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Determine the percent composition of hydrogen for the following: NaHCO3

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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)}}

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}

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

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\%

So, the percentage composition of H in NaHCO₃ is approximately 1.20%.

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