The electrolysis of water is used to generate g of H 2 (g). If the hydrogen was generated at a rate of amps over a period of hours, calculate Avogadro’s number (the number of atoms of H in 1 gram) 1 amp is defined as the number of coulombs per second. There are 6.24 x electrons in a coulomb. It takes 1 electron to make 1 H atom = 3.01 x at H x 1 at Hx 6.24 x el 1 el 1 coul x coul s hx 60 min 1 h x 60 s 1 min g = 6.02 x at H

C C N N O O S S H H Need to determine the empirical formula We have a chemical compound We burn the compound in O 2 and measure the amount of CO 2, H 2 0, N 2, and SO 2 produced. From the mass of CO 2 we can calculate moles of C and grams C C N N O O S S H H From the mass of H 2 O we can calculate moles of H and grams From the mass of N 2 we can calculate moles of N and grams From the mass of SO 2 we can calculate moles of S and grams How do we get moles of O ? moles gggg 5.43 g = g =Total g minus = moles

Now for the hard part Lets get ready to rumble!

A compound was known to contain C, H, N, O, and S. When a 5.43 g sample was burned the products were 8.43 g CO 2, 1.15 g H 2 O, g N 2, and 3.07 g of SO 2. Determine the empirical formula of the compound. Mass of O = g Mass of O = 5.43 g – Mass of O = Total C H N S O – Mass of C H N S Mass of C H N S = g 8.43 g CO 2 x 1 mole x 1 mole C = mole C x 12.0 g = g C 44.0 g 1 mole CO 2 1 mole 1.15 g H 2 O x 1 mole x 2 mole H = mole H x 1.01 g = g H g 1 mole H 2 O 1 mole 3.07 g SO 2 x 1 mole x 1 mole S = mole S x 32.1 g = g S 64.1 g 1 mole SO 2 1 mole g N 2 x 1 mole x 2 mole N = mole N x 14.0 g = g N 28.0 g 1 mole N 2 1 mole x 1 mole 16.0 g = moles O

Empirical Formula or Mole Ratio mole C mole H mole N moles O mole S moles Empirical FormulaC 12 H 8 N 2 O 4 S 3 =6=6 =4=4 =1=1 =2=2 =1.5 x2=12 x2=8x2=8 x2=2x2=2 x2=4x2=4 x2=3x2= moles