A sample containing only `CaCo_(3)` and `MgCo_(3)` is ignited to CaO and MgO. The mixture of oxides produced weight exactly half as much as the original sample. Calculate the percentages of `CaCO_(3)` and `MgCO_(3)` (by mass) in the sample.

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Molar Masses
Calculate the molar masses of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \).
- Molar mass of \( \text{CaCO}_3 \):
\[ \text{Ca} = 40 \, \text{g/mol}, \quad \text{C} = 12 \, \text{g/mol}, \quad \text{O} = 16 \, \text{g/mol} \times 3 = 48 \, \text{g/mol} \]
\[ \text{Molar mass of } \text{CaCO}_3 = 40 + 12 + 48 = 100 \, \text{g/mol} \]

- Molar mass of \( \text{MgCO}_3 \):
\[ \text{Mg} = 24.3 \, \text{g/mol}, \quad \text{C} = 12 \, \text{g/mol}, \quad \text{O} = 16 \, \text{g/mol} \times 3 = 48 \, \text{g/mol} \]
\[ \text{Molar mass of } \text{MgCO}_3 = 24.3 + 12 + 48 = 84.3 \, \text{g/mol} \]

### Step 2: Set Up the Mass Relationship
Let the mass of \( \text{CaCO}_3 \) in the sample be \( x \) grams. Then, the mass of \( \text{MgCO}_3 \) will be \( 100 - x \) grams (assuming the total mass of the sample is 100 grams).

### Step 3: Calculate the Mass of Oxides Produced
When \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) are ignited, they decompose as follows:
- \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)
- \( \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \)

The masses of the oxides produced from \( x \) grams of \( \text{CaCO}_3 \) and \( 100 - x \) grams of \( \text{MgCO}_3 \) are:
- Mass of \( \text{CaO} \) produced from \( x \) grams of \( \text{CaCO}_3 \):
\[ \text{Moles of } \text{CaCO}_3 = \frac{x}{100} \quad \text{(since molar mass is 100 g/mol)} \]
\[ \text{Mass of } \text{CaO} = \frac{x}{100} \times 56 = \frac{56x}{100} = 0.56x \]

- Mass of \( \text{MgO} \) produced from \( 100 - x \) grams of \( \text{MgCO}_3 \):
\[ \text{Moles of } \text{MgCO}_3 = \frac{100 - x}{84.3} \]
\[ \text{Mass of } \text{MgO} = \frac{100 - x}{84.3} \times 40.3 = \frac{40.3(100 - x)}{84.3} \]

### Step 4: Set Up the Equation
According to the problem, the total mass of the oxides produced is half of the original sample mass (50 grams):
\[ 0.56x + \frac{40.3(100 - x)}{84.3} = 50 \]

### Step 5: Solve the Equation
Multiply through by 84.3 to eliminate the fraction:
\[ 84.3 \times 0.56x + 40.3(100 - x) = 50 \times 84.3 \]
\[ 47.408x + 4030 - 40.3x = 4215 \]
Combine like terms:
\[ 7.108x + 4030 = 4215 \]
\[ 7.108x = 4215 - 4030 \]
\[ 7.108x = 185 \]
\[ x = \frac{185}{7.108} \approx 26.0 \, \text{grams} \]

### Step 6: Calculate the Mass of \( \text{MgCO}_3 \)
\[ \text{Mass of } \text{MgCO}_3 = 100 - x = 100 - 26.0 = 74.0 \, \text{grams} \]

### Step 7: Calculate the Percentages
- Percentage of \( \text{CaCO}_3 \):
\[ \text{Percentage of } \text{CaCO}_3 = \left( \frac{x}{100} \right) \times 100 = \left( \frac{26.0}{100} \right) \times 100 = 26.0\% \]

- Percentage of \( \text{MgCO}_3 \):
\[ \text{Percentage of } \text{MgCO}_3 = \left( \frac{100 - x}{100} \right) \times 100 = \left( \frac{74.0}{100} \right) \times 100 = 74.0\% \]

### Final Result
- Percentage of \( \text{CaCO}_3 \): **26.0%**
- Percentage of \( \text{MgCO}_3 \): **74.0%**