A mixture of rubidium carbonate and other inert compounds was heated to a high t

Question

A mixture of rubidium carbonate and other inert compounds was heated to a high temperature and the rubidium carbonate decomposed to give rubidium oxide and carbon dioxide.

A 3.3968-g sample of this mixture resulted in 3.1423 g of material after being heated at a high temperature. What was the mass percent (% w/w) of rubidium carbonate in the original sample?

I got this question wrong, and the answer is apparently 39.31%, but I don't know how to solve this problem. If someone could explain how to get to that answer, that would be great!

Explanation / Answer

Ans. Step 1: The loss in mass of sample is solely due to formation of gaseous CO2 during thermal decomposition of Rb2CO3.

So,

            Mass of CO2 released = Initial mass of sample – Mass of sample after drying

                                                = 3.3968 g – 3.1423 g

                                                = 0.2545 g

            Moles of CO2 released = Mass / Molar mass

                                                = 0.2545 g / (44.0098 g/ mol0

                                                = 0.0057828 mol

# Step 2: According to the stoichiometry of balanced reaction, 1 mol CO2 is released by 1 mol Rb2CO3.

So,

            Moles of Rb2CO3 in sample = 0.0057828 mol = Moles of CO2 released

# Mass of Rb2CO3 in sample = Moles x Molar mass

                                                = 0.0057828 mol x (230.9448 g/ mol)

                                                = 1.3355 g

Now,

            % Rb2CO3 in sample = (Mass of Rb2CO3 / Mass of sample) x 100

                                                = (1.3355 g / 3.3968 g) x 100

                                                = 39.316 %

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