## General Chemistry: Principles and Modern Applications (10th Edition)

$\%\,C=23.68\%$ $\%\,H=3.186\%$ $\%\,N=13.81\%$ $\%\,P=18.32\%$ $\%\,O=41.01\%$
We find from the molar masses that: $\%\,C=\frac{10\,mol\,C\times\frac{12.01\,g\,C}{1\,mol\,C}}{507.18\,g\,ATP}\times100\%=23.68\%$ $\%\,H=\frac{16\,mol\,H\times\frac{1.01\,g\,H}{1\,mol\,H}}{507.18\,g\,ATP}\times 100\%=3.186\%$ $\%\,N=\frac{5\,mol\,N\times\frac{14.0067\,g\,N}{1\,mol\,N}}{507.18\,g\,ATP}\times 100\%=13.81\%$ $\%\,P=\frac{3\,mol\,P\times\frac{30.97\,g\,P}{1\,mol\,P}}{507.18\,g\,ATP}\times 100\%=18.32\%$ $\%\,O=\frac{13\,mol\,O\times\frac{15.999\,g\,O}{1\,mol\,O}}{507.18\,g\,ATP}\times 100\%=41.01\%$