Question

A 0.5927-g sample was dissolved, and the ${\mathrm{Ca}}^{2+}$and ${\mathrm{Ba}}^{2+}$ions present were precipitated as ${\mathrm{BaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}$and ${\mathrm{CaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}$. The oxalates were then heated in a thermogravimetric apparatus leaving a residue that weighed 0.5127 g in the range of 320°C–400°C and 0.4362 g in the range of 580°C–620°C. Calculate the percentage Ca and percentage Ba in the sample.

Short answer

The percentage of calcium is $7.1\%$ and the percentage of Barium is data-custom-editor="chemistry" $38.8\%$ in the sample.

Step-by-step solution

Step 1. Given information

A 0.5927-g sample was dissolved, and the ${\mathrm{Ca}}^{2+}$and ${\mathrm{Ba}}^{2+}$ions present were precipitated as ${\mathrm{BaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}$and ${\mathrm{CaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}$. The oxalates were then heated in a thermogravimetric apparatus leaving a residue that weighed 0.5127 g in the range of 320°C–400°C and 0.4362 g in the range of 580°C–620°C.

Step 2. Write the chemical equations

Write the chemical equation for dehydration of Calcium oxalates and Barium oxalate.

$$\begin{gathered} {\mathrm{CaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}\stackrel{∆}{\to }{\mathrm{CaC}}_2{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{BaC}}_2{\mathrm{O}}_4.{\mathrm{H}}_2\mathrm{O}\stackrel{∆}{\to }{\mathrm{BaC}}_2{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Write the chemical equation for the conversion of Calcium oxalate to Calcium carbonate and Barium oxalate to Barium carbonate.

$$\begin{gathered} {\mathrm{CaC}}_2{\mathrm{O}}_4\stackrel{320°-400°\mathrm{C}}{\to }{\mathrm{CaCO}}_3+{\mathrm{CO}}_2 \\ {\mathrm{BaC}}_2{\mathrm{O}}_4\stackrel{320°-400°\mathrm{C}}{\to }{\mathrm{BaCO}}_3+{\mathrm{CO}}_2 \end{gathered}$$

Write the chemical equation for the conversion of Calcium carbonate to Calcium oxide and Barium carbonate to Barium oxide.

$$\begin{gathered} \mathrm{CaC}{\mathrm{O}}_3\stackrel{580°-620°\mathrm{C}}{\to }\mathrm{CaO}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{BaC}{\mathrm{O}}_3\stackrel{580°-620°\mathrm{C}}{\to }\mathrm{BaO}+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Step 3. Calculate the mass of Calcium oxalate and carbonate 

Write the expression for mass of Calcium oxalate.

$m_{{\mathrm{CaC}}_2{\mathrm{O}}_4}=\frac{m_{\mathrm{Ca}}}{M_{\mathrm{Ca}}}\left(M_{{\mathrm{CaC}}_2{\mathrm{O}}_4}\right).......\left(1\right)$

Substitute data-custom-editor="chemistry" $40\mathrm{g}/\mathrm{mol}$for data-custom-editor="chemistry" $M_{\mathrm{ca}}$and data-custom-editor="chemistry" $128.1\mathrm{g}/\mathrm{mol}$for $M_{{\mathrm{CaC}}_2{\mathrm{O}}_4}$ in the Equation (1).

data-custom-editor="chemistry" $$\begin{gathered} m_{{\mathrm{CaC}}_2{\mathrm{O}}_4}=\frac{m_{\mathrm{Ca}}}{40\mathrm{g}/\mathrm{mol}}\left(128.1\mathrm{g}/\mathrm{mol}\right) \\ =3.19m_{\mathrm{Ca}} \end{gathered}$$

Write the expression for mass of Calcium carbonate.

data-custom-editor="chemistry" $m_{{\mathrm{CaCO}}_3}=\frac{m_{\mathrm{Ca}}}{M_{\mathrm{Ca}}}\left(M_{{\mathrm{CaCO}}_3}\right)......\left(2\right)$

Substitute data-custom-editor="chemistry" $40\mathrm{g}/\mathrm{mol}$for $M_{\mathrm{Ca}}$and $100\mathrm{g}/\mathrm{mol}$for data-custom-editor="chemistry" $M_{{\mathrm{CaCO}}_3}$in the Equation (2).

$$\begin{gathered} m_{{\mathrm{CaCO}}_3}=\frac{m_{\mathrm{Ca}}}{40\mathrm{g}/\mathrm{mol}}\left(100\mathrm{g}/\mathrm{mol}\right) \\ =2.497m_{\mathrm{Ca}} \end{gathered}$$

Step 4. Calculate the mass of Barium oxalate and carbonate

Write the expression for the mass of Barium oxalate.

$m_{{\mathrm{BaC}}_2{O}_4}=\frac{m_{\mathrm{Ba}}}{M_{\mathrm{Ba}}}\left(M_{{\mathrm{BaC}}_2{\mathrm{O}}_4}\right)......\left(3\right)$

Substitute $137.3\mathrm{g}/\mathrm{mol}$for $M_{\mathrm{Ba}}$and $225.4\mathrm{g}/\mathrm{mol}$for $M_{{\mathrm{BaC}}_2{\mathrm{O}}_4}$in the Equation (3).

$$\begin{gathered} m_{{\mathrm{BaC}}_2{O}_4}=\frac{m_{\mathrm{Ba}}}{137.3\mathrm{g}/\mathrm{mol}}\left(225.4\mathrm{g}/\mathrm{mol}\right) \\ =1.642m_{\mathrm{Ba}} \end{gathered}$$

Write the expression for the mass of Barium carbonate.

role="math" localid="1646037139327" $m_{{\mathrm{BaCO}}_3}=\frac{m_{\mathrm{B}a}}{M_{\mathrm{B}a}}\left(M_{{\mathrm{BaCO}}_3}\right)......\left(4\right)$

Substitute $137.3\mathrm{g}/\mathrm{mol}$for $M_{\mathrm{Ba}}$and $197.4\mathrm{g}/\mathrm{mol}$for $M_{{\mathrm{BaCO}}_3}$in the Equation (4).

role="math" localid="1646037256051" $$\begin{gathered} m_{{\mathrm{BaCO}}_3}=\frac{m_{\mathrm{B}a}}{137.3\mathrm{g}/\mathrm{mol}}\left(197.4\mathrm{g}/\mathrm{mol}\right) \\ =1.438m_{\mathrm{Ba}} \end{gathered}$$

Step 5. Calculate the mass of Calcium and Barium 

Write the expression for the total mass of Calcium oxalate and Barium oxalate.

$m_o=m_{{\mathrm{CaC}}_2{\mathrm{O}}_4}+m_{{\mathrm{BaC}}_2{\mathrm{O}}_4}......\left(5\right)$

Substitute $0.5127\mathrm{g}$for $m_o$$3.19m_{\mathrm{Ca}}$for $m_{{\mathrm{CaC}}_2{\mathrm{O}}_4}$, $1.642m_{\mathrm{Ba}}$for $m_{{\mathrm{BaC}}_2{\mathrm{O}}_4}$in the Equation (5).

localid="1646038203833" $0.5127\mathrm{g}=3.19m_{\mathrm{Ca}}+1.642m_{\mathrm{Ba}}......\left(6\right)$

Write the expression for the total mass of Calcium carbonate and Barium carbonate.

localid="1646038218480" $m_c=m_{{\mathrm{CaCO}}_3}+m_{{\mathrm{BaCO}}_3}......\left(7\right)$

Substitute $0.4362\mathrm{g}$for $m_c$, $2.497m_{\mathrm{Ca}}$for $m_{{\mathrm{CaCO}}_3}$and $1.438m_{\mathrm{Ba}}$for $m_{{\mathrm{BaCO}}_3}$in the Equation (7).

localid="1646038171003" $0.4362\mathrm{g}=2.497m_{\mathrm{Ca}}+1.438m_{\mathrm{Ba}}......\left(8\right)$

From Equation (6) and (8), the mass of Calcium is $0.042\mathrm{g}$and the mass of Barium is $0.230\mathrm{g}$.

Step 6. Calculate the percentage of Calcium 

Write the expression for a percentage of calcium.

$$\begin{gathered} p_{\mathrm{Ca}}=\left(\frac{m_{\mathrm{Ca}}}{m_o}\right)\times 100 \\ =\left(\frac{0.042\mathrm{g}}{0.5127\mathrm{g}}\right)\times 100 \\ =7.1\% \end{gathered}$$

Step 7. Calculate the percentage of Barium

Write the expression for the percentage of Barium.

$$\begin{gathered} p_{\mathrm{Ba}}=\left(\frac{m_{\mathrm{Ba}}}{m_o}\right)\times 100 \\ =\left(\frac{0.230\mathrm{g}}{0.5127\mathrm{g}}\right)\times 100 \\ =38.8\% \end{gathered}$$