Chemical Formulas Chapter 7.

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Presentation transcript:

Chemical Formulas Chapter 7

A diatomic molecule contains only two atoms A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds H2 H2O NH3 CH4 A diatomic molecule contains only two atoms H2, N2, O2, Br2, HCl, CO A polyatomic molecule contains more than two atoms O3, H2O, NH3, CH4 2.5

cation – ion with a positive charge An ion is an atom, or group of atoms, that has a net positive or negative charge. cation – ion with a positive charge If a neutral atom loses one or more electrons it becomes a cation. Na 11 protons 11 electrons Na+ 11 protons 10 electrons anion – ion with a negative charge If a neutral atom gains one or more electrons it becomes an anion. Cl- 17 protons 18 electrons Cl 17 protons 17 electrons 2.5

A monatomic ion contains only one atom Na+, Cl-, Ca2+, O2-, Al3+, N3- A polyatomic ion contains more than one atom OH-, CN-, NH4+, NO3- 2.5

Do You Understand Ions? How many protons and electrons are in ? Al 27 13 3+ 13 protons, 10 (13 – 3) electrons How many protons and electrons are in ? Al 78 34 2- 34 protons, 36 (34 + 2) electrons 2.5

2.5

2.6

An empirical formula shows the simplest A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance H2O molecular empirical H2O C6H12O6 CH2O O3 O N2H4 NH2 2.6

The ionic compound NaCl ionic compounds consist of a combination of cations and an anions the formula is always the same as the empirical formula the sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero The ionic compound NaCl 2.6

Formula of Ionic Compounds 2 x +3 = +6 3 x -2 = -6 Al2O3 Al3+ O2- 1 x +2 = +2 2 x -1 = -2 CaBr2 Ca2+ Br- 1 x +2 = +2 1 x -2 = -2 Na2CO3 Na+ CO32- 2.6

2.6

2.7

Chemical Nomenclature Ionic Compounds often a metal + nonmetal anion (nonmetal), add “ide” to element name BaCl2 barium chloride K2O potassium oxide Mg(OH)2 magnesium hydroxide KNO3 potassium nitrate 2.7

Transition metal ionic compounds indicate charge on metal with Roman numerals FeCl2 iron(II) chloride 2 Cl- -2 so Fe is +2 FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride Cr2S3 3 S-2 -6 so Cr is +3 (6/2) chromium(III) sulfide 2.7

Molecular compounds nonmetals or nonmetals + metalloids common names H2O, NH3, CH4, C60 element further left in periodic table is 1st element closest to bottom of group is 1st if more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom last element ends in ide 2.7

Molecular Compounds HI hydrogen iodide NF3 nitrogen trifluoride SO2 sulfur dioxide N2Cl4 dinitrogen tetrachloride TOXIC! NO2 nitrogen dioxide N2O dinitrogen monoxide Laughing Gas 2.7

2.7

molecular mass (amu) = molar mass (grams) Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO2 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3

Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 60 g C3H8O x 8 mol H atoms 1 mol C3H8O x 6.022 x 1023 H atoms 1 mol H atoms x = 72.5 g C3H8O 5.82 x 1024 atoms H 3.3

Heavy Light KE = 1/2 x m x v2 v = (2 x KE/m)1/2 F = q x v x B 3.4

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5

g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O g CO2 mol CO2 mol C g C 6.0 g C = 0.5 mol C g H2O mol H2O mol H g H 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O 3.6