Answer to Question #103654 in General Chemistry for James

Question

mass of unknown salt (Li2CO3, Na2CO3, or K2CO3) 1.00 g
mass of filter paper 0.65 g
mass of filter paper + precipitate 2.02 g

Calculate the number of moles of the precipitate recovered from this experiment. Show your work below, and round your answer to the proper number of significant figures.

The formula that is most likely to represent the identity of the unknown salt used in this experiment is ( Li2CO3 Na2CO3 K2CO3 )

Show your work below for the calculation, based on laboratory data, that provides the best evidence that the formula you circled is likely to be the correct answer.

Answer 1

When solving this task, you need to understand what kind of sediment we got. Most likely it was either calcium carbonate СaCO₃ or barium carbonate BaCO₃. We will calculate for both probable precipitation.

• m_prec= 2.02g - 0.65g = 1.37g

If precipitate is calcium carbonate СaCO3:

$$n(CaCO_3) = m/M = 1.37g:100g/mol = 0.0137mol$$

$$n(salt) = 0.0137mol$$

$$M(salt) = m/n = 1.00g/0.0137mol = 72.99g/mol$$

$$A_r(Me) = 6.5g/mol \approx 7g/mol$$

$$Me = Li$$

• Salt: Li₂CO₃
• If precipitate is barium carbonate BaCO3:

$$n(BaCO_3) = 1.37g : 197 g/mol = 0.007 mol$$

$$M(salt) = 1.00g/0.007mol = 143g/mol$$

$$A_r(Me) = 41.5g/mol \approx 39 g/mol$$

$$Me = K$$

• Salt: K₂CO₃