Reduced and cyanoethylated human ceruloplasmin form I was freed of sialic acid and digested by trypsin. A glycopeptide fraction isolated by gel filtration was separated into three carbohydrate‐containing fractions by ion exchange chromatography on Dowex I. Each of these was further separated by column electrophoresis to yield a unique glycopeptide. With a stoichiometry of two chains per mole, these peptides account reasonably well for all the carbohydrate in the molecule. In a corresponding preparation from ceruloplasmin form II with smaller carbohydrate content, two of the glycopeptides were, present in greatly reduced amounts. The three glycopeptides were obtained in a pure state from sialic acid‐containing ceruloplasmin form I by gel filtration, ion exchange chromatography, and column electrophoresis. Two different kinds of oligosaccharide chains were found attached to the purified peptides. The first contained 6 hexose, 5 glucosamine, 2 sialic acid, and 0.5 fucose residues; the other about 9 hexose, 8 glucosamine, 4 sialic acid, and 0.5 fucose residues. According to the composition of the protein, the larger chain accounts only for a small part of the carbohydrate content. On the basis of the results, it is proposed that human ceruloplasmin has six carbohydrate chains, three on each of two identical or nearly identical halves of the molecule. The carbohydrate chains are found in two different degrees of completion. Form II of the protein lacks 2–3 of these chains, having two particular sites partly unsubstituted.