1189 Words5 Pages

The ULTIMATE Chemistry Semester 2 Study Guide!
Iron: Fe+2 is ferrous
Fe+3 is ferric
Copper: Cu+1 is cuprous
Cu+2 is cupric
Lead: Pb+2is plumbous
Pb+4 is plumbic
Tin: Sn+2 is stannous
Sn+4 is stannic
Periodic Table TRENDS * As you move right along the periodic table… decreasing atomic radius, increasing ionization energy, increasing electronegativity, increasing electron affinity, constant shielding * As you move down the periodic table… increasing atomic radius, decreasing ionization energy, decreasing electronegativity, increasing shielding
WTF is that stuff? * Atomic radius is the distance between the nuclei of atoms when they are involved in a chemical bond. As you increase the number of protons in the nucleus*…show more content…*

Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040*…show more content…*

93 g/mol? Not we get to use it! Yay! 93 g/mol / 31.06 g/mol = 3 (this is the multiplier) Multiply that whole number through the subscripts of the empirical formula. 3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate. Mole conversions (DRAW) Nomenclature Help: B: -3 C: -2 N: -1 Cl: -1 Br: -1 I: -2 S, Se, Te: -2 P, As: -3 Si: -4 PER adds one

Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040

93 g/mol? Not we get to use it! Yay! 93 g/mol / 31.06 g/mol = 3 (this is the multiplier) Multiply that whole number through the subscripts of the empirical formula. 3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate. Mole conversions (DRAW) Nomenclature Help: B: -3 C: -2 N: -1 Cl: -1 Br: -1 I: -2 S, Se, Te: -2 P, As: -3 Si: -4 PER adds one

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