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The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

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Presentation on theme: "The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,"— Presentation transcript:

1 The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 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 10 18 electrons in a coulomb. It takes 1 electron to make 1 H atom = 3.01 x 10 23 at H x 1 at Hx 6.24 x 10 18 el 1 el 1 coul x 3.000 coul s 4.46625 hx 60 min 1 h x 60 s 1 min 0.5000 g = 6.02 x 10 23 at H

2 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

3 Now for the hard part Lets get ready to rumble!

4 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, 0.450 g N 2, and 3.07 g of SO 2. Determine the empirical formula of the compound. Mass of O = 1.0151 g Mass of O = 5.43 g – 4.4149 Mass of O = Total C H N S O – Mass of C H N S Mass of C H N S =4.4149 g 8.43 g CO 2 x 1 mole x 1 mole C = 0.1916 mole C x 12.0 g = 2.299 g C 44.0 g 1 mole CO 2 1 mole 1.15 g H 2 O x 1 mole x 2 mole H = 0.1276 mole H x 1.01 g = 0.1289 g H 18.02 g 1 mole H 2 O 1 mole 3.07 g SO 2 x 1 mole x 1 mole S = 0.04789 mole S x 32.1 g = 1.537 g S 64.1 g 1 mole SO 2 1 mole 0.450 g N 2 x 1 mole x 2 mole N = 0.03214 mole N x 14.0 g = 0.4500 g N 28.0 g 1 mole N 2 1 mole x 1 mole 16.0 g = 0.06344 moles O

5 Empirical Formula or Mole Ratio 0.1916 mole C 0.1276 mole H 0.03214 mole N 0.06344 moles O 0.04779 mole S 0.03214 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=3 0.03214 moles


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