Stoichiometry

Stoichiometry

Moles. One mole of substance contains 6.02 x 10^23 (L) of particles.
1. n (moles)
  1. 1. Write the balanced Equation, 2. List the measurements that is given and a symbol for the one to be calculated. 3. Determine the conversion factors (conversion to moles) and list them under the correct formula. 4. Calculate n. 5. Multiply n by the mole ratio (unknown coefficient over known coefficient). 6. Calculate what you want.
    1. Limiting Reactant: Limits the amount of product that can be formed. The reaction will stop when all of the lim reactant is consumed.
      1. Excess Reactant: Remains when a reaction stops when the lim reactant is completely consumed.
      2. 1. Calculate n for both reactants. 2. Divide n by stoichiometry coefficients and identify the smaller number. 3. Perform all calculations with initial n (not the n divided by the coefficients) of the limiting reactant.
  2. Gas Stoichiometry: When the gas is at STP, 1 mol = 22.7 dm^3
    1. When the gas is not at STP, use PV = nRT
2. N (number of particles)
  1. n = N / L
  2. N = n x L
3. m (Molar Mass)
  1. m = n x M
  2. n = m / M
States
1. Solid: Particles tighlty packed, strong inter-particle forces, particles vibrate in position, fixed shape and volume.
2. Liquid: Particles are more spaced, inter-molecular forces are weaker, particles can slide over each other, no fixed shape, fixed volume.
3. Gas: particles spread out, inter-particle forces are negligible, particles move more freely, no fixed shape or volume.
4. Temperatura is a measure of the average kinetic energy of the particles of a substance. States of matter at a given T and P are determined by the strength of the forces that may exist between particles. Higher Temperature (T) means higher Kinetic Energy (KE)
Empirical Formula: the simplest whole number ratio of the elements in a compounds.
1. Finding the empirical formula: 1. Start with the number of grams of each element (if percentages are given, assume total mass is 100). 2. Convert the mass of each element to moles using its molar mass. 3. Divide each mole value by the smallest number of moles calculated. 4. Round to the nearest whole number.
Molecular Formula: shows all the atoms in a molecule, it is a multiple of the empirical formula.
1. Calculating molecular formula: the empirical formula and the molar mass of the compound are needed. Find the molar mass of the molecule with the empirical formula. Divide the larger molar mass (the one from the molecular formula) by the smaller one (the one from the empirical formula) and you will find a whole number. Multiply all the subscripts of the empirical formula by that number.
Relative Atomic Mass (Ar): (weighted average of one atom of the element) / (1/12 mass of one atom of Carbon - 12). Ratio, therefore unit less.
1. Relative Formula Mass (Mr): (sum of the weighted average of the masses of the atoms in a formula unit) / (1/12 mass of one atom of Carbon - 12)
Percent Composition: (mass of element) / (total mass of compound) x 100%
Chemical Reactions
1. Balancing: Determine the correct formulas for all the reactants. 2. Being balandinc, if possible wtith the simplest metal, or the most complex species. Leave elements like H2 or O2 last. Check again.
2. Synthesis Reaction: when two or more simple compounds combine to form a complicated one.
  Nota:
  - Example: Mg + O --> MgO
3. Decomposition Reaction: opposite of a synthesis reaction, complex molecules breaks down to makes simples ones. They require heat or energy to decompose.
  Nota:
  - Example: 2NaCL --> 2NA + Cl2
4. Single Replacement ReactionL one element trades place with another element in a compound.
  Nota:
  - Examples: Metal and H: Mg + 2HCl --> MgCl2 + H2 Halogens: F2 + KCl --> FCl + K Activity Series of Metals: top ones are more reactive so they replace the ones below it.
5. Double Replacement Reaction: when the anions and cations of two different compounds switch places, forming two entirely different compounds.
  Nota:
  - Example: Pb(NO3)2 + K2CrO4 --> PbCrO4 + 2KNO3
6. Combustion Reaction: generally with oxygen (oxidation), products are CO2 (complete combustion) and H2O, when there is not enough O2, an incomplete combustion happesn, products are CO, C, and H20.
  Nota:
  - Example: 2CH4 + 3O2 --> 2CO2 + 4H2O
Percentage Uncertainty = absolute uncertainty / measured value x 100%
1. Percentage Error: (Accepted Value - Experimental Value) / Accepted Value x 100%
  1. %U > %E --> Random Error
  2. %U < %E --> Systematic Error
Solutions
1. Solute: less abundant component. Solvent: more abundant component in which the solute is dissolved.
2. Concentration is in mol dm^-3 and it is C = n/V. n is in moles and V is in dm^3
  1. Parts per million = mass of solute / total mass of solution
  2. C1V1 = C2V2
  3. n1 + n2 = n3 --> C1V1 + C2V2 = C(V1+V2)
3. Concentrated solutions have relatively large quantities of dissolved solute. Dilute solution has small quantity os dissolved solute. Solubility of solids increases with temperature and solubility of gases decreases with increasing temperature.

Siguiente

Descripción

Resumen del Recurso

Similar

	Creado por rdumontheil hace más de 9 años