Measuring Moles & Masses in Chemistry: Avogadro's Number & Formulas, Study notes of Italian

Download Measuring Moles & Masses in Chemistry: Avogadro's Number & Formulas and more Study notes Italian in PDF only on Docsity! I. Measuring Matter • Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance. • As you know, atoms and molecules are extremely small. There are so many of them in even the smallest sample that it’s impossible to actually count them. I. Measuring Matter • That’s why chemists created their own counting unit called the mole. • The number 6.022 136 7 x 1023 is called Avogadro’s number in honor of the Italian physicist and lawyer Amedeo Avogadro who, in 1811, determined the volume of one mole of a gas. A. Measuring Moles • In your textbook, Avogadro’s number will be rounded to three significant figures— 6.02 x 1023. • If you write out Avogadro’s number, it looks like this. A. Measuring Moles 602 000 000 000 000 000 000 000 • One-mole quantities of three substances are shown, each with a different representative particle. A. Measuring Moles • The representative particle in a mole of water is the water molecule. • Suppose you want to determine how many particles of sucrose are in 3.50 moles of sucrose. You know that one mole contains 6.02 x 1023 representative particles. B . Converting Moles to Particles • Therefore, you can write a conversion factor, Avogadro’s number, that relates representative particles to moles of a substance. • You can find the number of representative particles in a number of moles just as you found the number of roses in 3.5 dozen. B. Converting Moles to Particles • For sucrose, the representative particle is a molecule, so the number of molecules of sucrose is obtained by multiplying 3.50 moles of sucrose by the conversion factor, Avogadro’s number. • There are 2.11 x 1024 molecules of sucrose in 3.50 moles. B. Converting Moles to Particles • Multiply the number of zinc atoms by the conversion factor that is the inverse of Avogadro’s number. B. Converting Particles to Moles 4.50x1024 atoms Zn X 1 mole Zn 6.02x1023 atoms Zn = = 7.48 mol Zn • The relative scale of atomic masses uses the isotope carbon-12 as the standard. II. Mass and the Mole • Each atom of carbon-12 has a mass of 12 atomic mass units (amu). • The atomic masses of all other elements are established relative to carbon-12. • For example, an atom of hydrogen-1 has a mass of 1 amu. II. Mass and the Mole • The mass of an atom of helium-4 is 4 amu. • Therefore, the mass of one atom of hydrogen-1 is one-twelfth the mass of one atom of carbon-12. • The mass of one atom of helium-4 is one- third the mass of one atom of carbon-12. II. Mass and the Mole • The mass in grams of one mole of any pure substance is called its molar mass. • The molar mass of any element is numerically equal to its atomic mass and has the units g/mol. A. Converting Mass to Moles • A roll of copper wire has a mass of 848 g. • How many moles of copper are in the roll? • Use the atomic mass of copper given on the periodic table to apply a conversion factor to the mass given in the problem. A. Converting Mass to Moles 848 g Cu X 1 mole Cu 63.5 g Cu = 13.4 mol Cu B. Converting Moles to Mass 0.625 mol Ca X 40.1 g Ca 1 mol Ca = 25.1 g Ca C. Converting Mass to Number of Particles • Calculate the number of atoms in 4.77 g lead. • To find the number of atoms in the sample, you must first determine how many moles are in 4.77 g lead. Question 1 Calculate the number of molecules in 15.7 mol carbon dioxide. Answer 9.45 x 1024 molecular CO2 Question 4 A chemist needs 0.0700 mol selenium for a reaction. What mass of selenium should the chemist use? Answer 5.53g Se Question 5 Calculate the number of moles in 17.2 g of benzene (C6H6). Answer 0.220 mol C6H6 C. Converting Mass to Number of Particles • According to data from the periodic table, the molar mass of lead is 207.2 g/mol. Apply a conversion factor to convert mass to moles. 4.77 g Pb X 1 mole Pb 207.2 g Pb III. Empirical and Molecular Formulas • Recall that every chemical compound has a definite composition—a composition that is always the same wherever that compound is found. • The composition of a compound is usually stated as the percent by mass of each element in the compound. A. Percent composition The percent of an element in a compound can be found in the following way. Calculating Percent Composition • Example: Two iron oxide samples were analyzed. The first one had a mass of 36 g and was found to contain 28 g of Fe and 8 g of O. The other one had a mass of 160 g and was found to contain 112 g of Fe and 48 g of O. Are the samples the same compound? Calculating Percent Composition Are the two compounds the same? No, they are not! Calculating Percent Composition As a check, be sure that the percentages add up to 100%. In this case, the percentages add up to 100.000%. Calculating Percent Composition Example: What is the percent composition of Na2CO3? 2(23) Na x 100 106 total 12 x 100 106 total = 43% Na 3(16) x 100 106 total = 11% C = 45% O B. Empirical Formula from Percent Composition • The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula. • Because percent means “parts per hundred parts,” assume that you have 100 g of the compound. B. Empirical Formula from Percent Composition • Follow these rules for Empirical Formulas: A. Find or start with g of each element B. Find moles of each element using Periodic Table C. Divide by smallest moles to get whole #’s D. Multiply if necessary to get whole # Express percent by mass in grams. Find the number of moles of each element. Examine the mole ratio.  | mol O 44.77 gO X ———_ ECTS I Ts) = 2.798 mol O = 1.399 mol C • The results show the following relationship. Rule C. To obtain the simplest whole- number ratio of moles, divide each number of moles by the smallest number of moles. Rule C. To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles. • Notice that the molecular formula for acetic acid (C2H4O2) has exactly twice as many atoms of each element as the empirical formula (CH2O). • The molecular formula for a compound is always a whole-number multiple of the empirical formula. • Divide molar mass of the compound by the molar mass of the empirical formula which should be a whole number. Write the empirical formula. Determine the integer that relates the empirical and molecular formulas. Multiply the subscripts by n. Write the molecular formula. • In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula. • Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula as shown in the following example problem. Rules for Molecular Formula Start by determining the empirical formula for the compound. iene moles of C (in 100 g) = 41.39 g€ x PRN = 3.446 mol C Bete es! moles of H (in 100 g) = 3.47 gH X ———— = 3.442 mol H 1.008 gH 1 mol O moles of O (in 100 g) = 55.14 g0 X Eee = 3.446 mol O The numbers of moles of C, H, and O are nearly equal, so it is not necessary to divide through by the smallest value. You can see by inspection that the smallest whole-number ratio is 1C : 1H : 1O, and the empirical formula is CHO. Next, calculate the molar mass represented by the formula CHO. Here, the molar mass is the sum of the masses of one mole of each element. Calculate the percent composition of aluminum oxide (Al2O3). Question 1 52.93% Al; 47.07% O Answer Determine the percent composition of magnesium nitrate, which has the formula Mg(NO3)2. Question 2 16.39% Mg; 18.89% N; 64.72% O Answer The composition of acetic acid is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Calculate the empirical formula for acetic acid. Question 3 CH2O Answer