3.11 Earth’s population is about 6.5 billion. Suppose that every person on Earth participates in a process of counting identical particles at the rate of 2 particles per second. How many years would it take to count 6.0x1023 particles? Assume that there are 365 days in a year. 12/9/2018

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 12/9/2018

Heavy Light An instrument for obtaining molar mass (m/z) 12/9/2018

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% 12/9/2018

Typical % composition problem 3.39) Tin (Sn) exists in Earth’s crust as SnO2. Calculate the percent composition by mass of Sn and O in SnO2. Sn: 78.77%, O: 21.23% 12/9/2018

Remember empirical formula? An empirical formula shows the simplest whole-number ratio of the atoms in a substance H2O molecular empirical C6H12O6 CH2O O3 O N2H4 NH2 12/9/2018

Elemental analysis One of the first things that chemists want to know when they isolate or synthesize a new compound is its composition. What elements are present? What are their relative amounts? Combustion analysis: Generally a sample is sent to an analytical laboratory where a carefully weighed amount is completely burned in an oxygen atmosphere. The carbon dioxide and water that are formed are collected and weighed. Knowing the mass of the initial sample and the masses of CO2 and H2O formed allows the percent composition of the sample to be determined in a straightforward manner. 12/9/2018

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 2molH 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 12/9/2018

Determining a molecular formula Elemental analysis gives empirical formula Mass spectrometry gives molar mass If you know the empirical formula and the molar mass, then you have the information necessary to determine the molecular formula of a compound! 12/9/2018