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3 years ago

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A compound is found to contain 26.73 % phosphorus, 12.09 % nitrogen, and 61.18 % chlorine by mass. What is the empirical formula

for this compound

1 answer:

Mrrafil [7]3 years ago

4 0

Answer:

$PNCl_2$

Explanation:

Hello!

In this case, when determining empirical formulas by knowing the by-mass percent, we first must assume the percentages as masses so we can compute the moles of each element:

$n_P=\frac{26.73g}{30.97g/mol}=0.863mol\\\\n_N=\frac{12.09g}{14.01g/mol}=0.863mol\\\\n_C_l=\frac{61.18g}{35.45g/mol}=1.726mol$

Now, for the determination of the subscript of each element in the empirical formula, we divide the moles by the fewest moles (P or N):

$P=\frac{0.863mol}{0.863mol}=1\\\\N= \frac{0.863mol}{0.863mol}=1\\\\Cl=\frac{1.726mol}{0.863mol}2$

Thus, the empirical formula is:

$PNCl_2$

Regards!

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Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

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By rearranging the terms, we get the value of 'n'

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$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

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