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Chapter 8 Chemical Composition Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. . Moles and Avogadro's number Molar Mass Convert between particles,moles, and mass Mass percent of elements in compounds Empirical formulas Calculating molecular formulas Nylon, aspartame, Kevlar (bulletproof vests), PVC, Teflon All originated in chemist's laboratory Once they make it – they must determine what it is ◦ What is it's composition? ◦ What is it's chemical formula? Recall that matter is composed of atoms, molecules, and ions. • • These particles are much, much smaller than grains of sand, and an extremely large number of them are in a small sample of a substance. Obviously, counting particles one by one is not practical. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. . Since atoms are so small, we need a quantity with a larger amount We use the mole (mol) o o 1 mole = 6.02 X 1023 representative particles (atoms, molecules, formula units) It is known as Avogadro’s number We usually are dealing with three types of representative particles o Molecules o Atoms o H2O and H2 are molecules, covalently bonded Al and Na are atoms, not bonded Formula units CaCl2 and NaOH are formula units, ionicly bonded 1. How many particles can be found in 10 grams of neon? 2. If you had a bottle that contained 5.69x1024 particles of water, how many grams of water does the bottle hold? 3. How many particles are in 50 g of Ca(OH)2? 10 g Ne X 1 mol X 20.17 g Ne 6.02 x 1023 atoms Ne = 1 mole = 2.98 x 1023 Ne atoms 5.69 x 1024 molecules H2O X ________1 mol 6.02 x 1023 molecules H2O = 170.32 g H2O X 18.02 g H2O 1 mol = 50 g Ca(OH)2 1 mol 6.02 x 1023 f.u. Ca(OH)2 X X = 74.10 g Ca(OH)2 1 mol = 4.06 x 1023 f.u. Ca(OH)2 ◦ The mole allows chemists to count the number of representative particles in a substance. ◦ The atomic mass of an element expressed in grams is the mass of a mole of the element. ◦ To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. . Percent composition of a compound is the relative amount of each element in the compound. Which of the following compounds contains the highest percent of Iron? Iron III Acetate, Iron II Hydroxide or Iron II nitride. Sometimes, percent composition can include compounds called hydrates Hydrates are compounds that bind water molecules to their structure BaCl2 • 6H2O MgSO4 • 3H2O Name -> barium chloride hexahydrate Name -> Find the percentage of water in barium chloride hexahydrate H2O MgSO4 7H2O %H2O = MM =24.31 + 32.06 + 4(16) + 7(18.02) = 246.51 g/mol 7(18.02 g/mol) 246.51g/mol = 51.17 % H2O x 100 Find the percent composition of water in potassium sulfate tetrahydrate. K2SO4 4H2O %H2O = MM =2(39.10) + 32.06 + 4(16) + 4(18.02) = 246.34 g/mol 4(18.02 g/mol) 246.34 g/mol = 29.26 % H2O x 100 Empirical formula is the simplest whole number ratio between elements in a compound You are given % of an element. Change the % to grams (the number does not change). Example 1: You have 26.56% K, 35.41% Cr, and 38.03% O . Therefore, you have: 26.56 g K, 35.41 g Cr, and Now change Grams to Moles 26.56 g K x 1 mol K 39.10 g K = x 1 mol Cr 52 g Cr = 0.6810 mol Cr 38.03 g O x 1 mol O 16 g O = 2.3769 mol O 35.41 g Cr 0.6793 mol K Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles. 0.6793 mol K 0.6793 mol = 1 0.6810 mol Cr 0.6793 mol = 1 2.3769 mol O 0.6793 mol = 3.5 Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. 0.6793 mol K 0.6793 mol = 0.6810 mol Cr 0.6793 mol = 2.3769 mol O 0.6793 mol = 1 x2 = 2 1 x2 = 2 3.5 x 2 Answer = = 7 K2Cr2O7 1. Change % to grams 2. Change grams to moles 3. Divide all moles by the lowest number of moles 4. If all the numbers are not whole numbers, you must multiply the lowest number possible to get a whole number. Example 2: 63.52% Fe and 36.48% S Find the Empirical Formula. Calculate the empirical formula with the following: 56.4% potassium, 8.7% carbon, 34.9% oxygen Answer: K2CO3 Molecular Formula – A multiple of an empirical formula. Can also be the same as empirical in some cases Example: Empirical formula N2O – 2:1 ratio Molecular Formula of N2O could be: N4O2 – A multiple of 2 2(N2O) = N4O2 N6O3 – A multiple of 3 3(N2O) = N6O3 How to determine the multiple you need: Molecular Formula Mass = Multiple Empirical Formula Mass Then multiply the multiple to the Empirical Formula Example 1: Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. First calculate the empirical formula mass. Use the formula Molecular Formula Mass = Multiple Empirical Formula Mass Multiply the formula subscripts by this multiple to determine the molecular formula. Example 2 What is the molecular formula of a compound with the empirical formula CClN and a molar mass of 184.5 g/mol. Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62.0 g/mole, and the empirical formula is CH3O