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

The sample contains Na₂CO₃ and NaHCO₃. The total mass of the sample is 2 g. When the sample is heated to 300°C, Na₂CO₃ does not decompose, but NaHCO₃ decomposes according to the following reaction:

$2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)$

The mass loss when the sample is heated is due to the evolution of gaseous products, CO₂ and H₂O, from the decomposition of NaHCO₃. The given mass loss is 0.248 g.

Let's calculate 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 Molar mass of CO₂ = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol Molar mass of H₂O = 2 * 1.01 (H) + 16.00 (O) = 18.02 g/mol

According to the stoichiometry of the decomposition reaction, 2 moles of NaHCO₃ decompose to produce 1 mole of CO₂ and 1 mole of H₂O. The mass of 2 moles of NaHCO₃ is $2 \times 84.01 = 168.02$ g. The mass of 1 mole of CO₂ is 44.01 g. The mass of 1 mole of H₂O is 18.02 g. The total mass of gaseous products (CO₂ + H₂O) from the decomposition of 168.02 g of NaHCO₃ is $44.01 + 18.02 = 62.03$ g. This means that for every 168.02 g of NaHCO₃ that decomposes, the mass loss is 62.03 g.

Let $m_{\text{NaHCO}_3}$ be the mass of NaHCO₃ in the 2 g sample. The mass loss of 0.248 g is caused by the decomposition of $m_{\text{NaHCO}_3}$ grams of NaHCO₃. We can set up a proportion based on the stoichiometry: $\frac{\text{Mass of NaHCO}_3 \text{ decomposed}}{\text{Mass loss}} = \frac{\text{Molar mass of 2 NaHCO}_3}{\text{Molar mass of CO}_2 + \text{Molar mass of H}_2\text{O}}$ $\frac{m_{\text{NaHCO}_3}}{0.248 \text{ g}} = \frac{168.02 \text{ g}}{62.03 \text{ g}}$

Now, we can solve for $m_{\text{NaHCO}_3}$: $m_{\text{NaHCO}_3} = 0.248 \text{ g} \times \frac{168.02}{62.03}$ $m_{\text{NaHCO}_3} = \frac{0.248 \times 168.02}{62.03}$ $m_{\text{NaHCO}_3} = \frac{41.66896}{62.03} \approx 0.67178$ g

The total mass of the sample is 2 g, and it contains Na₂CO₃ and NaHCO₃. Mass of Na₂CO₃ + Mass of NaHCO₃ = Total mass Mass of Na₂CO₃ + 0.67178 g = 2 g Mass of Na₂CO₃ = $2 - 0.67178 = 1.32822$ g

The question asks for the mass percentage of Na₂CO₃ in the given mixture. Mass percentage of Na₂CO₃ = $\frac{\text{Mass of Na₂CO}_3}{\text{Total mass of sample}} \times 100$ Mass percentage of Na₂CO₃ = $\frac{1.32822 \text{ g}}{2 \text{ g}} \times 100$ Mass percentage of Na₂CO₃ = $0.66411 \times 100 = 66.411$ %

Rounding to one decimal place, the mass percentage of Na₂CO₃ is 66.4%.