to determine the percentage of each element present in the salt Potassium Hydrogen Phosphate Pentahydrate (K2HPO4 * 5H_2O).
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Welcome back Candido — missed you this week.
The question asks to determine the percentage of each element present in the salt Potassium Hydrogen Phosphate Pentahydrate ($\text{K}_2\text{HPO}_4 \cdot 5\text{H}_2\text{O}$).
The given atomic masses are: $\text{K}=39$, $\text{P}=31$, $\text{O}=16$, $\text{H}=1$.
Step 1: Calculate the molar mass of $\text{K}_2\text{HPO}_4 \cdot 5\text{H}_2\text{O}$.
The formula contains:
• 2 atoms of Potassium (K)
• 1 atom of Phosphorus (P)
• 4 atoms of Oxygen (O) from $\text{HPO}_4$
• 1 atom of Hydrogen (H) from $\text{HPO}_4$
• 5 molecules of water ($\text{5H}_2\text{O}$), which means $5 \times 2 = 10$ atoms of Hydrogen and $5 \times 1 = 5$ atoms of Oxygen.
Total number of atoms:
• K: 2
• P: 1
• O: $4 + 5 = 9$
• H: $1 + 10 = 11$
Molar mass:
$$ M = (2 \times 39) + (1 \times 31) + (9 \times 16) + (11 \times 1) $$
$$ M = 78 + 31 + 144 + 11 $$
$$ M = 264 \text{ g/mol} $$
Step 2: Calculate the mass contribution of each element.
• Mass of K: $2 \times 39 = 78 \text{ g}$
• Mass of P: $1 \times 31 = 31 \text{ g}$
• Mass of O: $9 \times 16 = 144 \text{ g}$
• Mass of H: $11 \times 1 = 11 \text{ g}$
Step 3: Calculate the percentage of each element.
Percentage of an element = $\frac{\text{mass of element}}{\text{molar mass of compound}} \times 100\%$
Percentage of K:
$$ \% \text{K} = \frac{78 \text{ g}}{264 \text{ g}} \times 100\% $$
$$ \% \text{K} = 29.545\% $$
$$ \% \text{K} \approx 29.55\% $$
Percentage of P:
$$ \% \text{P} = \frac{31 \text{ g}}{264 \text{ g}} \times 100\% $$
$$ \% \text{P} = 11.742\% $$
$$ \% \text{P} \approx 11.74\% $$
Percentage of O:
$$ \% \text{O} = \frac{144 \text{ g}}{264 \text{ g}} \times 100\% $$
$$ \% \text{O} = 54.545\% $$
$$ \% \text{O} \approx 54.55\% $$
Percentage of H:
$$ \% \text{H} = \frac{11 \text{ g}}{264 \text{ g}} \times 100\% $$
$$ \% \text{H} = 4.166\% $$
$$ \% \text{H} \approx 4.17\% $$
The percentages of the elements are:
• Potassium (K): $\boxed{29.55\%}$
• Phosphorus (P): $\boxed{11.74\%}$
• Oxygen (O): $\boxed{54.55\%}$
• Hydrogen (H): $\boxed{4.17\%}$
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