To find the percent purity of MgCO₃, we can follow these steps:
### Step 1: Write the decomposition reaction
The decomposition of magnesium carbonate (MgCO₃) can be represented as:
\[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \]
### Step 2: Calculate the molar mass of MgCO₃
Using the given atomic weights:
- Molar mass of Mg = 24 g/mol
- Molar mass of C = 12 g/mol
- Molar mass of O = 16 g/mol
The molar mass of MgCO₃ is:
\[ \text{Molar mass of MgCO}_3 = 24 + 12 + (16 \times 3) = 24 + 12 + 48 = 84 \, \text{g/mol} \]
### Step 3: Calculate the theoretical yield of MgO
From the balanced reaction, 1 mole of MgCO₃ produces 1 mole of MgO. Thus, from 10 g of MgCO₃, we can find out how much MgO would theoretically be produced if the sample were pure.
First, calculate the number of moles of MgCO₃ in 10 g:
\[ \text{Moles of MgCO}_3 = \frac{10 \, \text{g}}{84 \, \text{g/mol}} = 0.119 \, \text{mol} \]
Since 1 mole of MgCO₃ produces 1 mole of MgO, the moles of MgO produced would also be 0.119 mol.
Now, calculate the mass of MgO produced from this amount:
\[ \text{Mass of MgO} = \text{Moles of MgO} \times \text{Molar mass of MgO} \]
The molar mass of MgO is:
\[ 24 + 16 = 40 \, \text{g/mol} \]
Thus,
\[ \text{Mass of MgO} = 0.119 \, \text{mol} \times 40 \, \text{g/mol} = 4.76 \, \text{g} \]
### Step 4: Calculate the loss due to impurities
We know that the actual mass of MgO obtained is 4 g. The difference between the theoretical yield and the actual yield gives us the loss due to impurities:
\[ \text{Loss of MgO} = 4.76 \, \text{g} - 4 \, \text{g} = 0.76 \, \text{g} \]
### Step 5: Calculate the theoretical yield of CO₂
Using the same logic, we can calculate the theoretical yield of CO₂. From the balanced equation, 1 mole of MgCO₃ produces 1 mole of CO₂. Thus, the moles of CO₂ produced would also be 0.119 mol.
The molar mass of CO₂ is:
\[ 12 + (16 \times 2) = 12 + 32 = 44 \, \text{g/mol} \]
Thus, the mass of CO₂ produced would be:
\[ \text{Mass of CO}_2 = 0.119 \, \text{mol} \times 44 \, \text{g/mol} = 5.24 \, \text{g} \]
Now, we find the loss of CO₂:
\[ \text{Loss of CO}_2 = 5.24 \, \text{g} - 4.4 \, \text{g} = 0.84 \, \text{g} \]
### Step 6: Calculate the total loss due to impurities
The total loss due to impurities in the sample is:
\[ \text{Total loss} = 0.76 \, \text{g (MgO)} + 0.84 \, \text{g (CO₂)} = 1.6 \, \text{g} \]
### Step 7: Calculate the percent purity of MgCO₃
The mass of pure MgCO₃ can be calculated as:
\[ \text{Mass of pure MgCO}_3 = 10 \, \text{g} - 1.6 \, \text{g} = 8.4 \, \text{g} \]
Now, the percent purity is calculated as:
\[ \text{Percent purity} = \left( \frac{\text{Mass of pure MgCO}_3}{\text{Total mass of sample}} \right) \times 100 \]
\[ \text{Percent purity} = \left( \frac{8.4 \, \text{g}}{10 \, \text{g}} \right) \times 100 = 84\% \]
### Final Answer
The percent purity of MgCO₃ is **84%**.
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