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son4ous [18]
3 years ago
15

If 65.0 mL of 0.150 M HCl ( aq ) is needed to neutralize all the NH 3 ( g ) from a 2.25 g sample of organic material, calculate

the mass percentage of nitrogen in the sample.
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
6 0

<u>Answer:</u> The mass percentage of nitrogen in the sample is 6.04 %

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of HCl solution = 0.150 M

Volume of solution = 65.0 mL = 0.065 L    (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.150M=\frac{\text{Moles of HCl}}{0.065L}\\\\\text{Moles of HCl}=(0.150mol/L\times 0.065L)=9.75\times 10^{-3}mol

The chemical equation for the reaction of HCl and ammonia follows:

HCl+NH_3\rightarrow NH_4Cl

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 9.75\times 10^{-3}mol of HCl will react with = \frac{1}{1}\times 9.75\times 10^{-3}mo=9.75\times 10^{-3}mol of ammonia

1 mole of ammonia contains 1 mole of nitrogen and 3 moles of hydrogen element

Moles of nitrogen in ammonia = 9.75\times 10^{-3}mol

  • To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

Moles of nitrogen = 9.75\times 10^{-3}mol

Molar mass of nitrogen = 14 g/mol

Putting values in above equation, we get:

9.75\times 10^{-3}mol=\frac{\text{Mass of nitrogen}}{14g/mol}\\\\\text{Mass of nitrogen}=(9.75\times 10^{-3}mol\times 14g/mol)=0.136g

  • To calculate the mass percentage of nitrogen in the sample, we use the equation:

\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100

Mass of sample= 2.25 g

Mass of nitrogen = 0.136 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen}=\frac{0.136g}{2.25g}\times 100=6.04\%

Hence, the mass percentage of nitrogen in the sample is 6.04 %

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<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

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C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

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In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

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The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

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