By definition, the percent composition of an atom in a compound is its mass percentage in the formula.
That is, if we have 1 mol of NaHCO₃, we have also 1 mol of H (because there is only on H for each molecule).
So, we calculate the mass of this 1 mol of NaHCO₃ and the mass of 1 mol of H and calculate the percentage.
In equations, we want the following:
![C_H=\frac{m_H}{m_{NaHCO_(3)}}](https://img.qammunity.org/2023/formulas/chemistry/college/q5auz71y4iv5kwn2zrruxbdjekozl4gdo6.png)
Since these are ratios, we doesn't matter if we talk about 1, 2 or any number of moles, but 1 mol is easier because the molecular and atomi masses are for 1 mol.
The molecular mass of NaHCO₃ is:
![\begin{gathered} M_(NaHCO_3)=M_(Na)+M_H+M_C+3\cdot M_O \\ M_(NaHCO_3)\approx(22.990+1.008+12.011+3\cdot15.999)g/mol \\ M_(NaHCO_3)\approx84.006g/mol \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/1ncnckgsu18lbjajt3c5habvqlte75hd0y.png)
Which means that we have approximately 84.006 grams of NaHCO₃ in 1 mol of it.
The atomic mass of H is:
![M_H\approx1.008g/mol](https://img.qammunity.org/2023/formulas/chemistry/college/y8jtulxcvfifczklbr5ovqyjxe1ujsq2sw.png)
Which means that we have approximately 1.008 grams of H in 1 mol of it.
Now, we can take the percentage of mass of H:
![C_H\approx\frac{1.008g_{}}{84.006g}\cdot100\%\approx1.20\%](https://img.qammunity.org/2023/formulas/chemistry/college/eaexwgzvadhy8p61vavr9xh2kylmkpvox2.png)
So, the percentage composition of H in NaHCO₃ is approximately 1.20%.