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3 years ago

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200 grams of an organic sample which contains only carbon, hydrogen, and oxygen is analyzed and found to contain 97.30 grams of

carbon, 16.22 grams of hydrogen and the remainder oxygen. What is the empirical formula for the compound?

2 answers:

faltersainse [42]3 years ago

4 0

Answer:

The Empirical formula of an organic compound is= $C_3H_6O_2$

Explanation:

Mass of an organic compound = 200 g

Mass of carbon = 97.30 g

moles of carbon = $\frac{97.30 g}{12 g/mol}=8.108 mol$

Mass of hydrogen = 16.22 g

Moles of hydrogen = $\frac{16.22 g}{1 g/mol}=16.220 mol$

Mass of oxygen = 200 g - 97.30 g - 16.22 g = 86.48 g

Moles of oxygen = $\frac{86.48 g}{16 g/mol}=5.405 mol$

Divide the all the moles of element with smallest value of moles.

Carbon = $\frac{8.108 mol}{5.405 mol}=1.5$

Hydrogen = $\frac{16.220 mol}{5.405 mol}=3$

Oxygen = $\frac{5.405 mol}{5.405 mol}=1$

Empirical formula = $C_{1.5}H_3O_1$

Converting them into whole number ratios.

Empirical formula = $C_3H_6O_2$

ipn [44]3 years ago

3 0

The empirical formula for 200 grams of an organic sample which contains carbon,hydrogen, and oxygen is C3H6O

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Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

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