Answer :
To determine the molecular formula of a compound given its empirical formula and its molar mass, follow these steps:
1. Calculate the Empirical Formula Mass:
- The empirical formula is given as CsH4N4O7. Use the atomic masses of each element to find the mass of the empirical formula.
- Cesium (Cs) has an atomic mass of 132.91 g/mol.
- Hydrogen (H) has an atomic mass of 1.01 g/mol.
- Nitrogen (N) has an atomic mass of 14.01 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
- Calculate the total mass:
[tex]\[ \text{Empirical Formula Mass} = (1 \times 132.91) + (4 \times 1.01) + (4 \times 14.01) + (7 \times 16.00) \][/tex]
[tex]\[ \text{Empirical Formula Mass} = 132.91 + 4.04 + 56.04 + 112.00 \][/tex]
[tex]\[ \text{Empirical Formula Mass} = 304.99 \text{ g/mol} \][/tex]
2. Determine the Ratio of the Molar Mass to the Empirical Formula Mass:
- The molar mass of the compound is given as 464.26 g/mol.
- Divide the molar mass by the empirical formula mass to find the ratio:
[tex]\[ \text{Ratio} = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} \][/tex]
[tex]\[ \text{Ratio} = \frac{464.26 \text{ g/mol}}{304.99 \text{ g/mol}} \][/tex]
[tex]\[ \text{Ratio} \approx 1.52 \][/tex]
3. Round the Ratio to the Nearest Whole Number:
- In this case, the ratio rounds to 2.
4. Determine the Molecular Formula:
- Multiply the subscripts in the empirical formula by this ratio to get the molecular formula.
- For Cs: [tex]\(1 \times 2 = 2\)[/tex]
- For H: [tex]\(4 \times 2 = 8\)[/tex]
- For N: [tex]\(4 \times 2 = 8\)[/tex]
- For O: [tex]\(7 \times 2 = 14\)[/tex]
- Therefore, the molecular formula is Cs2H8N8O14.
So, the final molecular formula of the compound is Cs2H8N8O14.
1. Calculate the Empirical Formula Mass:
- The empirical formula is given as CsH4N4O7. Use the atomic masses of each element to find the mass of the empirical formula.
- Cesium (Cs) has an atomic mass of 132.91 g/mol.
- Hydrogen (H) has an atomic mass of 1.01 g/mol.
- Nitrogen (N) has an atomic mass of 14.01 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
- Calculate the total mass:
[tex]\[ \text{Empirical Formula Mass} = (1 \times 132.91) + (4 \times 1.01) + (4 \times 14.01) + (7 \times 16.00) \][/tex]
[tex]\[ \text{Empirical Formula Mass} = 132.91 + 4.04 + 56.04 + 112.00 \][/tex]
[tex]\[ \text{Empirical Formula Mass} = 304.99 \text{ g/mol} \][/tex]
2. Determine the Ratio of the Molar Mass to the Empirical Formula Mass:
- The molar mass of the compound is given as 464.26 g/mol.
- Divide the molar mass by the empirical formula mass to find the ratio:
[tex]\[ \text{Ratio} = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} \][/tex]
[tex]\[ \text{Ratio} = \frac{464.26 \text{ g/mol}}{304.99 \text{ g/mol}} \][/tex]
[tex]\[ \text{Ratio} \approx 1.52 \][/tex]
3. Round the Ratio to the Nearest Whole Number:
- In this case, the ratio rounds to 2.
4. Determine the Molecular Formula:
- Multiply the subscripts in the empirical formula by this ratio to get the molecular formula.
- For Cs: [tex]\(1 \times 2 = 2\)[/tex]
- For H: [tex]\(4 \times 2 = 8\)[/tex]
- For N: [tex]\(4 \times 2 = 8\)[/tex]
- For O: [tex]\(7 \times 2 = 14\)[/tex]
- Therefore, the molecular formula is Cs2H8N8O14.
So, the final molecular formula of the compound is Cs2H8N8O14.
