Calculate the percentage composition of (NH4)CO3

%N = 29.16 %

%H = 8.39 %

%C = 12.50 %

%O = 49.95%

Explanation:

Step 1: Data given

Molar mass of (NH4)2CO3 = 96.086 g/mol

Molar mass of N = 14.007 g/mol

Molar mass of H = 1.008 g/mol

Molar mass of C = 12.011 g/mol

Molar mass of O = 15.999 g/mol

Step 2: Calculate the percentage of the elements

% N = (2*14.007 / 96.086) *100 %

%N = 29.16 %

%H = (8*1.008 / 96.086) *100%

%H = 8.39 %

%C = (12.011 / 96.086) *100%

%C = 12.50 %

%O = (3*15.999/96.086) *100%

%O = 49.95%

i think that the answer is c