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What is the empirical formula for a compound if a sample contains [tex][tex]$3.72 \, g$[/tex][/tex] of [tex]P[/tex] and [tex][tex]$21.28 \, g$[/tex][/tex] of [tex]Cl$[/tex]?

A. [tex]PCl_5[/tex]
B. [tex]PCl_3[/tex]
C. [tex]P_2Cl_{10}[/tex]
D. [tex]P_2Cl_5[/tex]


Sagot :

To determine the empirical formula of the compound, we can follow these steps:

1. Find the number of moles of each element in the sample:

- For Phosphorus (P):
[tex]\[ \text{Mass of P} = 3.72 \text{ g} \][/tex]
The molar mass of P is [tex]\(30.97 \text{ g/mol}\)[/tex].
[tex]\[ \text{Moles of P} = \frac{\text{Mass of P}}{\text{Molar mass of P}} = \frac{3.72}{30.97} \approx 0.120 \text{ moles} \][/tex]

- For Chlorine (Cl):
[tex]\[ \text{Mass of Cl} = 21.28 \text{ g} \][/tex]
The molar mass of Cl is [tex]\(35.45 \text{ g/mol}\)[/tex].
[tex]\[ \text{Moles of Cl} = \frac{\text{Mass of Cl}}{\text{Molar mass of Cl}} = \frac{21.28}{35.45} \approx 0.600 \text{ moles} \][/tex]

2. Determine the simplest whole number ratio of moles of each element:

- Find the ratio of moles of P to moles of P:
[tex]\[ \text{Ratio of P} = \frac{\text{Moles of P}}{\text{Moles of P}} = \frac{0.120}{0.120} = 1.0 \][/tex]

- Find the ratio of moles of Cl to moles of P:
[tex]\[ \text{Ratio of Cl} = \frac{\text{Moles of Cl}}{\text{Moles of P}} = \frac{0.600}{0.120} \approx 5.0 \][/tex]

3. Determine the empirical formula:

The simplest whole number ratio of P to Cl is approximately 1:5. Therefore, the empirical formula of the compound is:

[tex]\[ \boxed{PCl_5} \][/tex]

Thus, the empirical formula for the compound given the sample contents of [tex]\(3.72 \text{ g}\)[/tex] of [tex]\(P\)[/tex] and [tex]\(21.28 \text{ g}\)[/tex] of [tex]\(Cl\)[/tex] is [tex]\(PCl_5\)[/tex].
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