Question

Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-g sample is burned, and 3.10 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Answer 1

Answer – We are given, mass of sample = 5.50 g , mass of CO₂ formed = 3.10 g

When we burn the sample then there is formed CO₂ from the sugar only. So the combustion reaction of sugar is as follow –

C₁₂H₂₂O₁₁ + 12 O₂ ----> 12 CO₂ + 11 H₂O

We need to calculate the mole of CO₂

Moles of CO₂ = 3.10 g / 44.0 g.mol^-1

                     = 0.0705 moles

Moles of C₁₂H₂₂O₁₁ form the moles of CO₂

From the above balanced equation

12 moles of CO₂ = 1 moles of C₁₂H₂₂O₁₁

So, 0.0705 moles of CO₂ = ?

= 0.0705 moles of CO₂ * 1 moles of C₁₂H₂₂O₁₁ / 12 moles of CO₂

= 0.00587 moles of C₁₂H₂₂O₁₁

Now we need to convert this moles to mass

Mass of C₁₂H₂₂O₁₁ = 0.00587 moles * 342.30 g/mol

                              = 2.01 g of C₁₂H₂₂O₁₁

We know the,

Mass of sample = mass of C₁₂H₂₂O₁₁ + mass of NaCl

So, mass of NaCl = mass of sample – mass of C₁₂H₂₂O₁₁

                           = 5.50 g – 2.01 g

                            = 3.49 g

So mass percent of NaCl = mass of NaCl / mas of sample * 100 %

                                      = 3.49 g / 5.50 g * 100 %

                                      = 63.5 % of table salt.