CHEM 101 - TOPIC 2 atoms and elements

atom attraction

• e- and e- repulse

• p+ and p+ repulse

• p+ and e- attract

Protons(Atomic #), Neutrons, Mass #, Atomic Mass

Atomic Number (Z) = number of protons (p+) in nucleus

• An element with 6 protons must be carbon (no other atom)

Chemical Symbol = one or two letter abbreviation (118 of them) (C for carbon)

Neutrons (n0) – similar in mass to protons (p+) but no charge

Mass number (A) = #p+ + #n0 = whole number (always)

• Carbon atom with 6 neutrons is C-12 or 12C (A = 6p+ + 6n0=12)

Atomic mass = mass of one atom in units of amu or u

• Includes mass of all proton, neutrons and electrons.

• Generally not an integer (exception is 12C with a mass of 12 u)

Mass in amu or u (atomic mass units or unified atomic mass units)

• Same (very close). Historically 16O and 12C standards used

Charge of atoms

Neutral Atom: #p+ = #e- net charge = 0

Cation: #p+ > #e- net charge = positive

Anion: #p+ < #e- net charge = negative

NEON THREE ISOTOPES

• same atomic # and same # protons

• different # neutrons

• each isotope has different mass

Percent Natural Abundance or Natural Abundance

90.48% suggests 10,000 atoms of Ne, 9048 are 20Ne0.27% suggests 10,000 atoms of Ne, 27 are 21Ne9.25% suggests 10,000 atoms of Ne, 925 are 22Ne

Atomic Mass neon

• Saw that Neon has 3 different isotopes

  • (varying number of neutrons(n0)).

• Each isotope has a mass # (A) = whole number (P+ & no) and each isotope has atomic mass = number with decimals

• A Neon sample has all three isotopes in the percentages given (exp determined) = natural abundances.

• What mass should we use for a sample of Neon?

  • One with highest natural abundance? 19.9424 u

  • The simple average of the three? 20.9769 u

  • We use 20.180 u, the weight average atomic mass or more simply, the atomic mass (see in periodic table)

FORMULA: Atomic mass = Σ(fractional abundance of isotope)_n x (mass of isotope)_n

Atomic masses

• Atoms too small to find masses using balances.

• Use a relative system

• Define mass of one isotope

• Measure all the rest relative to this isotope

• Carbon-12 (6 neutrons and 6 protons) was picked as the relative standard.

• ¹²C assigned a mass of 12 u
  (u = unified atomic mass unit)

• Not initially sure of the mass of a unified atomic mass unit (u). 

Property of masses

• Charged masses deflected by magnetic field

• Amount of deflection directly related to mass

• Measure magnetic field strength needed to deflect an atom the same amount and you will know the relative mass

• i.e., 12C deflected by a magnetic field strength of 1.0000000. 13C needs magnetic field of 1.0836129 for same deflection.

Mass Spectrometry

• Masses and abundances of isotopes are measured using a mass spectrometer

• Atoms or molecules are ionized, then accelerated down a tube

• Their path is bent by a magnetic field, separating them by mass

ONLY 3 WAY TO ASK A QUESTION ABOUT ATOMIC MASS

• FINDING ATOMIC MASS

  • Copper is 69.17% Cu-63 with mass of 62.9396 u and the rest is Cu-65 with mass of 64.9278 u, find copper’s atomic mass

• FINDING PERCENT ABUNDANCE

  • An atom has an atomic mass of 273.303 u and is made up of two isotopes. One isotope has a mass of 271.125 u, the other Has a mass of 280.121 u. Determine the percent abundances

• FINDING MASS OF AN ISOTOPE

  • An atom has an atomic mass of 375.322 u and is made up of two isotopes. One isotope has a mass of 373.025 u and is 62.11% abundant Determine the mass of the second isotope.

  (ANSWERS ON SLIDES)

Atomic mass = Σ(fractional abundance of isotope)_n x (mass of isotope)_n

• first question ask atomic mass

• second question ask fractional abundance of isotope

• third question ask mass

Molar Mass: Counting Atoms by Weighing Them

The Mole: A Chemist’s “Dozen”

• The value of the mole is originally defined as the number of atoms in exactly 12 grams of ¹²C.

• By definition: 12 g ¹²C =1 mol ¹²C atoms

Measured value = 6.022×10^{23 12}C atoms

Converting between Number of Moles and Number of Atoms

1 mol atoms / 6.022×10^{23 12}C atoms or 6.022×10^{23 12}C atoms / 1 mol atoms

conversions

g A ←(periodic table)— mol A ←(formula or reaction)— mol B ←(periodic table)—> g B

6.022×10^{23 12}C atoms A / 1 mol A

Atoms/molecules A <—> Atoms/molecules B