Answer:
The correct answer is O=49.94 approx
Step-by-step explanation:
To calculate the correct percent composition of (NH4)2CO3, first calculate the molar mass of the compound,
Molar mass of N: 14.01 g/mol
Molar mass of H: 1.01 g/mol
Molar mass of C: 12.01 g/mol
Molar mass of O: 16.00 g/mol
Molar mass of (NH4)2CO3: (2 x 14.01 g/mol) + (8 x 1.01 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol) = 96.09 g/mol
Then we can calculate the percent composition of each element in (NH4)2CO3:
Percent composition of N: (2 x 14.01 g/mol) / 96.09 g/mol x 100% = 29.14%
Percent composition of H: (8 x 1.01 g/mol) / 96.09 g/mol x 100% = 8.41%
Percent composition of C: 12.01 g/mol / 96.09 g/mol x 100% = 12.50%
Percent composition of O: (3 x 16.00 g/mol) / 96.09 g/mol x 100% = 50.95%
Therefore, the correct percent composition for (NH4)2CO3 is:
N = 29.14%
H = 8.41%
C = 12.50%
O = 50.95%