Em tham khảo nha :
\(\begin{array}{l}
{M_{NaHC{O_3}}} = 23 + 1 + 12 + 16 \times 3 = 84dvC\\
\% Na = \dfrac{{{M_{Na}}}}{{{M_{NaHC{O_3}}}}} \times 100\% = \dfrac{{23}}{{84}} \times 100\% = 27,38\% \\
\% H = \dfrac{{{M_H}}}{{{M_{NaHC{O_3}}}}} \times 100\% = \dfrac{1}{{84}} \times 100\% = 1,19\% \\
\% C = \dfrac{{{M_C}}}{{{M_{NaHC{O_3}}}}} \times 100\% = \dfrac{{12}}{{84}} \times 100\% = 14,29\% \\
\% O = \dfrac{{3{M_O}}}{{{M_{NaHC{O_3}}}}} \times 100\% = \dfrac{{3 \times 16}}{{84}} \times 100\% = 57,14\% \\
\end{array}\)
