\(M_{HNO_3}=1+14+16.3=63\left(\dfrac{g}{mol}\right)\\ \%m_H=\dfrac{1.100\%}{63}=1,6\%\\ m_N=\dfrac{14.100\%}{63}=22,2\%\\ m_O=100\%-\left(1,6\%+22,2\%\right)=76,2\%\\ M_{CaCO_3}=40+12+16.3=100\left(\dfrac{g}{mol}\right)\\ \%m_{Ca}=\dfrac{40.100\%}{100}=40\%\\ \%m_C=\dfrac{12.100\%}{100}=12\%\\ \%m_O=100-\left(40\%+12\%\right)=48\%\)