1. Calculate the percent composition for each element in CH 206 Glucose.

%C = 40

%H = 6.67

%O = 53.33

Given

Glucose  C₆H₁₂O₆

Required

The percent composition

Solution

MW Gulucose = 180 g/mol

Ar C = 12 g/mol

Ar H = 1 g/mol

Ar O = 16 g/mol

%C = 6.12/180 x 100% = 40%

%H = 12.1/180 x 100% = 6.67%

%O = 6.16/180 x 100% = 53.33%

a chemical reaction

explanation:

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .