (the analyte) to determine the unknown concentration.
• Indicators, substances that change color with pH, are used to identify the endpoint of the titration.
• At the endpoint of the titration, the number of moles of acid is equal to the number of moles of base.

The relationship between the moles of acid and base at the equivalence point is given by:
 

How to determine the concentration of an acid solution:

1. Add a strong base solution with known concentration to the acid solution
2. Stop adding the base exactly when all the acid has been neutralized; the indicator should change color.
3. Determine the volume of base added
4. Determine the concentration of the acid solution using the titration formula: M_aV_a = M_bV_b.