Chemical Quantities in Reactions: Law of Conservation of Mass, Mole Relationships, and Mass Calculations

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Chemical Quantities in Reactions

Law of Conservation of Mass

The Law of Conservation of Mass states that in an ordinary chemical reaction, matter cannot be created or destroyed. This principle ensures that the total mass of reactants equals the total mass of products in a chemical reaction.

No change in total mass occurs during a reaction.
The mass of the products is equal to the mass of the reactants.
Example: The reaction of silver (Ag) with sulfur (S) to form silver sulfide (Ag2S) demonstrates this law.

Mass of reactants and product in Ag and S reaction

In the reaction: 2Ag(s) + S(s) → Ag2S(s), the mass of the reactants (Ag and S) is the same as the mass of the product (Ag2S).

Equation	Reactants	Products
Atoms/Formula Units	2 Ag atoms + 1 S atom	1 Ag2S formula unit
Avogadro's Number of Atoms	2 × 6.022 × 1023 Ag atoms + 1 × 6.022 × 1023 S atoms	1 × 6.022 × 1023 Ag2S formula units
Moles	2 mol Ag + 1 mol S	1 mol Ag2S
Mass (g)	2 × 107.9 g Ag + 1 × 32.1 g S	1 × 247.9 g Ag2S
Total Mass (g)	247.9 g	247.9 g

Table showing conservation of mass in Ag and S reaction

Example: Combustion of methane (CH4) with oxygen produces carbon dioxide and water. The total mass of reactants equals the total mass of products.

Equation	Reactants	Products
Moles	1 mol CH4 + 2 mol O2	1 mol CO2 + 2 mol H2O
Mass (g)	16.04 g CH4 + 64.00 g O2	44.01 g CO2 + 36.03 g H2O
Total Mass (g)	80.04 g	80.04 g

Table showing conservation of mass in methane combustion

Mole Relationships in Chemical Equations

Balanced chemical equations provide quantitative relationships between reactants and products. The coefficients in a balanced equation represent the number of moles of each substance involved.

Mole–Mole Factor: A ratio of the moles of two substances in a balanced equation, used to convert between moles of different substances.
Example: In the reaction 2Fe(s) + 3S(s) → Fe2S3(s), the mole–mole factor between Fe and S is 2:3.

Iron and sulfur react to form iron(III) sulfide

Sample Problem: How many moles of sulfur are needed to react with 1.42 moles of iron?

Given	Need	Connect
1.42 mol Fe	moles of S	mole–mole factor

Problem setup for Fe and S reaction

Plan: Convert moles of Fe to moles of S using the mole–mole factor.

Conversion plan for Fe to S

Mole–Mole Factors:

Mole-mole factors for Fe and S

Calculation:

Calculation for moles of S

1.42 mol Fe × (3 mol S / 2 mol Fe) = 2.13 mol S

Calculating Moles of a Product

Propane (C3H8) reacts with oxygen to produce carbon dioxide and water. The mole–mole factor from the balanced equation allows calculation of product moles.

Given	Need	Connect
2.25 mol C3H8	moles of CO2	mole–mole factor

Problem setup for propane combustion

Plan: Convert moles of C3H8 to moles of CO2 using the mole–mole factor.

Conversion plan for C3H8 to CO2

Mole–Mole Factors:

Mole-mole factors for C3H8 and CO2

Calculation:

Calculation for moles of CO2

2.25 mol C3H8 × (3 mol CO2 / 1 mol C3H8) = 6.75 mol CO2

Mass Calculations for Chemical Reactions

Given the mass of a substance in a chemical reaction, the mass of another substance can be calculated using molar masses and mole–mole factors.

Convert mass of substance A to moles using its molar mass.
Use the mole–mole factor from the balanced equation to convert moles of A to moles of B.
Convert moles of B to mass using its molar mass.

Flow chart for mass-to-mass calculations

Sample Problem: How many grams of CO2 are produced when 54.6 grams of C2H2 is burned?

Given	Need	Connect
54.6 g C2H2	grams of CO2	molar masses, mole–mole factor

Problem setup for acetylene combustion

Plan: Convert grams of C2H2 to grams of CO2 using molar masses and mole–mole factor.

Conversion plan for C2H2 to CO2

Mole–Mole and Molar Mass Factors:

Mole-mole and molar mass factors for C2H2 and CO2

Calculation:

Calculation for grams of CO2

54.6 g C2H2 × (1 mol C2H2 / 26.04 g C2H2) × (4 mol CO2 / 2 mol C2H2) × (44.01 g CO2 / 1 mol CO2) = 185 g CO2

Mass Calculations: Grams of Reactant from Grams of Reactant

To find the mass of a reactant required to react with a given mass of another reactant, use the same conversion steps: grams → moles → moles → grams.

Given	Need	Connect
22.5 g C7H16	grams of O2	molar masses, mole–mole factor

Problem setup for heptane combustion

Plan: Convert grams of C7H16 to grams of O2 using molar masses and mole–mole factor.

Conversion plan for C7H16 to O2

Mole–Mole and Molar Mass Factors:

Mole-mole and molar mass factors for C7H16 and O2

Calculation:

Calculation for grams of O2

22.5 g C7H16 × (1 mol C7H16 / 100.2 g C7H16) × (11 mol O2 / 1 mol C7H16) × (32.00 g O2 / 1 mol O2) = 79.1 g O2

Learning Check: Mass of Potassium Required for KCl Production

Given the mass of KCl produced, calculate the mass of potassium required using the balanced equation and conversion factors.

Equation: 2K(s) + Cl2(g) → 2KCl(s)
Given: 36.0 g KCl
Need: grams of K

Problem setup for KCl production

Plan: Convert grams of KCl to grams of K using molar masses and mole–mole factor.

Conversion plan for KCl to K

Mole–Mole and Molar Mass Factors:

Mole-mole and molar mass factors for KCl and K

Calculation:

Calculation for grams of K

36.0 g KCl × (1 mol KCl / 74.55 g KCl) × (2 mol K / 2 mol KCl) × (39.10 g K / 1 mol K) = mass of K required

Summary: Mass Calculations for Chemical Reactions

To calculate the mass of a substance produced or required in a chemical reaction:

Use the molar masses of the substances.
Apply mole–mole factors from the balanced equation.
Convert grams to moles, use the mole ratio, then convert moles back to grams.

Additional info: These concepts are foundational for understanding stoichiometry, quantitative analysis, and predicting yields in chemical reactions.