Question

A solid sample initially contains a simple mixture of calcium oxalate (CaC₂O₄) and calcium carbonate (CaCO₃).
This TGA graph (thermogram) plots the mass, in mg, of solid remaining as the sample temperature is raised at a constant rate, in the temperature range 350 °C to 800 °C (approx.).

(Note that CaC₂O₄ decomposes around 500 °C and CaCO₃ decomposes around 700 °C.)
What is the mass% of calcium carbonate in the original mixture?
(Do NOT include the '%' character in your response.
Try to determine your answer to 3 sig figs.)

A solid sa plenilially contains a simple mixture of calcium uxain (LIL ) and Caicium carbonate (LLU). This TGA graph (thermogram) plots the mass, in mg, of solid remaining as the sample temperature is raised at a constant rate, in the temperature range 350 C to 800 °C (approx.). (Note that Caczo, decomposes around 500 °C and Caco, decomposes around 700 °C) What is the mass of calcium carbonate in the original mixture? (Do NOT include the character in your response. Try to determine your answer to 3 sigtigs.) 604 Sunt AS Incorrect. Tres 2/10 Previous Thes Increasing temperature -

Answer 1

mass % of calcium carbonate = 38.0 %

Explanation

From the given graph, initial mass of mixture = 8.85 g

First CaC₂O₄ decomposes according to the equation,

CaC₂O₄ (s) $\overset{\Delta }{\rightarrow}$ CaCO₃ (s) + CO (g)

After first heating, mass becomes constant at 7.65 g

mass of CO evolved = (initial mass of sample) - (mass of sample after first heating)

mass of CO evolved = (8.85 g) - (7.65 g)

mass of CO evolved = 1.20 g

moles CO evolved = (mass of CO evolved) / (molar mass CO)

moles CO evolved = (1.20 g) / (28.01 g/mol)

moles CO evolved = 0.0428 mol

moles CaC₂O₄ present = moles CO evolved

moles CaC₂O₄ present = 0.0428 mol

mass CaC₂O₄ present = (moles CaC₂O₄ present) * (molar mass CaC₂O₄)

mass CaC₂O₄ present = (0.0428 mol) * (128.1 g/mol)

mass CaC₂O₄ present = 5.49 g

mass CaCO₃ present in mixture = (initial mass of mixture) - (mass CaC₂O₄ present)

mass CaCO₃ present in mixture = (8.85 g) - (5.49 g)

mass CaCO₃ present in mixture = 3.36 g

mass percent CaCO₃ = (mass CaCO₃ present in mixture / initial mass of mixture) * 100

mass percent CaCO₃ = (3.36 g / 8.85 g) * 100

mass percent CaCO₃ = 38.0 %