Summary of Omvattende Fisika en Chemie Notas

Omvattende Fisika en Chemie Notas vir Studente Sukses

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Introduction

Stoichiometry is the part of chemistry that uses balanced chemical equations to predict amounts of reactants and products. It links masses, moles, volumes (for gases), and chemical formulas so you can calculate how much of a substance you need or will produce in a reaction.

Definition: Stoichiometry is the quantitative study of reactants and products in chemical reactions using mole relationships from balanced equations.

Key Concepts Broken Down

1. The Mole and Molar Mass

  • The mole is a counting unit: $1\ \mathrm{mol} = 6.02 \times 10^{23}$ particles.
  • Molar mass ($M$) is the mass of $1\ \mathrm{mol}$ of a substance in $\mathrm{g\cdot mol^{-1}}$. Use the periodic table to add atomic masses.

Definition: Molar mass is the mass in grams of one mole of a substance (units $\mathrm{g\cdot mol^{-1}}$).

Example: For $\ce{H2O}$, $M(\ce{H2O}) = 2(1.01) + 16.00 = 18.02\ \mathrm{g\cdot mol^{-1}}$.

2. Balanced Equations and Mole Ratios

  • A balanced chemical equation gives the mole ratio (coefficients) of reactants to products.
  • Use these ratios to convert between moles of different substances in the same reaction.

Definition: Mole ratio is the ratio of coefficients from a balanced chemical equation used to convert between moles of substances.

Example: For combustion of methane: $$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$ Mole ratio $\ce{CH4}:\ce{O2}:\ce{CO2}:\ce{H2O} = 1:2:1:2$.

3. Mass <-> Moles Conversions

Steps:

  1. Calculate molar mass $M$ of the substance.
  2. Convert mass $m$ (g) to moles: $$n = \frac{m}{M}$$
  3. Use mole ratio to find moles of target.
  4. Convert moles back to mass if needed: $$m = nM$$

Example: How many moles are in $24.5\ \mathrm{g}$ of $\ce{NaCl}$?
$$M(\ce{NaCl}) = 22.99 + 35.45 = 58.44\ \mathrm{g\cdot mol^{-1}}$$
$$n = \frac{24.5}{58.44} = 0.419\ \mathrm{mol}$$

4. Limiting Reactant and Theoretical Yield

  • The limiting reactant is used up first and determines maximum product amount.
  • Calculate moles of product possible from each reactant; the smallest value is the theoretical yield.

Definition: Limiting reactant is the reactant that runs out first in a chemical reaction, limiting product formation.

Example: For $$\ce{N2 + 3 H2 -> 2 NH3}$$ if you have $1\ \mathrm{mol}$ $\ce{N2}$ and $3\ \mathrm{mol}$ $\ce{H2}$, both are exactly stoichiometric; if $2\ \mathrm{mol}$ $\ce{H2}$, $\ce{H2}$ is limiting.

5. Conservation of Mass

  • Mass is conserved in chemical reactions: total mass of reactants equals total mass of products in a closed system.
  • Atoms are rearranged but not created or destroyed.

Definition: Law of Conservation of Mass states that the total mass of reactants equals the total mass of products in a chemical reaction.

6. Gas Volumes and Avogadro's Law

  • At the same temperature and pressure, equal moles of gases occupy equal volumes (Avogadro's law).
  • Standard conditions (STD): $0^{\circ}\mathrm{C} = 273\ \mathrm{K}$ and $1\ \mathrm{atm} = 101.3\ \mathrm{kPa}$.
  • Molar gas volume at STD: $$V_M = 22.4\ \mathrm{dm^3\cdot mol^{-1}}$$

Use for gas volume calculations at STD: $$V = n V_M$$

Example: Volume of $3.0\ \mathrm{mol\ N2}$ at STD:
$$V = 3.0 \times 22.4 = 67.2\ \mathrm{dm^3}$$

💡 Věděli jste?Fun fact: At the same temperature and pressure, 1 mol of helium gas and 1 mol of oxygen gas each occupy $22.4\ \mathrm{dm^3}$ at STD.

7. Percent Composition and Empirical Formula

  • Percent composition shows mass percent of each element in a compound:
    $$%\text{ element} = \frac{\text{mass of element in formula}}{\text{molar mass of compound}} \times 100%$$

Example: Percent S in $\ce{H2S}$:
$$M(\ce{H2S}) = 2(1.01) + 32.06 = 34.08\ \mathrm{g\cdot mol^{-1}}$$
$$%S = \frac{32.06}{34.08} \times 100% = 94.1%$$

  • To find an empirical formula from percent composition:
    1. Assume $100\ \mathrm{g}$ sample so percent equals grams.
    2. Convert grams to moles: $n = \frac{m}{M}$ for each element.
    3. Divide all mole amou
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Stoichiometry Essentials

Klíčové pojmy: Define mole: $1\ \mathrm{mol} = 6.02\times10^{23}$ particles, Calculate molar mass by summing atomic masses (units $\mathrm{g\cdot mol^{-1}}$), Convert mass to moles: $n = \dfrac{m}{M}$ and moles to mass: $m = nM$, Use balanced equation coefficients as mole ratios to relate substances, Identify the limiting reactant by comparing moles available to mole ratios, Theoretical yield found from limiting reactant; percent yield compares actual to theoretical, Gas volume at STD: $V_M = 22.4\ \mathrm{dm^3\cdot mol^{-1}}$, and $V = nV_M$, Percent composition: $\%\text{element} = \dfrac{\text{mass of element}}{\text{molar mass}}\times100\%$, Find empirical formula by assuming $100\ \mathrm{g}$, converting to moles, dividing by smallest mole, Mass is conserved: total mass of reactants equals total mass of products, Always balance equations before stoichiometric calculations, Check units and round only at the end

## Introduction Stoichiometry is the part of chemistry that uses balanced chemical equations to predict amounts of reactants and products. It links masses, moles, volumes (for gases), and chemical formulas so you can calculate how much of a substance you need or will produce in a reaction. > Definition: Stoichiometry is the quantitative study of reactants and products in chemical reactions using mole relationships from balanced equations. ## Key Concepts Broken Down ### 1. The Mole and Molar Mass - The **mole** is a counting unit: $1\ \mathrm{mol} = 6.02 \times 10^{23}$ particles. - **Molar mass** ($M$) is the mass of $1\ \mathrm{mol}$ of a substance in $\mathrm{g\cdot mol^{-1}}$. Use the periodic table to add atomic masses. > Definition: Molar mass is the mass in grams of one mole of a substance (units $\mathrm{g\cdot mol^{-1}}$). Example: For $\ce{H2O}$, $M(\ce{H2O}) = 2(1.01) + 16.00 = 18.02\ \mathrm{g\cdot mol^{-1}}$. ### 2. Balanced Equations and Mole Ratios - A balanced chemical equation gives the mole ratio (coefficients) of reactants to products. - Use these ratios to convert between moles of different substances in the same reaction. > Definition: Mole ratio is the ratio of coefficients from a balanced chemical equation used to convert between moles of substances. Example: For combustion of methane: $$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$ Mole ratio $\ce{CH4}:\ce{O2}:\ce{CO2}:\ce{H2O} = 1:2:1:2$. ### 3. Mass <-> Moles Conversions Steps: 1. Calculate molar mass $M$ of the substance. 2. Convert mass $m$ (g) to moles: $$n = \frac{m}{M}$$ 3. Use mole ratio to find moles of target. 4. Convert moles back to mass if needed: $$m = nM$$ Example: How many moles are in $24.5\ \mathrm{g}$ of $\ce{NaCl}$? $$M(\ce{NaCl}) = 22.99 + 35.45 = 58.44\ \mathrm{g\cdot mol^{-1}}$$ $$n = \frac{24.5}{58.44} = 0.419\ \mathrm{mol}$$ ### 4. Limiting Reactant and Theoretical Yield - The **limiting reactant** is used up first and determines maximum product amount. - Calculate moles of product possible from each reactant; the smallest value is the theoretical yield. > Definition: Limiting reactant is the reactant that runs out first in a chemical reaction, limiting product formation. Example: For $$\ce{N2 + 3 H2 -> 2 NH3}$$ if you have $1\ \mathrm{mol}$ $\ce{N2}$ and $3\ \mathrm{mol}$ $\ce{H2}$, both are exactly stoichiometric; if $2\ \mathrm{mol}$ $\ce{H2}$, $\ce{H2}$ is limiting. ### 5. Conservation of Mass - Mass is conserved in chemical reactions: total mass of reactants equals total mass of products in a closed system. - Atoms are rearranged but not created or destroyed. > Definition: Law of Conservation of Mass states that the total mass of reactants equals the total mass of products in a chemical reaction. ### 6. Gas Volumes and Avogadro's Law - At the same temperature and pressure, equal moles of gases occupy equal volumes (Avogadro's law). - Standard conditions (STD): $0^{\circ}\mathrm{C} = 273\ \mathrm{K}$ and $1\ \mathrm{atm} = 101.3\ \mathrm{kPa}$. - Molar gas volume at STD: $$V_M = 22.4\ \mathrm{dm^3\cdot mol^{-1}}$$ Use for gas volume calculations at STD: $$V = n V_M$$ Example: Volume of $3.0\ \mathrm{mol\ N2}$ at STD: $$V = 3.0 \times 22.4 = 67.2\ \mathrm{dm^3}$$ Fun fact: At the same temperature and pressure, 1 mol of helium gas and 1 mol of oxygen gas each occupy $22.4\ \mathrm{dm^3}$ at STD. ### 7. Percent Composition and Empirical Formula - **Percent composition** shows mass percent of each element in a compound: $$\%\text{ element} = \frac{\text{mass of element in formula}}{\text{molar mass of compound}} \times 100\%$$ Example: Percent S in $\ce{H2S}$: $$M(\ce{H2S}) = 2(1.01) + 32.06 = 34.08\ \mathrm{g\cdot mol^{-1}}$$ $$\%S = \frac{32.06}{34.08} \times 100\% = 94.1\%$$ - To find an **empirical formula** from percent composition: 1. Assume $100\ \mathrm{g}$ sample so percent equals grams. 2. Convert grams to moles: $n = \frac{m}{M}$ for each element. 3. Divide all mole amou