\(a)\%m_{Na}=\dfrac{23}{58,5}\cdot100\%\approx39,32\%\\ \%m_{Cl}=100\%-39,32=60,68\%\\ b)\%m_C=\dfrac{12}{44}\cdot100\%\approx27,27\%\\ \%m_O=100\%-27,27\%=72,73\%\\ c)\%m_{Ca}=\dfrac{40}{100}\cdot100\%=40\%\\ \%m_C=\dfrac{12}{100}\cdot100\%=12\%\\ \%m_O=100\%-40\%-12\%=48\%\)