Agung Budi P Stoichiometry

Agung Budi P:

Agung Budi P Xa.001 Stoichiometry

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Relative atomic mass (A r ) As the weighted mean mass of all the naturally occurring isotopes of an element relative to one twelfth of the mass of a carbon-12 atom. Not whole numbers No mass unit Also applies to relative molecular mass (M r )

Molar Mass (or atomic mass):

The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table), or add the atoms together for a molecule 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of O 2 molecules = 32.0 g Molar Mass (or atomic mass)

Molar Mass of Compounds:

Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 20 C a 40.08 17 Cl 35.45

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The Mole Similar to a dozen, except instead of 12, it’s 602,000,000,000,000,000,000,000 6.02 X 10 23 mol -1 (in scientific notation) It applies to all kinds of particles: atoms, particles, molecules, ions, electrons, formula units depending the way the question is asked, so be careful. This number is named in honor of Amedeo Avogadro (1776 – 1856)

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Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 10 23 Multiply by 6.02 X 10 23 Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table

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1. Molar mass ( Mr ) The mass of one mole of a substance called the molar mass (symbol Mr ). Magnitude molar mass of substance is relative atomic mass or relative molecular mass substances expressed in units of grams per mole . Molar mass = Mr substance or Ar (g / mol) The mass of a substance is a multiplication molarnya mass (g / mol) with mole substance (n).

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2. Molar volume ( Vm ) Volume of one mole of a substance in gas form is called the molar volume, which denoted by Vm . With the conversion of gas at the same temperature and pressure According to Avogadro's law, the ratio of the gases that same amount molnya have the same volume. Mathematically can be expressed as follows n1/v1=n2/v2 STP ATP T = 0˚C (273.15 K) T = 25˚C (298.15 K) P = 1 atm P = 1 atm Vm = 22.4 L/mol Vm = 24.4 L/mol =22.4 dm3/mol =24.4 dm3/mol

Percentage Yield:

Percentage Yield 9 Percent Yield = actual amount of product x 100 theoretical amount of product

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Determinationof Empirical Formulas and Molecular Formulas The chemical formula indicates the type of atom and the relative amount of eachelement elements contained in the substance. The number of elements contained the substance indicated by the index numbers. The chemical formula can be either empirical formula and molecular formula. "empirical formula, the formula which states the smallest ratio atomatom of the elements that make up the compound. " "molecular formula, the formula yamg state the number of atoms of elements that make up one molecule compounds. " Molecular formula = (Empirical formula) n Mr. Molecule = n X Formulas (Mr. Empirical formula )

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Determination of empirical formula and molecular formula of a compound can taken with the following steps. 1. Find the mass (percentage) of each element making up the compound, 2. Change to unit mole, 3. Mole ratio of each element is an empirical formula, 4. Find the molecular formula by: (Mr. empirical formula) n = Mr. molecular formula, n can be calculated, 5. Multiply n obtained from the calculation with the empirical formula

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Chemistry Counts Determination of the amount of reagents and reaction products involved in the reaction must be calculated in units of moles. That is, the units known to be converted into flats. This method is called the method mol approach . The step-by-step method you can approach these mole refer to the following chart. 1. Write the equation of the reaction and balance the matter in question. 2. Change all the known units of each substance into the mol. 3. Use coefficients to balance the number of reaction moles of reactant and product substances. 4. Change the unit of moles of the substance in question into the unit asked (L or g or particles, etc..).

Limiting Reactant:

Limiting Reactan t You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Limiting Reactant: Example:

Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

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Determining the Chemical Formula Hydrate (Water Crystal) Hydrates are solid crystalline compounds containing crystal water (H2O). Solid crystalline compound chemical formula is known. So basically determination of the formula determining the amount of hydrate is a crystalline watermolecule (H2O) or the value of x . In general, the formula hydrates can be written as follows. Solid crystalline compound chemical formula: x. H2O

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EXAMPLE HYDRAT A total of 5 g of copper (II) sulfate hydrate is heated until all water crystal evaporated. The mass of copper (II) sulfate formed solid 3.20 g. Determine the formula of the hydrate! ( Ar : Cu = 63.5; S = 32; O = 16; H = 1) Answer: Step-by-step formula hydrate determination: a. Suppose the formula hydrate CuSO4. x H2O. b. Write the reaction equation. c. Determine the moles of substances before and after the reaction. d. Calculate the value of x, using the mole ratio of CuSO4: mole H2O. CuSO4. xH2O (s) 􀁯 CuSO4 (s) + xH2O 5 g 3.2 g 1.8 g In comparison, CuSO4 mol: mol H2O = 0.02: 0.10. In comparison, CuSO4 mol: mol H2O = 1: 5. So, the formula hydrates of copper (II) sulfate is CuSO4. 5H2O.