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4 years ago

9

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molec

ular formulas of the compound?

1 answer:

anzhelika [568]4 years ago

4 0

Answer:

Empirical formula = NS

Molecular formula = N₄S₄

Explanation:

Since, the compound only contain nitrogen and sulfur,

Thus ,

Name of the element % moles Simplest ratio

S 69.6 69.6/32 = 2.17 1

N 100 - 69.6 = 30.4 30.4/14 = 2.17 1

<u>Hence, the empirical formula = NS</u>

The molar mass of the compound : 184 g/mol

So,

x ( 32 +14 ) = 184

<u>x = 4</u>

<u>So, Molecular formula = N₄S₄</u>

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A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250

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Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

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Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

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B. 5.00 mL of 0.00250 M $KSCN$

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Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

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$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

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Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

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0.0005 M 0.00125 M 0

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