# Chemical Calculations | General Chemistry 2

## Mass-Mole-Number Relationship

**Mole number n (in mol):**

The number of moles in a sample is given by

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

m = mass of the substance (in g)

M = molar mass of the substance (in g.mol^{-1})

$\mathrm{n}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$

N = number of particles in the substance

N_{A} = Avogadro’s number = 6.022 x 10^{23} mol^{-1}

Mass to moles:Calculate the number of mole in a 2.0 g sample of N

_{2}:

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 particlesCalculate the number of nitrogen atom N

_{N}in a 2.0 g sample of N_{2}:

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 10^{22}molecules⇒ N

_{N}= 2 x N_{N2}= 8.6 x 10^{22}atoms.

## Stoichiometry

**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 H

_{2}+ N_{2}→ 2 NH_{3}

- Molecular interpretation: 3 molecules of H
_{2}react with 1 molecule of N_{2}to form 2 molecules of NH_{3}.- Molar interpretation: 3 moles of H
_{2}react with 1 mole of N_{2}to form 2 moles of NH_{3}.

**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:**

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

## Limiting Reactant

**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

- Consider one of the starting reagents as the limiting reactant (for example A).
- 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: $\frac{{\mathrm{n}}_{\mathrm{A}}}{\mathrm{a}}$ = $\frac{{\mathrm{n}}_{\mathrm{B}}}{\mathrm{b}}$ - 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.

## Reaction Yield

**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 N_{2}reacts with excess H_{2}:

- Balanced equation: N
_{2}+ 3 H_{2}→ 2 NH_{3}- Molar mass of N
_{2} = 28.02 g.mol^{-1}.- Moles of N
_{2}= $\frac{28\mathrm{g}}{28.02\mathrm{g}.{\mathrm{mol}}^{-1}}$ = 1 mol.- According to the balanced equation, the mole ratio N
_{2}to NH_{3} is 1:2 ⇒ Moles of NH_{3}= 2 mol.- Molar mass of NH
_{3}= 17.03 g.mol^{-1}.- Theoretical yield of NH
_{3}= 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.

% yield = $\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}$ x 100

## Percent Composition by Mass

**Percent composition by mass:**

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

%X = n x $\frac{{\mathrm{M}}_{\mathrm{X}}}{{\mathrm{M}}_{\mathrm{compound}}}$ x 100

%X = percent composition of X

n = number of atoms X in a molecule of the compound

M_{x} = atomic mass of X (in amu or g.mol^{-1})

M_{compound} = molecular mass of the compound (in amu or g.mol^{-1})

**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 H_{2}O:%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 Al_{2}(SO_{4})_{3}:

- Find the molar mass of Al
_{2}(SO_{4})_{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 Al

_{2}(SO_{4})_{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 vs. Molality

**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.

*M* = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{V}}_{\mathrm{solution}}}$

*M* = molarity (in mol.L^{-1})

n_{solute} = moles of solute (in mol)

V_{solution} = volume of solution (in L)

**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.

*m* = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{m}}_{\mathrm{solution}}}$

*m* = molality (in mol.kg^{-1})

n_{solute} = moles of solute (in mol)

m_{solution} = masse of solution (in kg)

**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.

## Solution Dilution

**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:

n = *M* x V

n = number of moles (in mol)

*M* = molarity (in mol.L^{-1})

V = volume (in L)

**Dilution formula:**

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

C_{1}V_{1} = C_{2}V_{2}

C_{1} = initial concentration (stock solution)

V_{1} = initial volume (stock solution)

C_{2} = final concentration (diluted solution)

V_{2} = final volume (diluted solution)

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

- C
_{1}= 2.0 M; C_{2}= 0.5 M; V_{2}= 500 mL- C
_{1}V_{1}= C_{2}V_{2}⇒ (2.0 M) V_{1}= 0.5 M x 500mL- V
_{1}= $\frac{0.5\mathrm{M}\mathrm{x}500\mathrm{mL}}{2.0\mathrm{M}}$ = 125 mL

**Steps for dilution:**

- Determine the final concentration and volume needed.
- Calculate the volume of stock solution needed: use the dilution formula to calculate V
_{1}. - Add enough solvent to the stock solution to reach the final volume V
_{2}.

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 C
_{2}= C_{1}/ 10 ⇒ C_{1}= 10 C_{2}- Using the dilution formula: C
_{1}V_{1}= C_{2}V_{2}⇒ 10 C_{2}V_{1}= C_{2}V_{2}⇒ 10 V_{1}= V_{2}- To dilute a solution 10 times, add 10 times the volume of solvent to the original solution.

## Acid-Base Titrations

**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:

*M*_{a}V_{a} = *M*_{b}V_{b}

*M*_{a} = molarity of the acid (in mol.L^{-1})

V_{a} = volume of the acid (in L)

*M*_{b} = molarity of the base (in mol.L^{-1})

V_{b} = volume of the base (in L)

**How to determine the concentration of an acid solution:**

- Add a strong base solution with known concentration to the acid solution
- Stop adding the base exactly when all the acid has been neutralized; the indicator should change color.
- Determine the volume of base added
- Determine the concentration of the acid solution using the titration formula:
*M*V_{a}_{a}=*M*V_{b}_{b}.

### Check your knowledge about this Chapter

The relationship between the mass of a substance and the amount in moles is quantified by the molar mass (in g/mol), which is the mass in grams of one mole of the substance. To determine the number of moles from mass, you divide the mass of the substance by its molar mass, and to convert moles to mass, you multiply the number of moles by the molar mass.

To convert between the number of particles (atoms, molecules) and moles, you use Avogadro's number, which is 6.022 x 10^{23} particles per mole. If you know the number of particles in a sample, you can divide by Avogadro's number to find the number of moles. Conversely, to find the number of particles from the number of moles, you multiply the moles by Avogadro's number.

The coefficients in chemical equations directly represent the stoichiometric ratios of reactants and products in a reaction. They indicate the proportionate amounts of substances involved when using the mole unit.

In the balanced chemical equation: 2 H

_{2}+ O_{2}→ 2 H_{2}O, the coefficients suggest that:2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. This means that for every mole of oxygen consumed, twice as many moles of hydrogen are needed, and two moles of water are formed.

To perform a stoichiometric calculation in a chemical reaction, follow these steps:

- First, write down the balanced chemical equation for the reaction.
- Calculate the molar masses of the reactants and products.
- Convert the masses of the substances involved in the reaction to moles using their molar masses.
- Using the coefficients from the balanced equation, relate the moles of one substance to the moles of another substance.
- Identify the limiting reactant to determine the amount of products formed.
- Calculate the theoretical yield based on the moles of the limiting reactant and the mole ratio of the desired product to the limiting reactant.
- If given experimental data, calculate the percent yield by comparing the actual yield to the theoretical yield.

Stoichiometry is used to calculate the amount of product formed from the reactants by following these steps:

- Write the balanced chemical equation for the reaction.
- Convert the quantities of known substances (reactants or products) into moles using their molar masses.
- Use the mole ratio from the balanced equation to determine the number of moles of product that can form from the moles of reactant available.
- Convert the moles of product to grams (or another unit of measure) using the molar mass of the product.

This process relies on the law of conservation of mass and the principle that mole ratios in a balanced equation imply proportional relationships between reactants and products.

A limiting reactant is the reactant in a chemical reaction that is completely consumed first, thus limiting the extent of the reaction and determining the maximum amount of product that can be formed. The reaction stops when the limiting reactant is consumed, regardless of the amounts of other reactants present.

By comparing the mole ratio of the reactants used to the mole ratio of the reactants required by the balanced chemical equation, you can determine which reactant is the limiting one. Once identified, it is then used to calculate the theoretical yield of the product(s), knowing that no further product can form once the limiting reactant is expended.

To identify the limiting reactant in a chemical reaction, compare the molar ratios of the reactants used with those required by the balanced chemical equation:

- First, calculate the moles of each reactant present.
- Then, using the stoichiometry of the balanced equation, determine the amount of product(s) that each reactant could theoretically produce.
- The reactant that produces the least amount of product is the limiting reactant, as it will be consumed first, stopping the reaction, and therefore determining the maximum amount of product that can be formed.

Percent yield is calculated by dividing the actual yield (the experimentally obtained amount of product) by the theoretical yield (the maximum possible amount predicted by stoichiometry) and multiplying by 100.

#### Can you explain the concept of theoretical yield and its significance in percent yield calculations?

Theoretical yield is the maximum amount of product that could be formed in a reaction under ideal conditions. Percent yield relates the actual yield to the theoretical yield, providing insight into the efficiency of the reaction.

Factors that contribute to a lower than expected percent yield include incomplete reactions, side reactions, loss during transfer, and contaminants in the product.

Percent composition by mass refers to the percentage by mass of each element within a compound. To calculate it, one needs to divide the mass of each element in a molecule by the total molecular mass and then multiply by 100%. This calculation gives the contribution of each element to the overall mass of the molecule and is often used to determine the empirical formula or assess the purity of a compound.

The mass percent of an element in a compound, for example A in AB, is given by:

%A = n x $\frac{\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{A}}{\mathrm{molecular}(\mathrm{or}\mathrm{formula})\mathrm{mass}\mathrm{of}\mathrm{AB}}$ x 100%

n = number of atoms A in a molecule (or formula unit) of compound AB

Molarity is the concentration of a solute in a solution expressed as moles of solute per liter of solution, whereas molality is the concentration expressed as moles of solute per kilogram of solvent. Molarity is more suitable for reactions occurring in a solution volume, while molality is preferred in situations where temperature changes may affect the solution volume.

Molarity is calculated by dividing the moles of solute by the volume of the solution in liters, while molality is calculated by dividing the moles of solute by the mass of the solvent in kilograms. Molarity is in moles per liter (mol/L), and molality is in moles per kilogram (mol/kg).

The relationship between moles and the volume of a solution can be expressed using the formula for molarity:

n = *M* x V

n = number of moles (in mol)

*M* = molarity (in mol.L^{-1})

V = volume (in L)

The formula for dilution is M_{1} V_{1} = M_{2} V_{2}, where and M_{1} and V_{1} are the initial concentration and volume, and M_{2} and V_{2} are the final concentration and volume. This formula ensures the conservation of moles of solute before and after dilution. It is commonly used to calculate the volume or concentration needed when diluting a solution to achieve a desired concentration.

To prepare a dilute solution from a concentrated stock solution, follow these steps:

- Begin by calculating the volume of stock solution needed using the dilution equation
*C*, where_{1}V_{1}= C_{2}V_{2}*C*and_{1}*C*are the concentrations of the stock and dilute solutions, respectively, and_{2}*V*and_{1}*V*are the volumes of the stock and final solutions._{2} - Measure the calculated volume of the concentrated stock solution using appropriate laboratory equipment such as a pipette for accuracy.
- Transfer the stock solution into a volumetric flask or a graduated cylinder.
- Add solvent—commonly water for aqueous solutions—to the flask until it reaches the desired final volume, thereby achieving the required final concentration.
- Thoroughly mix the solution by inverting the flask several times or stirring with a glass rod to ensure homogeneity of the dilute solution.

An acid-base titration is performed to determine the concentration of an unknown solution by reacting it with a known solution of opposite acidity. The titration involves incremental addition of the titrant (known solution) until the reaction reaches completion, as indicated by a noticeable change in the pH of the solution.

An equivalence point in a titration is the moment at which the amount of titrant added is stoichiometrically equivalent to the amount of substance present in the sample. This means the reactants have reacted in their exact proportions according to the balanced chemical equation, with no excess of either reactant.

It is typically determined by using an indicator that changes color at a specific pH or by using a pH meter to detect a sudden change in the pH of the solution, which corresponds to the completion of the reaction.

The role of indicators in acid-base titrations is to provide a visual signal, usually a color change, to identify the endpoint of the titration. Indicators are chosen based on their pH transition range, which should coincide with the pH at which the reaction between the acid and base is stoichiometrically equal. The correct choice of an indicator ensures that the color change will happen at the equivalence point, thus allowing the experimenter to accurately determine when the titration is complete.

To calculate the concentration of an unknown solution in an acid-base titration:

- Write the balanced equation.
- Determine the mole ratio between the acid and base.
- Record the volume and concentration of the titrant (the solution of known concentration).
- Identify the equivalence point volume.
- Calculate moles of titrant.
- Use the mole ratio to find moles of the unknown.
- Divide the moles of the unknown by the volume of the solution to get the concentration.