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A 1.1000 gram carbonate sample chosen from Li2CO3, K2CO3, Na2CO3 and CaCO3 was reacted with H2SO4 and was found to lose 0.3497 gram of CO2. (1) Show the calculation of the % CO2 in the unknown carbonate sample. (2) Show the calculation of the % CO2 in each of the carbonate compounds and identify the unknown carbonate from the list. Atomic weights: C = 12.01, O = 16.00. MWs: Li2CO3 = 73.892, K2CO3 = 138.21, Na2CO3 = 105.99 and CaCO3 = 100.09

User Hosam Aly
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1 Answer

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Answer:

For 1: The percentage composition of
CO_2 in unknown carbonate sample is 31.79 %

For 2: The unknown carbonate sample is potassium carbonate.

Step-by-step explanation:

To calculate the percentage composition of carbon dioxide in sample, we use the equation:


\%\text{ composition of }CO_2=\frac{\text{Mass of }CO_2}{\text{Mass of sample}}* 100 .....(1)

  • For 1:

Mass of sample = 1.1000 g

Mass of carbon dioxide = 0.3497 g

Putting values in equation 1, we get:


\%\text{ composition of }CO_2=(0.3497g)/(1.1000g)* 100=31.79\%

Hence, the percentage composition of
CO_2 in unknown carbonate sample is 31.79 %

  • For 2:

We are given:

Mass of carbon dioxide =
[1* 12.01)+(2* 16.00)]=44.01g

  • For lithium carbonate:

Mass of lithium carbonate= 73.892 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:


\%\text{ composition of }CO_2=(44.01g)/(73.892g)* 100=59.55\%

The percent composition of carbon dioxide in lithium carbonate is 59.55 %

  • For potassium carbonate:

Mass of potassium carbonate= 138.21 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:


\%\text{ composition of }CO_2=(44.01g)/(138.21g)* 100=31.84\%

The percent composition of carbon dioxide in potassium carbonate is 31.84 %

  • For sodium carbonate:

Mass of sodium carbonate= 105.99 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:


\%\text{ composition of }CO_2=(44.01g)/(105.99g)* 100=41.52\%

The percent composition of carbon dioxide in sodium carbonate is 41.52 %

  • For calcium carbonate:

Mass of calcium carbonate = 100.09 g

Mass of carbon dioxide = 44.01 g

Putting values in equation 1, we get:


\%\text{ composition of }CO_2=(44.01g)/(100.09g)* 100=43.97\%

The percent composition of carbon dioxide in calcium carbonate is 43.97 %

As, the percent composition of carbon dioxide in the unknown sample matches the percent composition of carbon dioxide in potassium carbonate.

Hence, the unknown carbonate sample is potassium carbonate.

User Martin G
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