A student attempts to identify an unknown compund by using chemical reactions to

Question

A student attempts to identify an unknown compund by using chemical reactions to calculate mass relationships. After heating a sample weighing 0.4521g, the mass decreased to 0.3400g. When the product was reacted to form a chloride, the resulting mass was 0.3304g.

is this unknown a carbonate or hydrogen carbonate? give a reason

the unknown is one of what two compounds

write the overrall balanced equation

how many moles of chloride salt would be produced from one mole of the original unknown compound?

given this mole ratio, how many grams of the sodium chloride salt can be produced from one mole of the unknown sodium compound?
_______g unknown Na compound (1 mole)=______g sodium chloride

How many grams of the potassium chloride salt can be produced from one mole of the unknown potassium compound?

_______g unknown K compound (1 mole)=_______g potassium chloride

What is the theoretical mass ratio to the correct number of significant figures?
mass chloride/mass unknown compound

Na ratio=______:1.000 K ratio=______:1.000

What is the unknown sample mass ratio?(mass chloride/mass unknown compound) compare this with the answers to part e.

_______1.000

what is the unknown?

Explanation / Answer

We must note one point given in the question. The original compound gives either sodium chloride (NaCl) or potassium chloride (KCl) on treatment with a chloride salt. Therefore, the compound must be a Na or K containing compound.

Part A

The original compound loses in mass on heating. It is given that the original compound can only be a carbonate or a bicarbonate salt. Since carbonates are stable and do not decompose on heating, hence the compound is a bicarbonate. Since we can have only Na or K as the metal, the compound is either sodium bicarbonate (NaHCO3) or potassium bicarbonate (KHCO3).

Part B

The balanced chemical equations for the decomposition of NaHCO3 or KHCO3 are as below:

2 NaHCO3 (s) --------> Na2CO3 (s) + CO2 (g) + H2O (g)

2 KHCO3 (s) -------> K2CO3 (s) + CO2 (g) + H2O (g)

Part C

Let us denote a chloride salt as MCl2 where M is a divalent metal. We can write down the balanced chemical equations for the reactions of NaHCO3 or KHCO3 with MCl2 as below:

2 NaHCO3 + MCl2 -------> 2 NaCl + M(HCO3)2

2 KHCO3 + MCl2 -------> 2 KCl + M(HCO3)2

As per the stoichiometric reactions above,

2 moles NaHCO3 = 2 moles NaCl

2 moles KHCO3 = 2 moles KCl.

Therefore, 1 mole NaHCO3 or 1 mole KHCO3 = 1 mole NaCl or 1 mole KCl.

Therefore, 1 mole of chloride salt can be produced from 1 mole of the original bicarbonate salt.

Part D

We need to find out the molar masses of NaHCO3 and NaCl.

Molar mass of NaHCO3 = (22.989 + 1.008 + 12.01 + 3*15.9994) g/mol = 84.0052 g/mol.

Molar mass of NaCl = (22.989 + 35.453) g/mol = 58.442 g/mol.

Therefore, 84.0052 g Na compound (1 mole) = 58.442 g sodium chloride (ans).

Part E

We need to find out the molar masses of KHCO3 and KCl.

Molar mass of KHCO3 = (39.0983 + 1.008 + 12.01 + 3*15.9994) g/mol = 100.1145 g/mol.

Molar mass of KCl = (39.0983 + 35.453) g/mol = 74.5513 g/mol.

Therefore, 100.1145 g K compound (1 mole) = 74.5513 g potassium chloride (ans).

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