Question

A 2 g sample containing Na₂CO₃ and NaHCO₃ loses 0.248 g when heated to 300° C, the temperature at which NaHCO₃ decomposes to Na₂CO₃, CO₂ and H₂O. What is the mass percentage of Na₂CO₃ in the given mixture ?

Answer 1

66.4%

Answer 2

Let the mass of Na₂CO₃ in the mixture be $m_1$ grams and the mass of NaHCO₃ be $m_2$ grams. The total mass of the mixture is given as 2 g, so $m_1 + m_2 = 2$.

When the mixture is heated to 300°C, NaHCO₃ decomposes according to the equation: $2\text{NaHCO}_3\text{(s)} \rightarrow \text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{O}\text{(g)} + \text{CO}_2\text{(g)}$ Na₂CO₃ does not decompose at this temperature.

The loss in mass (0.248 g) is due to the evolution of gaseous H₂O and CO₂ from the decomposition of NaHCO₃.

We need the molar masses of the substances involved: Molar mass of NaHCO₃ = 22.99 (Na) + 1.01 (H) + 12.01 (C) + 3 * 16.00 (O) = 84.01 g/mol (using standard atomic weights rounded to two decimal places) Molar mass of H₂O = 2 * 1.01 (H) + 16.00 (O) = 18.02 g/mol Molar mass of CO₂ = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol

According to the stoichiometry of the decomposition reaction, 2 moles of NaHCO₃ produce 1 mole of H₂O and 1 mole of CO₂. Mass of 2 moles of NaHCO₃ = $2 \times 84.01 = 168.02$ g Mass of H₂O produced = $1 \times 18.02 = 18.02$ g Mass of CO₂ produced = $1 \times 44.01 = 44.01$ g Total mass of gaseous products (H₂O + CO₂) = $18.02 + 44.01 = 62.03$ g

So, 168.02 g of NaHCO₃ loses 62.03 g upon decomposition.

Let $m_2$ be the mass of NaHCO₃ in the original mixture. This mass of NaHCO₃ decomposes and causes a mass loss of 0.248 g. We can set up a proportion based on the stoichiometry: $\frac{\text{Mass of NaHCO}_3 \text{ decomposed}}{\text{Mass loss}} = \frac{\text{Molar mass of 2 NaHCO}_3}{\text{Molar mass of H}_2\text{O} + \text{Molar mass of CO}_2}$ $\frac{m_2}{0.248 \text{ g}} = \frac{168.02 \text{ g}}{62.03 \text{ g}}$

Now, we solve for $m_2$: $m_2 = 0.248 \times \frac{168.02}{62.03}$ $m_2 \approx 0.248 \times 2.708689$ $m_2 \approx 0.67176$ g

The mass of NaHCO₃ in the mixture is approximately 0.67176 g. The mass of Na₂CO₃ in the mixture is $m_1 = 2 - m_2$. $m_1 = 2 - 0.67176 = 1.32824$ g

The mass percentage of Na₂CO₃ in the given mixture is: Mass percentage of Na₂CO₃ = $\left( \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Total mass of mixture}} \right) \times 100$ Mass percentage of Na₂CO₃ = $\left( \frac{1.32824 \text{ g}}{2 \text{ g}} \right) \times 100$ Mass percentage of Na₂CO₃ = $0.66412 \times 100$ Mass percentage of Na₂CO₃ = $66.412 \%$

Rounding to a reasonable number of significant figures (based on the mass loss 0.248 g having 3 significant figures), we can report the answer to 3 significant figures. Mass percentage of Na₂CO₃ = $66.4 \%$