Question

NaCl is stable in the temperature range between 200 ^oC and 850 ^oC, but in that range, NaHCO₃decomposes by the reaction:

2 NaHCO₃ (s) → Na₂CO₃ (s) + H₂O (g) + CO₂ (g)

If a mixture of the two is heated to temperatures in the above range, the gaseous products will escape and the residue will contain the original NaCl and an amount of solid Na₂CO₃. The Na₂CO₃ is related stoichiometrically to the amount of NaHCO₃ which has decomposed. If the reaction has proceeded to completion (no remaining NaHCO₃), the moles, and therefore the weight of NaHCO₃originally in the sample, are able to be determined.

We heat the sample several times, weighing the residue after each heating.

We use the following criterion to establish that the reaction is complete:

The reaction is considered complete if the change in weight of the residue

is less that 5.0 mg (plus or minus) in two successive heatings.

The following data were collected in such a determination:

DATA	VALUE (g)
Weight of empty crucible	15.6016
Initial weight of crucible + sample	17.0532
Weight of crucible + residue after 1st heating	16.6005
Weight of crucible + residue after 2nd heating	16.588
Weight of crucible + residue after 3rd heating	16.5841
Weight of crucible + residue after 4th heating	16.5849

Some heatings may have been unnecessary. You may assume that the weight after the 4th heating is the final weight, even if prior heatings were unnecessary.

From the data, answer the following questions:

Q: Based on the amount of mass that was lost after the last heating, calculate the grams of NaHCO₃ in the original sample of the mixture. Calculate the percent of NaHCO₃ in the original sample before heating.

Answer 1

Ans. Initial Mass of mixture or sample = (Mass of crucible + Sample) – Mass of crucible

                                                            = 17.0532 g – 15.6016 g

                                                            = 1.4516 g

# Mass of sample after heating =

(Mass of crucible + Residue, 4^th heating) – Mass of Crucible

= 16.5849 g – 15.6061 g

= 0.9833 g

# Loss in mass after 4^th drying = Initial mass of sample – Mass of sample after 4^th drying

                                    = 1.4516 g – 0.9833 g

                                    = 0.4683 g

# Calculate Moles of NaHCO3 in sample:

H₂O and CO₂ are lost as gases. So, net loss in mass during drying must be equal to the mass lost in form of H₂O and CO₂.

See the stoichiometry of decomposition, 2 mol NaHCO₃ produces releases 1 mol each of CO₂ and H₂O.

Let the number of moles of NaHCO₃ in the mixture be X.

So,

            Moles of H₂O lost = 0.5X

            Moles of CO₂ lost = 0.5X

Now,

Mass of H₂O lost = Moles x Molar mass = 0.5X mol x (18.01 g/mol) = 9.005X g

Mass of CO₂ lost = 0.5X x (44.01 g/mol) = 22.005X g

Total mass lost in form of H₂O and CO₂ = 9.005X + 22.005X = 31.010X g

So, 31.010X g must be equal to 0.4683 g – the mass lost during drying.

            Or, 31.010X g = 0.4683 g

            Or, X = 0.4683 / 31.010 = 0.01510

Hence, X = 0.01510

Therefore, moles of NaHCO₃ in sample = X moles = 0.01510 mol

# Calculate % NaHCO₃ in original sample:

Mass of NaHCO₃ in sample = Moles of NaHCO₃ x Molar mass

                                                = 0.01510 mol x (84.006908 g/mol)

                                                = 1.2686 g

% NaHCO₃ in original sample = (Mass of NaHCO₃ / Initial mass of original sample) x 100

                                                = (1.2686 g / 1.4516 g) x 100

                                                = 87.396%

                                                = 87.40 %