Chapter 7

41. % nitrogen ¼ 12:04 g39:54 g ð100Þ¼ 30:45%% oxygen ¼39:54 g 12:04 g39:54 gð100Þ ¼ 69:55%empirical formula: moles of nitrogen ¼12:04 g N ¼ 0:8594 mol N14:01 g=molmoles of oxygen ¼27:50 g O ¼ 1:719 mol O16:00 g=mol O0:8594 molrelative number of nitrogen atoms ¼ ¼ 1:0000:8594 mol1:719 molrelative number of oxygen atoms ¼0:8594 mol ¼ 2:00empirical formula ¼ NO 2molecular formula: (molar mass of NO 2 ) x ¼ 92.02 g, 46.01 x ¼ 92.02, x ¼ 2The molecular formula is twice the empirical formula.molecular formula ¼ N 2 O 442. Total mass of C þ H þ O ¼ 30:21 g þ 5:08 g þ 40:24 g ¼ 75:53 g% carbon ¼ 30:21 g75:53 g ð100Þ¼ 40:0%% hydrogen ¼ 5:08 g75:53 g ð100Þ¼ 6:73%% oxygen ¼ 40:24 g75:53 g ð100Þ¼ 53:3%empirical formula: moles of carbon ¼30:21 g C ¼ 2:515 mol C12:01 g=molmoles of hydrogen ¼5:080 g H ¼ 5:03 mol H1:008 g=molmoles of oxygen ¼40:24 g O ¼ 2:515 mol O16:00 g=mol2:515 molrelative number of carbon atoms ¼2:515 mol ¼ 1:0005:03 molrelative number of hydrogen atoms ¼2:515 mol ¼ 2:00relative number of oxygen atoms ¼2:515 mol2:515 mol ¼ 1:000empirical formula ¼ CH 2 Omolecular formula: (molar mass of CH 2 O) x ¼ 180.18 g=mol,ð30:03 g=molÞx ¼ 180:18 g=mol,- Chapter 7 -180:18 g=molx ¼30:03 g=mol ¼ 6The molecular formula is six times the empirical formula.molecular formula ¼ C 6 H 12 O 6-66-