QuestionFind the molecular formula of $\mathrm{PNBr}_{2}$ with a molecular weight of 204.8 amu using atomic masses of P, N, Br.

Studdy Solution

STEP 1

Assumptions1. The empirical formula of the compound is $\mathrm{PNBr}_{}$ . The molecular weight of the compound is $204.8 \, \mathrm{amu}$
3. The atomic masses of phosphorus (), nitrogen (), and bromine (Br) are respectively $30.97 \, \mathrm{amu}$ , $14.01 \, \mathrm{amu}$ , and $79.90 \, \mathrm{amu}$

STEP 2

First, we need to find the molecular weight of the empirical formula. We can do this by adding up the atomic masses of each element in the empirical formula.
$Molecular\, weight\, of\, empirical\, formula = Atomic\, mass\, of\, + Atomic\, mass\, of\, N +2 \times Atomic\, mass\, of\, Br$

STEP 3

Now, plug in the given atomic masses for phosphorus, nitrogen, and bromine to calculate the molecular weight of the empirical formula.
$Molecular\, weight\, of\, empirical\, formula =30.97 \, \mathrm{amu} +14.01 \, \mathrm{amu} +2 \times79.90 \, \mathrm{amu}$

STEP 4

Calculate the molecular weight of the empirical formula.
$Molecular\, weight\, of\, empirical\, formula =30.97 \, \mathrm{amu} +14.01 \, \mathrm{amu} +2 \times79.90 \, \mathrm{amu} =204.78 \, \mathrm{amu}$

STEP 5

Now that we have the molecular weight of the empirical formula, we can find the ratio of the molecular weight of the compound to the molecular weight of the empirical formula. This ratio will give us the number of empirical formula units in the molecular formula.
$Ratio = Molecular\, weight\, of\, compound / Molecular\, weight\, of\, empirical\, formula$

STEP 6

Plug in the values for the molecular weight of the compound and the molecular weight of the empirical formula to calculate the ratio.
$Ratio =204.8 \, \mathrm{amu} /204.78 \, \mathrm{amu}$

STEP 7

Calculate the ratio.
$Ratio =204. \, \mathrm{amu} /204.78 \, \mathrm{amu} \approx1$

STEP 8

Since the ratio is approximately1, this means there is one empirical formula unit in the molecular formula. Therefore, the molecular formula of the compound is the same as the empirical formula.
The molecular formula of the compound is $\mathrm{PNBr}_{2}$ .

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