Question

1. complete table 2 by recording the data from the experiment in the video that determines the amount of sodium bicarbonate in the effervescent tablet.

table 2: sodium bicarbonate in an effervescent tablet

mass of vinegar and cup (g)	mass of tablet (g)	mass of vinegar, cup, and tablet - before reaction (g)	mass after reaction (g)	mass of co₂ (g)

1. use the mass of co₂ from table 2 and the following balanced chemical equation to calculate the mass of nahco₃ in the tablet (the chemical equation simplifies nahco₃ to hco₃⁻ - use the molar mass of nahco₃ in your calculations). show your work for full credit.

1. use your answer from question 4 as the experimental value to calculate the percent error (equation 1) for this experiment. assume that the reported amount of nahco₃ on the package (1.92 g) is the accepted value. show your work for full credit.

percent error = (|experimental value - accepted value| / accepted value) × 100

Explanation:

Step1: Determine molar masses

The molar mass of $CO_2$ is $M_{CO_2}=12.01 + 2\times16.00=44.01\ g/mol$, and the molar mass of $NaHCO_3$ is $M_{NaHCO_3}=22.99+1.01 + 12.01+3\times16.00 = 84.01\ g/mol$.

Step2: Write the chemical - reaction and mole ratio

The reaction between vinegar (acetic acid $CH_3COOH$) and $NaHCO_3$ is $CH_3COOH+NaHCO_3
ightarrow CH_3COONa + H_2O+CO_2\uparrow$. The mole ratio of $NaHCO_3$ to $CO_2$ is $1:1$.

Step3: Calculate moles of $CO_2$

Given the mass of $CO_2$ from the table is $m_{CO_2}=0.99\ g$. Using the formula $n=\frac{m}{M}$, the number of moles of $CO_2$ is $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{0.99\ g}{44.01\ g/mol}\approx0.0225\ mol$.

Step4: Calculate mass of $NaHCO_3$

Since the mole ratio of $NaHCO_3$ to $CO_2$ is $1:1$, the number of moles of $NaHCO_3$ is $n_{NaHCO_3}=n_{CO_2}=0.0225\ mol$. Then, using $m = n\times M$, the mass of $NaHCO_3$ is $m_{NaHCO_3}=n_{NaHCO_3}\times M_{NaHCO_3}=0.0225\ mol\times84.01\ g/mol\approx1.89\ g$.

Step5: Calculate percent error

The accepted value of $NaHCO_3$ is $m_{accepted}=1.92\ g$, and the experimental value is $m_{experimental}=1.89\ g$. Using the percent - error formula $\text{Percent Error}=\frac{\vert m_{experimental}-m_{accepted}\vert}{m_{accepted}}\times100\%$, we have $\text{Percent Error}=\frac{\vert1.89 - 1.92\vert}{1.92}\times100\%=\frac{0.03}{1.92}\times100\%\approx1.56\%$.

Answer:

The mass of $NaHCO_3$ in the tablet is approximately $1.89\ g$, and the percent error is approximately $1.56\%$.