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# Stoichiometry Part I

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Atomic Mass up to Molecular Formula

Atomic Mass up to Molecular Formula

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• 1. Ariane B. Rosos Chemistry 1 SY 2009 - 2010
• 2.
• The quantitative relationships between the substances involved in a chemical reaction, established by the equation for the reaction
• 3.
• ATOM
• Smallest particle of an element
• MOLECULE
• Smallest unit particle of a pure substance
• ION
• An atom or group of bonded atoms with electrical charge because of an excess or deficiency of electrons
• ELEMENT
• Pure substance; CANNOT be broken down into 2 or more pure substances by chemical means
• COMPOUND
• Pure substance; CAN be broken down into 2 or more pure substances by chemical means
• 4.
• Atomic Mass = used to numerically indicate the mass of an atom in its ground state, it is expressed in the non SI unit of u
• u = refers to unified atomic mass unit (formerly known as atomic mass unit or amu)
• 1 amu = 1/12 the mass of carbon-12 atom ,
• therefore the mass of C-12 atom is made EQUAL to 12 amu
•
• Carbon-12 atom is an isotope of carbon
•
• 1 amu = 1.66 x 10 -24 g
• 5.
• Mass of e = 1/1800 of mass of p and n so it is negligible making the equation
• Atomic mass of 12 C = mass of p + mass of n
subatomic particle charge Mass (amu) Neutron None 1.0087 ≈ 1 Proton Positive 1.0073 ≈ 1 Electron Negative 5.486 x 10 -4 ≈ 0
• 6. Atomic Mass Average Atomic Mass For carbon it is 12 u not 12.01 u *Used to relate the fact that the numerical value assigned to each element in the periodic table reflects the average abundances of the atoms that compose a naturally occurring element *Related to isotopes *For carbon it is 12.01 u *Chemists often will use the term “atomic mass” when they are actually referring to average atomic mass of an atom.
• 7.
• Solve for the Average Atomic Mass of the element Boron
• Average atomic mass
• = ∑ (mass x percent abundance)
• Where ∑ means “sum”
Isotope Mass (u) Percent Abundance 11 B 11.009305 80.1 10 B 10.012937 19.9
• 8.
• Have different meaning from atomic mass but synonymous with each other, although of different historical origins
• Relative Atomic Mass
• “ Ratio” of the average mass of the atom to the unified atomic mass; dimensionless
• Atomic Weight
• The name Dalton used in the early 19 th century to numerically describe the weight of atoms relative to each other
• 9.
• The average mass of the molecules of a binary compound (non metal-non metal)
• Unit: u
• Ex.: The molecular mass of carbon dioxide gas, CO 2 is 28.01 u.
• 10.
• The average mass of the molecules of an ionic compound (metal-non metal)
• Unit: u
• Ex.: The formula mass of barium chloride, BaCl 2 is 208.2 u.
• 11.
• Used to describe the number of particles (atoms, molecules, etc.) that make up sample of matter
• “ One mole is the amount of any substance that contains the same number of units as the number of atoms in exactly 12 grams of carbon-12.”
• 1 mol of any substance = 6.02 x 10 23 units of that substance,
• where 6.02 x 10 23 Avogadro’s Number, N
• 12.
• Mass of 1 mole of substance
• SI unit: g/mol
• Provides a bridge between mass and amount
• 13. GRAMS MOLES FORMULA UNITS Use molar mass Use N
• 14. Atomic mass Molar mass Unit: g/mol Unit: amu; u Atomic level Macroscopic level Numerically equivalent Comparison of Atomic and Molar Mass Total mass of p + and n o Mass of 1 mole
• 15.
• Hydrogen
• Atomic mass: 1 atom H = 1.008 u
• 1 atom =
• Molar mass: 1 mole H =
• 16.
• Avogadro’s number, N = 6.0221415 × 10 23 particles
• 1 u = 1.660538782 × 10 −24 grams
• 0.999999913 ≈ 1
• 17.
• 18.
• Shows the relative number of atom of each element in the compound
• 19.
• A compound is found out to contain 20.0% carbon, 2.2% hydrogen, and 77.8% chlorine. Determine EF of the compound.
• 20.
• Shows the actual number of atom of each element in the compound
• 21.
• MF = (EF)n
• where n is an integer
• therefore ,
• n = M MF
• M EF
• where M is molar mass
• 22.
• A compound with the empirical formula C 2 H 5 has a molar mass of 58.12 g/mol. Find the molecular formula of the compound.
• 23.
• The molar mass of a compound that is 54.6% carbon, 9.0% hydrogen and 36.4% oxygen is 88 g/mol. Find MF of the compound.
• 24.
• Thursday, 16 Sept 2009
• Bring:
• Scientific calculator
• Black/blue ball pen
• Intermediate paper
• 25.
• Stoichiometry I - Part I and Part II
• (up to page 8)
• Intermediate paper; show complete solution and express final answers in correct significant figures.
• Consultation (optional) - Tuesday, 15 September, 12nn-2:30pm or through email: [email_address]
• Deadline: Wednesday, 16 September 2009
• 12 noon, Chemistry Unit
• Answer key: available 2pm (Wed) @ the bulletin board in front of chemistry unit 