QuestionFind the empirical formula and molecular formula for a compound with 7.522% phosphorus and 92.47% iodine, given molecular weight 411.7 amu.

Studdy Solution

STEP 1

Assumptions1. The compound contains7.522% phosphorus and92.47% iodine by mass. . The empirical formula is the simplest, most reduced ratio of elements in a compound.
3. The molecular weight of the compound is411.7 amu.
4. The molecular formula is a multiple of the empirical formula and its mass is equal to the molecular weight.

STEP 2

To find the empirical formula, we first assume that we have100g of the compound. This means that we have7.522g of phosphorus and92.47g of iodine.

STEP 3

Next, we convert these masses to moles. The molar mass of phosphorus () is about30.97 g/mol, and the molar mass of iodine () is about126.9 g/mol.
$moles\, of\, = \frac{7.522g}{30.97g/mol}$ $moles\, of\, I = \frac{92.47g}{126.9g/mol}$

STEP 4

Calculate the moles of phosphorus and iodine.
$moles\, of\, = \frac{7.522}{30.97} =0.243\, moles$ $moles\, of\, I = \frac{92.47}{126.9} =0.728\, moles$

STEP 5

To find the empirical formula, we need to find the ratio of moles of phosphorus to iodine. We do this by dividing each by the smallest number of moles calculated.
$ratio\, of\,\, to\, I = \frac{0.243\, moles}{0.243\, moles} \frac{0.728\, moles}{0.243\, moles}$

STEP 6

Calculate the ratio of phosphorus to iodine.
$ratio\, of\,\, to\, I =13$

STEP 7

So, the empirical formula of the compound is PI3.

STEP 8

Next, we find the molecular formula. First, we need to calculate the molar mass of the empirical formula. The molar mass of PI3 is the molar mass of phosphorus plus three times the molar mass of iodine.
$molar\, mass\, of\, PI3 =30.97g/mol +3 \times126.g/mol$

STEP 9

Calculate the molar mass of PI3.
$molar\, mass\, of\, PI3 =30.97 +3 \times126.9 =411.67g/mol$

STEP 10

Now, we divide the molecular weight of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the molecular formula.
$number\, of\, units = \frac{411.7g/mol}{411.67g/mol}$

STEP 11

Calculate the number of empirical formula units in the molecular formula.
$number\, of\, units = \frac{411.7}{411.67} =$

STEP 12

So, the molecular formula of the compound is the same as the empirical formula, PI.
The empirical formula for this compound is PI. The molecular formula for this compound is PI.

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