eRification of a Compound by Mass Relationships 1. A student attempts to identif

Question

eRification of a Compound by Mass Relationships 1. A student attempts to identify an unknown compound by the method used in this expeni- ment. She finds that when she heated a sample weighing 0.5015 g the mass went down appreciably, to 0.3432 g. When the product was converted to a chloride, the mass went up, to 0.3726 g. a. Is the sample a carbonate? Give your reasoning. b. What are the two compounds that might be in the unknown? and Write the chemical equation for the overall reaction that would occur when the origi- nal compound was converted to a chloride. If the compound is a hydrogen carbonate, use the sum of Equations 1 and 2. If the sample is a carbonate, use Equation 2. Write the equation for a sodium salt and then for a potassium salt. c. d. How many moles of the chloride salt would be produced from one mole of original compound? e. How many grams of the chloride salt would be produced from one molar mass of original compound? g Na,Co, g NaC1 g Molar masses: NaHCO, g original compound g chloride If a sodium salt, chloride g on f. What is the theoretical mass ratio, mass chloride/mass original compound? :1 If she has a K salt, If she has an Na salt, (continued)

Explanation / Answer

Ans a)

Initial weight of the sample =0.5015 g

After heating the sample weight decreased. That means that some the compound decomposes and some gas is released from this.

Now if we will look at the decomposition properties of the carbonates, they decompose at very high temperature. So although the temperature is not mentioned, for normal heating the carbonate will not decompose. So for the answer, a is No it was not carbonate compound.

Answer b)

So as for the bicarbonates, they decompose easily by heating (around 80degree temp.) the unknown might possible for bicarbonate.

The two possible bicarbonate might be Sodium bicarbonate or Potassium bicarbonate.

Answer c)

If the unknown sample is Sodium bicarbonate

Eq-1        2 NaHCO3Heat = Na2CO3 + CO2 + H2O

Eq-2           Na2CO3 + 2HCl = 2NaCl   + CO2 + H2O   (Lets say for formation of chloride HCl is being used)

Summing equation 1 and 2

   NaHCO3 + HCl = NaCl   + CO2 + H2O

If the unknown sample is Potassum bicarbonate

Eq-3            2KHCO3   Heat = K2CO3 + CO2 + H2O

Eq-4               K2CO3 + 2HCL = 2KCl + CO2 +H2O

Summing 3 and 4

KHCO3 + HCl = KCl + CO2 +H2O  

Answer d)

Moles of chloride salt is produce from 1 mole of original compound.

If we will looking at the summed equation for both the salt sodium and potassium one mole of original salt is reacting to form one mole of the chloride salt.

Answer e)

Molar mass of NaHCO3 = 84 g

Molar mass of Na2CO3 = 105 g

Molar mass of NaCl      =58 g

Molar mass of KHCO3 =100 g

Molar mass of K2CO3 = 138 g

Molar Mass of KCl=74.5

Now we will calculate the moles and product weight.

For the case of Sodium salt

0.5015 g of NaHCO3 = (0.5015/84 )= 0.00597 mole of NaHCO3

Now from the equation 1 we can see that 2 moles of NaHCO3 is reacting to form 1 mole of Na2CO3

Now 0.00597 moles of NaHCO3 will form (0.00597/2) = 0.00298 mole of Na2CO3 salt. [ Mole = (Weight / Molecular weight) ]

Now 0.00298 mole of Na2CO3 = 0.00298 * 105 = 0.3134 g of Na2CO3

So, again from the equation 2 we can see that one mole of Na2CO3 reacts to form 2 moles of chloride salt.

So, 0.00298 moles of sodium carbonate reacts to form (0.00298 * 2) =0.00596 mole of NaCl salt

Now 0.00596 moles of NaCl = 0.00596 * 58 = 0.3456 g of NaCl.

If we will calculate for Potassium Salt

0.5015 g of KHCO3 = (0.5015/100 )= 0.00501 mole of KHCO3

Now from the equation 3 we can see that 2 moles of KHCO3 is reacting to form 1 mole of K2CO3

Now 0.00501 moles of NaHCO3 will form (0.00501/2) = 0.0025 mole of K2CO3 salt. [ Mole = (Weight / Molecular weight) ]

Now 0.0025 mole of K2CO3 = 0.0025 * 138 = 0.346 g of K2CO3

So, again from the equation 4, we can see that one mole of K2CO3 reacts to form 2 moles of the chloride salt.

So, 0.0025 moles of Potassium carbonate reacts to form (0.0025 * 2) =0.005 mole of KCl salt

Now 0.00596 moles of KCl = 0.005 * 74.5 = 0.3725 g of KCl.

Answer f)

Theoretical mass ratio for sodium Salt = ( mass of chloride / Mass of Original Compound)

                                =   (0.3456/0.5015)

                              = 0.68

Theoretical mass ratio for Potassium Salt = ( mass of chloride / Mass of Original Compound)

                                =   (0.3725/0.5015)

                              = 0.74

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