Fe2 Co3 3

Can you calculate the percentage composition of FE2 (CO3) 3? 3

Please suggest a solution

m molar Fe2 (CO3) 3 = 291.718 g / mol

% Fe = 2 x 55,847 x 100 / 291,718 = 38.29

% C = 3 x 12,011 x 100 / 291,718 = 12.35

% O = 9 x 15,999 x 100 / 291,718 = 49.36

Fe2 Co3 3

% N times m = m N / m connections x 100% then ...% N = (2x14.01) / (2x14.01 + 8x1.008 + 1x32.07 + 4x16.00) x 100% = 21.20%. 4. Sigfigs ******* Do you understand? #s The atom number of each atom is xm atom (from table ic) Is.

Fe2 Co3 3