9: Decomposition Of Baking Soda
- Page ID
- 506197
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The purpose of this experiment is to:
- Demonstrate decomposition of a compound.
- To experimentally determine the stoichiometry of the thermal decomposition of baking soda, \(\ce{(NaHCO3)}\).
- Estimate the percentage yield of sodium carbonate from a decomposition reaction.
INTRODUCTION
Stoichiometry is based on the law of conservation of mass, which states that the total mass of a reactant is equal to the total mass of products, where relationships among the quantities of reactants and products typically form a mole ratio, stoichiometric coefficients. Stoichiometric coefficients are also helpful in determining the mole ratio between the reactants and products. The mole ratio is important because it enables chemists to calculate the number of moles of product formed from a given number of reactants, or the number of moles required to produce a certain product.
When two reactants are mixed, one may be used up completely, while the other may remain in excess. The amount of product formed will be limited by the reactant that is used up first. A balanced chemical equation helps students determine the amount of product formed (actual yield), which reactant runs out first (limiting reagent), and which remains (excess reagent).
The thermal decomposition (decomposition by heating) of sodium bicarbonate \(\ce{(NaHCO3)}\) is a common chemical reaction. Sodium bicarbonate, more commonly known as baking soda, is widely listed as an ingredient in baked goods recipes. As the food item is being cooked or baked, the baking soda undergoes decomposition, releasing gas and causing the food item to “rise” and have a “light” texture.
In this experiment, we will use stoichiometry to determine the percent yield of sodium carbonate (Na2CO3) formed. When heated, baking soda, \(\ce{NaHCO3}\) (sodium bicarbonate or sodium hydrogen carbonate) decomposes, yielding Sodium carbonate \(\ce{(Na2CO3)}\), water (H2O), and carbon dioxide (CO2).
\[ \text{2 NaHCO}_3\,(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3\,(s) + \text{H}_2\text{O}\,(g) + \text{CO}_2\,(g) \]
Decomposition by heating is called thermal decomposition. During the thermal decomposition of baking soda, \(\ce{NaHCO3}\), the products water (H2O) and Carbon dioxide (CO2) are released as gases, and the sodium carbonate is left as a solid. The amount (mass) of \(\ce{(Na2CO3)}\) obtained is the actual yield.
The theoretical yield, the maximum amount of \(\ce{(Na2CO3)}\) that could be produced under the given conditions, is calculated using the mass of baking soda used and mole ratios from the balanced chemical equation, as shown in Example 1.
Percent Yield of Sodium Carbonate, \(\ce{(Na2CO3)}\), obtained by decomposition of Baking Soda, \(\ce{NaHCO3}\)
Example 1:
A student heated 3.255 g of baking soda (Sodium bicarbonate),\(\ce{NaHCO3}\), in a lab to produce 1.875 g of sodium carbonate, \(\ce{(Na2CO3)}\). Calculate the percent yield.
Solution:
\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
To find the percent yield, we need to calculate the theoretical yield, as the actual yield is given as the mass of product obtained after the chemical reaction.
The theoretical yield in this case is the amount of Na2CO3 produced from 3.255 g of baking soda, using the molar masses of baking soda and sodium carbonate and the coefficients of the balanced chemical equation as a mole ratio.
\[\text{2 NaHCO}_3\,(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3\,(s) + \text{H}_2\text{O}\,(g) + \text{CO}_2\,(g) \]
\[\text{g NaHCO}_3 \xrightarrow{\text{using molar mass}} \text{mol NaHCO}_3 \xrightarrow{\text{using mole ratio}} \text{mol Na}_2\text{CO}_3 \xrightarrow{\text{using molar mass}}\text{g Na}_2\text{CO}_3\]
\[\begin{aligned}\text{Molar mass of NaHCO}_3 &= (1 \times \text{Na}) + (1 \times \text{H}) + (1 \times \text{C}) + (3 \times \text{O}) \\&= (1 \times 22.99\,\text{g}) + (1 \times 1.01\,\text{g}) + (1 \times 12.01\,\text{g}) + (3 \times 16.00\,\text{g}) \\
&= 84.01\,\text{g/mol} \\
\\\text{Molar mass of Na}_2\text{CO}_3 &= (2 \times \text{Na}) + (1 \times \text{C}) + (3 \times \text{O}) \\
&= (2 \times 22.99\,\text{g}) + (1 \times 12.01\,\text{g}) + (3 \times 16.00\,\text{g}) \\
&= 105.99\,\text{g/mol}
\end{aligned}\]
The balanced chemical equation indicates that two moles of \(\ce{NaHCO3}\) yield one mole of Na2CO3. Using the molar masses of \(\ce{NaHCO3}\) and Na2CO3, we can calculate the theoretical yield.
\[\begin{aligned} \text{Theoretical yield} &= 3.255\,\text{g NaHCO}_3 \times \left( \frac{1\,\text{mol NaHCO}_3}{84.01\,\text{g NaHCO}_3} \right) \times \left( \frac{1\,\text{mol Na}_2\text{CO}_3}{2\,\text{mol NaHCO}_3} \right) \times \left( \frac{105.99\,\text{g Na}_2\text{CO}_3}{1\,\text{mol Na}_2\text{CO}_3} \right) \\ &= 2.053\,\text{g Na}_2\text{CO}_3 \end{aligned} \]
\[\begin{aligned} \text{Percent Yield} &= \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \\ &= \left( \frac{1.875\,\text{g}}{2.053\,\text{g}} \right) \times 100 \\ &= 91.330\% \end{aligned} \]