Answer :
To calculate the empirical formula of the compound, we'll follow these steps:
1. Find the moles of each element:
We'll assume we have 100 grams of the fertilizer, making the mass percentages equivalent to grams.
- Calcium (Ca): [tex]\( 17.1 \text{ g} \)[/tex]
- Hydrogen (H): [tex]\( 1.7 \text{ g} \)[/tex]
- Phosphorus (P): [tex]\( 26.5 \text{ g} \)[/tex]
- Oxygen (O): [tex]\( 54.7 \text{ g} \)[/tex]
2. Convert the masses to moles:
Use the atomic masses of each element to convert grams to moles.
- [tex]\( \text{Molar mass of Ca} = 40.08 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of Ca} = \frac{17.1 \text{ g}}{40.08 \text{ g/mol}} \approx 0.426 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of H} = 1.01 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of H} = \frac{1.7 \text{ g}}{1.01 \text{ g/mol}} \approx 1.683 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of P} = 30.97 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of P} = \frac{26.5 \text{ g}}{30.97 \text{ g/mol}} \approx 0.856 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of O} = 16.00 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of O} = \frac{54.7 \text{ g}}{16.00 \text{ g/mol}} \approx 3.419 \text{ mol} \][/tex]
3. Find the simplest whole-number ratio:
We need to divide the moles of each element by the smallest number of moles calculated.
- The smallest number of moles is [tex]\( 0.426 \text{ mol} \)[/tex] (for Ca).
- For Ca:
[tex]\[ \frac{0.426 \text{ mol}}{0.426 \text{ mol}} = 1 \][/tex]
- For H:
[tex]\[ \frac{1.683 \text{ mol}}{0.426 \text{ mol}} \approx 3.95 \approx 4 \][/tex]
- For P:
[tex]\[ \frac{0.856 \text{ mol}}{0.426 \text{ mol}} \approx 2.01 \approx 2 \][/tex]
- For O:
[tex]\[ \frac{3.419 \text{ mol}}{0.426 \text{ mol}} \approx 8.03 \approx 8 \][/tex]
4. Write the empirical formula:
From the simplest whole-number ratio, we determine that the empirical formula is [tex]\( \text{CaH}_4\text{P}_2\text{O}_8 \)[/tex].
5. Suggest the formula of the anion:
By analyzing the empirical formula, we see that each formula unit contains an anion part. Notice the ratio of elements commonly found in phosphate compounds.
- Calcium is typically [tex]\( \text{Ca}^{2+} \)[/tex].
- Phosphorus and oxygen are commonly found in phosphate ([tex]\( \text{PO}_4^{3-} \)[/tex]).
Combining this, a plausible anion could be [tex]\( \text{H}_2\text{PO}_4^- \)[/tex] or [tex]\( \text{HPO}_4^{2-} \)[/tex], but since we have 8 oxygens and 2 phosphorus and the simplest ratio we derived sums up roughly, [tex]\( \text{H}_2\text{PO}_4^- \)[/tex] is preferred.
To sum up, the empirical formula of the compound is [tex]\( \text{CaH}_4\text{P}_2\text{O}_8 \)[/tex], and a suggested formula for the anion in the fertilizer is [tex]\( \text{H}_2\text{PO}_4^- \)[/tex].
1. Find the moles of each element:
We'll assume we have 100 grams of the fertilizer, making the mass percentages equivalent to grams.
- Calcium (Ca): [tex]\( 17.1 \text{ g} \)[/tex]
- Hydrogen (H): [tex]\( 1.7 \text{ g} \)[/tex]
- Phosphorus (P): [tex]\( 26.5 \text{ g} \)[/tex]
- Oxygen (O): [tex]\( 54.7 \text{ g} \)[/tex]
2. Convert the masses to moles:
Use the atomic masses of each element to convert grams to moles.
- [tex]\( \text{Molar mass of Ca} = 40.08 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of Ca} = \frac{17.1 \text{ g}}{40.08 \text{ g/mol}} \approx 0.426 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of H} = 1.01 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of H} = \frac{1.7 \text{ g}}{1.01 \text{ g/mol}} \approx 1.683 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of P} = 30.97 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of P} = \frac{26.5 \text{ g}}{30.97 \text{ g/mol}} \approx 0.856 \text{ mol} \][/tex]
- [tex]\( \text{Molar mass of O} = 16.00 \text{ g/mol} \)[/tex]
[tex]\[ \text{Moles of O} = \frac{54.7 \text{ g}}{16.00 \text{ g/mol}} \approx 3.419 \text{ mol} \][/tex]
3. Find the simplest whole-number ratio:
We need to divide the moles of each element by the smallest number of moles calculated.
- The smallest number of moles is [tex]\( 0.426 \text{ mol} \)[/tex] (for Ca).
- For Ca:
[tex]\[ \frac{0.426 \text{ mol}}{0.426 \text{ mol}} = 1 \][/tex]
- For H:
[tex]\[ \frac{1.683 \text{ mol}}{0.426 \text{ mol}} \approx 3.95 \approx 4 \][/tex]
- For P:
[tex]\[ \frac{0.856 \text{ mol}}{0.426 \text{ mol}} \approx 2.01 \approx 2 \][/tex]
- For O:
[tex]\[ \frac{3.419 \text{ mol}}{0.426 \text{ mol}} \approx 8.03 \approx 8 \][/tex]
4. Write the empirical formula:
From the simplest whole-number ratio, we determine that the empirical formula is [tex]\( \text{CaH}_4\text{P}_2\text{O}_8 \)[/tex].
5. Suggest the formula of the anion:
By analyzing the empirical formula, we see that each formula unit contains an anion part. Notice the ratio of elements commonly found in phosphate compounds.
- Calcium is typically [tex]\( \text{Ca}^{2+} \)[/tex].
- Phosphorus and oxygen are commonly found in phosphate ([tex]\( \text{PO}_4^{3-} \)[/tex]).
Combining this, a plausible anion could be [tex]\( \text{H}_2\text{PO}_4^- \)[/tex] or [tex]\( \text{HPO}_4^{2-} \)[/tex], but since we have 8 oxygens and 2 phosphorus and the simplest ratio we derived sums up roughly, [tex]\( \text{H}_2\text{PO}_4^- \)[/tex] is preferred.
To sum up, the empirical formula of the compound is [tex]\( \text{CaH}_4\text{P}_2\text{O}_8 \)[/tex], and a suggested formula for the anion in the fertilizer is [tex]\( \text{H}_2\text{PO}_4^- \)[/tex].
