Chapter 3 Flashcards
(44 cards)
covalent bonds
sharing electrons
ionic bonds
transferring electrons
molecular compound
nonmetal atoms with covalent bonds
empirical formula
shows types of atoms present in the simplest whole-number ration ex// P2O5; not P4O10
molecular formula
based on the actual molecule ex// P4O10; not P2O5
structural formula
-order in which atoms are bonded together -what types of bonds ex// O=C=O
organic compounds
made up of C and H -may contain O or N
line angle formula/line structure
hexagon shaped things
ionic compound
made of positive and negative ions joined together by electrostatic forces of attraction
-metal atoms lose electrons, nonmetals gain
cation
positive ion
anion
negative ion
formula unit
smallest electrically neutral collection of ions
(ratio of atoms is the same as in the chem formula)
monatomic
a single ionized atom
polyatomic
2+ atoms
formula mass
mass of a forumla unit in atomic mass units
ex// Mg(NO3)2 = 24.3050 u + 2[14.0067 u + 3)15.9994 u)]
= 148.3148 u
(also known as molecular mass and atomic mass)
molar mass
mass of 1 mole of compound
ex// 1 mol H2O = 18.0153 g H2O = 6.022 x 10^23 molecules
allotropy
element exists in 1+ molecular forms
ex// oxygen exists in O2 and O3
determining percent composition
- determine the molar mass of the compound (denominator)
- determine the contribution of the given element to the molar mass (numerator)
- formulate the ratio of the mass of the given element to the mass of the compound as a whole
- multiply the ratio by 100% to obtain mass percent
mass % = [(# of atoms of element per formula unit) x molar mass of element]/molar mass of compound x 100%
determining formula of an unknown compound
- take 100 g of the masses; given mass percent
- convert masses of the elements in the 100 g sample to amounts in moles
- write tentative formula
- attempt to convert subscripts to smalles whole number by dividing each of the subscripts by smallest one
- then multiply if still not whole number
*gives empirical formula*
combustion analysis
sample of a compound is burned in a stream of oxygen gas
To Determine Empirical Formula
- determine mass of carbon in x g CO2 by first convert to mol, then g.
- do same for H2O for hydrogen
- take total mass of sample and subract mass of H and C
- convert grams of O to mols
- plug into CHO formula
- divide and multiply subscripts for formula
oxidation state
number of an element that corresponds with how many electrons gained/lost
eg. NaCl
Na has a +1 oxidation state
Cl has a -1 oxidation state
eg. MgCl2
Mg: +2 state
Cl: -1 state (b/c 2 of them)
rules for applying oxidation states
*rules to be applied in numerical order and whenever two rules contradict, go by the lower numbered rule*
- OS of an individual atom in a free element is 0.
- total of the OS of all atoms in the
a. neutral species, like isolated atoms, molecules, and formula units, is 0
b. an ion is equal to the charge on the ion - in their compounds, the group 1 metals have an OS of +1 and group 2 metals have an OS of +2 (eg. OS of K in KCl is +1, OS of Mg is +2 in MgBr2
- in its compounds, OS of fluorine is -1
- in its compounds, OS of hydrogen is usually +1
- in its compounds, OS of oxygen is usually -2
- in binary (2-element) compounds w/ metals, group 17 elements have an OS of -1; group 16, -2, group 15, -3
organic compound
compound formed by carbon and hydrogen
(orgo)
-mostly molecular; few are ionic
binary compounds
formed by 2 elements